MATH 55 NOTES AARON LANDESMAN

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1 MATH 55 NOTES AARON LANDESMAN Contents 1. Set Theory Defining Sets Equivalence Relations 2 2. Category Theory Categories Functors Natural Transformations /29/ Metric Spaces Topological Spaces The Intermediate Value Theorem /3/ /5/ Heine-Borel /12/ /19/ /24/ /26/ /3/ /5/ Inverse Function Theorem Step 2: As long as Df is invertible, then f is open Step 3: Show g is differentiable /10/ Differential Equations /12/ /26/ /31/15 and 4/2/ Integration Fubini s theorem Partitions of Unity A is compact A = i A i with A i compact and A i int(a i+1 ) A is open A general /7/ Case Case Case /9/ Stokes Theorem /14/ /16/

2 2 AARON LANDESMAN Case 1: f(z) = 1, z 0 = 0, and S is the circle of radius 1 centered at Case 2: General f, with z 0 = Case 3: General Case /21/ /23/ Meromorphic Functions /28/ Defining Sets. 1. Set Theory Remark 1.1. For the purposes of this class, we will think of a set as an object specified by its elements, and there is a membership operation, denoted a A if a is an element of A. Example 1.2. (Examples of sets) R Z {+1,-1} Here, we specify sets by their elements, although this is sort of cheating. Example 1.3. (Examples of Subsets) Y X Q R {0} R R R R {2, 3, 5, 6} R Definition 1.4. For X 1, X 2 X, define the intersection X 1 X 2 = {x X x X 1, x X 2 }. Example 1.5. X = students at 55, X 1 = wearing t-shirts, X 2 = wearing red. Definition 1.6. The union X 1 X 2 = {x X x X 1, or x X 2 }, For Y X the complement Ȳ = {x X x / Y }. We also notate Ȳ as Y \ X, Y X. For a bit more of a formal introduction, here are some of the axioms of set theory, although these weren t mentioned in class. Axiom 1.7. (1) A = B x, x A x B) (2) A set = {x P (x)} is a set. (3) A, B sets = {A, B} (4) A = {A i } = A = A i = {x B A, x B}. (5) (Power Set Axiom) A a set, then P(A) = {B B A} is a set. (6) (Infinity) N = {0, 1, 2,...}, the natural numbers (7) (Choice) For any set B A, there exists a choice function c : P(A) \ { } A such that c(b) B. The following lemma is included for pedagogical purposes, to show how solutions should be written up for homework. Lemma 1.8. For Y X, Ȳ = Y. Proof. x Ȳ x / Ȳ x Y

3 MATH 55 NOTES 3 Lemma 1.9. Proof. (X 1 X 2 ) = X 1 X 2. x (X 1 X 2 ) x / X 1 X 2 x / X 1 or x / X 2 x X 1 or x X 2 Lemma X 1 X 2 = X 1 X 2. Proof. Homework. x X 1 X 2 Definition Define the power set of X, denoted Subset(X) as the set of all subsets of X. Sometimes this is also notated as P(X). Example The set X = {1,..., n} has 2 n elements. Definition A map (of sets) f : X Y is an assignment x X of an element f(x) Y. Example X = { students of 55 }, Y = Z. Define f i : X Y by f 1 (student) = age, f 2 (student) = height, f 3 = $. X = Y = R, take functions f 1 (r) = r + 1, f 2 (r) = 2r, f 3 (r) = r 3, f 4 (r) = 0. Can X 1 X 2, f(x 1 ) = f(x 2 )? Yes! (As long as the function is not injective). Definition A map f : X Y is injective (a monomorphism) if x 1 x 2 = f(x 1 ) f(x 2 ). A map is surjective (an epimorphism) if y Y, x X f(x) = y. Example Consider f : Z Z, n 2n. epimorphism. This is a monomorphism but not an Definition For f : X Y, g : Y Z, we can define the composition, g f : X Z, x f(g(x)). Example Let X = R, Y = R, f(x) = x 2, Z = R, g(y) = y 3. Then, g f : R R, (g f)(x) = (x 2 ) 3. Example Again, X = Y = Z = R, let f(x) = x + 1, g(y) = y 2, (g f)(x) = (x + 1) 2. Then, f g(x) = x Equivalence Relations. Definition For X a set, an equivalence relation is a relation satisfying (1) (symmetry) X 1 X 2 = X 2 X 1 (2) (reflexivity) X X (3) (transitivity) X 1 X 2, X 2 X 3 = X 1 X 3. Technically, we haven t defined a relation. Formally, one can write a relation r : X X {0, 1} and say x y if and only if r(x, y) = 1. Example Let X = { students }, and define x y if x lives in the same dorm as y. x 1 x 2 if x 1 x 2 is divisible by 2. (non-example) Let X = { people } and x 1 x 2 if they are friends on Facebook is not an equivalence relation.

4 4 AARON LANDESMAN Definition For X a set, an equivalence relation, as subset Y X is called a coset (w.r.t. ) if (1) Y (2) x Y, x x = x Y (3) x 1, x 2 Y = x 1 x 2. Example In the case of X = Z with x y if x y is divisible by 2, the cosets are set of even integers and the set of odd integers. A coset is essentially the same as an equivalence class. Proposition For all x X, there exists a unique coset x such that x x. Proof. First, let us show uniqueness. Suppose Y 1, Y 2, are two cosets with x Y 1, x Y 2. It suffices to show Y 1 = Y 2, that is a Y 1 = a Y 2, as the converse is analogous. If a Y 1 by the third condition of coset, a x and by the second property a Y 2. Next we show existence. We do this by constructing x. We claim that in fact x = {y X y x}. We just have to check that this is indeed a coset and that x x. But, x x by reflexivity of equivalence relations. Now, we check the three definitions of coset. First, x because x x. Second, x 1 x, x 2 x 1 implies x 2 x by transitivity, and so x 2 x. Third, if x 1 x, x 2 x, then x 1 x, x x 2 and by transitivity, x 1 x 2. Definition For X a set, an equivalence relation on X we define the quotient X/ Subset(X), where X/ is the set of cosets. We will show this satisfies a universal property Categories. 2. Category Theory Definition 2.1. A category C is a collection of Objects such that for any two object c 1, c 2 Ob(C ) we can assign a set Hom C (c 1, c 2 ). There exists a map Hom(a, b) Hom(b, c) Hom(a, c), (f, g) g f, which is associative. We can express associativity as a commutative diagram: Hom(c 1, c 2 ) Hom(c 2, c 3 ) Hom(c 3, c 4 ) Hom(c 1, c 2 ) Hom(c 2, c 4 ) Hom(c 1, c 3 ) Hom(c 3, c 4 ) Hom(c 1, c 4 ) Additionally, there exists particular elements id c Hom C (c, c) for all c such that id c f = f, g id c = g. Example 2.2. (1) Let C = Set. The objects are sets and the maps are maps of sets. (2) Let C = Groups. The objects are all groups and the maps are maps are group homomorphisms. (3) Let C = Ring. The objects are rings, and the maps are ring maps. (4) Let C = Rmod. The objects are all modules over a fixed ring R. The maps are maps of R modules. (5) Let C = V ect. The objects are vector spaces and the maps are maps of vector spaces. (6) Let C = GRep. The objects are G-representations. The maps are G equivariant maps of representations. (7) Let =Ab. This is a special case of Rmod where the ring is Z. In other words, the objects are abelian groups and the maps are group homomorphisms. (8) Let X be set. Define C to be the category with Ob(C ) = {x X} and let {, if x 1 x 2 Hom C (x 1, x 2 ) =. { } if x 1 = x 2

5 MATH 55 NOTES 5 (9) For M an arbitrary monoid, we let define C M so that Ob(C M ) = { } and Hom CM (, ) = M Functors. Definition 2.3. For C 1, C 2 two categories, a covariant functor F : C 1 C 2 is a map F : Ob(C 1 ) Ob(C 2 ) and for all c 1, c 1 Ob(C 1 ) we have maps Hom C1 (C 1, C 1 ) Hom C2 (F(C 1), F(C 1 )). Additionally, we require F(id c ) = id F(c) and F(f g) = F(f) F(g). We can similarly define a contravariant functor between F : C 1 C 2 is a map F : Ob(C 1 ) Ob(C 2 ) and for all Hom C1 (C 1, C 1 ) Hom C2 (F(C 1 ), F(C 1)). ) Additionally, we must have F(id c ) = id F(c) and F(f g) = F(g) F(f). Remark 2.4. Observe covariant functors preserve the order of composition and contravariant functors reverse the order of composition. Example 2.5. (1) The identity functor id : C C, c c. (2) The functor Ab Grp sending an abelian group to itself as a group. (3) The trivial functor Grp Ab, G {0}. (4) The abelianization functor Grp Ab, G G/[G, G]. (5) The forgetful functor Grp Set, G G where we send G to the underlying set of G. (6) The group ring functor Grp Ring, G k[g]. (7) The functor of points (a.k.a. Yoneda Functor) h c : C Set, defined by c Hom(c, c), and defined on morphisms by Hom C (c, c ) Hom Set (Hom C (c, c ), Hom C (c, c )), f (g f g). (8) Fix a group H. The functor Grp Grp, G G H, f f id. (9) The free Abelian group functor Set Ab, S Z S. (10) Given φ : R 1 R 2. The restriction functor R 2 Mod R 1 Mod, M M where the former is viewed as an R 2 module and the latter is an R 1 module. There is also a tensoring-up functor R 1 Mod R 2 Mod, M R 2 R1 M Natural Transformations. Definition 2.6. A natural transformation between two functors F : C 1 C 2, G : C 1 C 2 is an assignment c 1 C 1, an element T c1 Hom C2 (F (c 1 ), G(c 1 )). such that for any f : a b F (a)t a G(a) F (f) G(f) F (b) T b G(b) commutes. We denote the set of natural transformations between F and G by Nat(F, G). Theorem 2.7. (Yoneda s Lemma) Let C be arbitrary. Let c, c Ob(C). Consider the set of all natural transformations Nat(h c, h c ). Then, this set is isomorphic to Hom C (c, c ). Proof. Given T : h c h c. We can define a map Nat(h c, h c Hom C (C, C ) by T T c (id C ). For the reverse map, given φ Hom C (c, c ) we define a natural transformation as follows. T c1 : h c (c 1 ) h c (c 1 ), f f φ. That is, T c1 : Hom C (c, c 1 ) Hom C (c, c 1 ), f f φ. Naturality will follow from associativity of morphisms in categories Metric Spaces. 3. 1/29/15 Definition 3.1. A metric space is a set X together with a metric d : X X R such that for all x, y, z X,

6 6 AARON LANDESMAN (1) d(x, y) = 0 x = y (2) d(x, y) = d(y, x) (3) d(x, y) + d(y, z) d(x, z) Definition 3.2. A map X Y of metric spaces is continuous if for all δ > 0, ɛ > 0 so that f(b(x, ɛ)) B(f(x), δ). Exercise 3.3. For X = R n, the function d(x, y) = max i {1,...,n} (x i y i ) defines a metric. { 1, if x y Exercise 3.4. Let X be the metric given by d(x, y) = Then, any map 0, otherwise X Y is continuous Topological Spaces. Definition 3.5. A topological space is a set X together with a collection of open sets {U} such that (1) is open (2) X is open (3) if U α are open then α U α is open (4) If U i are open, then n i=1 U i are open. Note that we require infinite unions of open sets to be open, but only finite intersections. Example 3.6. All subsets of x are open. This gives X the discrete topology Definition 3.7. For X a metric space, we define the metric topology by a set U is open if and only if x U, r > 0, such that B(x, r) U. Proposition 3.8. This defines a topological space. Proof. We need to check the axioms. The first two are automatic. For the third, if x α U α then for any x α U α we can find α with x U α and then there is an open ball containing x contained in U α, hence also contained in α U α. For the fourth axiom, if x i U i, then if B(x, r i ) U i, then B(x, min i r i ) is a ball contained in i U i. Lemma 3.9. The metric d(x, y) = δ x,y then every subset is open. Proof. Observe B(x, 1 2 ) = {x} and unioning such sets gives any set. Lemma Let X be a metric space. Then for all x, r > 0, we have B(x, r) is open. Proof. Take y B(x, r), and set t = d(x, y) with t < r. Choose ɛ with t + ɛ < r. Then B(y, ɛ) is open, contained in B(x, r). Definition A subset Y X is closed if X \ Y is open. Definition (Very Important!) A map of sets f : X Y is continuous if U Y open implies f 1 (U) X is open. Proposition A map f : X Y between metric spaces. Then, X is continuous, in the sense of a map of metric spaces, if and only if f is continuous, in the sense of a map of topological spaces. Proof. First, suppose f is continuous, in the sense of a map metric spaces. Let U Y be an open set. We have a cover of U by open balls B(y, r) U. It suffices to show f 1 (B(y, r)) is open, since f 1 (U) = y U B(y, r y ). Then, it suffices to show that for each x f 1 (B(y, r), there is an open set containing x, contained in f 1 (B(y, r)), but this is precisely the definition of metrically continuous. Conversely, suppose f is continuous in the sense of topological spaces. We know B(f(x), r) is open, and hence it s preimage is open. But, B(x, r ) f 1 (B(f(x), r)), as we saw above.

7 MATH 55 NOTES The Intermediate Value Theorem. Theorem Let A R be bounded above. Then, there exists a least upper bound. That is, there is y R with a y, but for any y < y, there is a A with a > y. Proof. Remark We can also take Theorem 3.14 as part of the definition of real numbers. Theorem Let f : [0, 1] R be continuous. Say f(0) < 0, f(1) > 0. Then, there exists a [0, 1] so that f(a) = 0. Proof. Let A = {x f(x) < 0}. Using Theorem 3.14, the set A has a least upper bound. Call it y. Then, we claim f(y) = 0. If f(y) < 0, using continuity, the preimage of V = [, 0) is open, and since y lies in the preimage of V, there is an open set y U f 1 (U), with U open. In particular, taking a U, a > y, we obtain with a bigger than y, but a A. This is possible, as by metric continuity, we can take U to be an open ball. Hence, y is not an upper bound. The case f(y) > 0 is analogous, except in this case, we produce an upper bound for A less than y. Remark Note this is false if we replace R by Q. For instance, this fails for f(x) = x Review from last time: (1) Metric Spaces (2) Topological Spaces (3) Metric Topology 4. 2/3/15 Definition 4.1. A map X Y is a homeomorphism if f is bijective and f, f 1 are continuous. Remark 4.2. The notion of homeomorphism corresponds to isomorphism in the category of topological spaces. Example 4.3. A continuous bijective map need not have continuous { inverse. For example, x, if x [0, 1] set X = [0, 1] (2, 3] [0, 2]. Defining f : X Y by x x 1, if x (2, 3]. Definition 4.4. The discrete topology, is that in which all sets are open. The trivial (or indiscrete) topology is that in which the only open sets are, X. Example 4.5. Yet another example is given by X X, x x, where we give the domain the discrete topology, meaning all sets are open, and the range has the trivial topology. Then, this map is always continuous, but if X > 1, its inverse is not continuous. Exercise 4.6. Show that the three topologies given on X in the previous lecture all have the same topology. Definition 4.7. A topological space is disconnected if there exist U 1, U 2 open with U 1, U 2 and U 1 U 2 = X, but U 1 U 2 =. A topological space is connected if it is not disconnected. Definition 4.8. Given Y X, with X a topological space, we can define the subspace topology on Y, so that a set U Y is open if U = V Y for V X open. Example 4.9. The space X = [0, 1] (2, 3], with the subspace topology from R is disconnected. To see this, note that [0, 1] is open because it is ( 1 2, 3 2 ) X. We can similarly see (2, 3] is open, and so X is disconnected. Theorem The closed subset [0, 1] R is connected, where R is given the metric topology, and [0, 1] has the subspace topology.

8 8 AARON LANDESMAN Proof. Assume [0, 1] = U 1 U 2 with U 1 U 2 =. We can assume 1 / U 1. Then, in fact the least upper bound of U 1 must be less than 1, because there is some open set containing 1, which is contained in U 1. Let y = LUB(U 1 ). We know y < 1. There are now two cases: either y U 1 or y U 1. Let us assume, for now y 1. If y U 1, then there exists r with (y r, y + r) U 1 implying y + r 2 U 1. So, y + r 2 > y, implying y was not the upper bound of U 1. If y / U 1, then y U 2, and so there exists r so that (y r, y + r) U 2, and we can take y r 2 is also an upper bound for U 1 implying y was not a least upper bound for U 1. Corollary (Intermediate Value Theorem) Any f : [0, 1] R, continuous with f(0) < 0, f(1) > 0, then there exists a [0, 1] with f(a) = 0 Proof. Let U 1 = (, 0), U 2 = (0, ), if there does not exist such an a, then f 1 (U 1 ) f 1 (U 2 ) covers [0, 1], implying [0, 1] is disconnected, a contradiction. Definition A sequence (in X) is a map N X. We often identify this function with the set of points in its image, indexed by natural numbers. Definition Let X be a topological space. A sequence of points x n x if for all x U X open, there exists N such that for all n N, x n U. Example If X has the trivial topology, then any sequence converges to all points of X. Definition A topological space is Hausdorff if for any x x there are U, U with x U, x U so that U U. Lemma If X is a Hausdorff space (e.g. a metric space), and x n x, x n x, then x = x. Proof. Suppose the sequence converges to two distinct points x, x. If x U, x U take N, N as in the definition of convergence for x, x. Then, n N = x n U, n N. Taking n max(n, N ) we obtain a contradiction, as x n U U =. Lemma Any Metric space X is Hausdorff. Proof. For x, y X, say d(x, y) = r. containing x, y which do not intersect. Then, B(x, r/4), B(y, r/4) define two open sets Definition Let X be a set. Define the cofinite topology on X, by letting all closed sets be those that are of the form (1) (2) X (3) {x 1,..., x n } Example If X is an infinite set with the cofinite topology, then X is not Hausdorff. Indeed, U 1 U 2 if both U 1, U 2. Theorem Let x n = 1 n. Then, x n 0. Proof. Given r, take N > 1 r. Then, for all n N, we have x n = 1 n < 1 N r. Definition Let x n : N X, y n N X be a sequence. Then, y m is a subsequence of x n if there exists an increasing map φ : N N so that the following diagram commutes N φ y n x n N X Remark Equivalently, but less abstractly, we can write y m = x im with i 1 < i 2 <.

9 MATH 55 NOTES 9 Definition Let x n be a sequence in X. Then, x is a partial limit of x n if there exists a subsequence x in that converges to x. Lemma Let x n be a sequence such that (1) x n x if and only if for all U, x U, then for all but finitely many n N, we have X n U. (2) Assuming further that X is a first countable space (e.g. a metric space), X is a partial limit if and only if for all r, there exist infinitely many n so that x n B(x, r). Proof. (1) Pick an N so that for all n > N, x n in U. Then, there are only finitely many x n / U, namely, there are at most N such x i. Conversely, choose N to be the maximum over all N with x n / U. (2) Let x be a partial limit. We want to know that for all U, there are infinitely indices i with x i U. Let x in be a convergent subsequence so that for all U, there exists N so that for all n N, we have x in U. By first countability, we will construct a chain U 4 U 3 U 2 U 1. Take x i1 U 1. There exists i 2 > i 1 so that x i2 U 2. This produces a subsequence x in so that x in U n. Then, x in x. To see this, for any U, x U, by first countability, we have some U N U. Then x in U N for n N, hence x in U for n N. So, x is a partial limit of x n. Definition Let X be a metric space. Let x n be a sequence. Then, x n is Cauchy if for all r there exists N so that for any n, n N, we have d(x n, x n ) < r. Definition Let X be a metric space. A subset A X is bounded if x, R with A B(x, R). Lemma (1) Every Cauchy sequence is bounded. (2) If a Cauchy sequence has a partial limit x then it converges to x. Proof. (1) Take r = 1. There exists N so that for all n, n N we have d(x n, x n ) < 1. So, choosing a large enough ball around x N, with radius bigger than 1, so that the first N terms lie in that ball, then all of the sequence will lie in that ball. (2) Let x in x. Pick r. We want and N so that for all n N, then x n B(x, r). Pick N given by the definition of Cauchy, so that for any n, n N, we have d(x n, x n ) < r 2. Then, there exists k > N, so that d(x i k, x) < r/2. Then, d(x, x n ) d(x, x k ) + d(x k, x n ) < r/2 + r/2 = r. Definition A space X is complete if every Cauchy sequence converges. Lemma If x n converges, then x n is Cauchy. Proof. Follows from the triangle inequality. Theorem (Heine-Borel) Every bounded sequence in R has a partial limit. Proof. Let A = {a R there exist infinitely many indices n such that x n a}. First, A because x n is bounded from below. Second, A is bounded above, because x n are bounded above. Let y = LUB(A). We will show that for all r there are infinitely many n so that x n (y r, y + r). Suppose not. Then, there are infinitely many n so that x n y + r. Then, y + r/2 A, contradicting that y = LUB(A). Suppose that all but finitely many x n are y r. Then, y r/2 would also be an upper bound for A, and so y is not the least upper bound. Theorem Every Cauchy sequence in R converges. Proof. Every Cauchy sequence is bounded, as we saw above. If x is a partial limit of our sequence, then since our sequence is Cauchy, it actually converges to x.

10 10 AARON LANDESMAN Review: 5. 2/5/15 (1) Cauchy sequences (2) Convergent = Cauchy Convergent (3) X is complete if every Cauchy sequence is convergent (4) R is complete (5) (Weierstrass) every sequence in [0, 1] has a convergent subsequence. Definition 5.1. A subset X X is dense if for any U X, X U. Lemma 5.2. If X is a metric space, X B(x, r) X. X is dense if and only if for all x, r then Proof. If X is dense, then we know it intersects every open set, in particular it intersects any ball. Conversely, if X intersects every ball non-trivially, we know every open set contains a ball, so it intersects every open set non-trivially. Theorem 5.3. Let X be a metric space. (1) There exists i : X X so that (a) X is complete (b) i is an isometric embedding (c) Im(i) is dense. (2) If f : X Y an isometric embedding, there exists f so that X f f Y X (3) If f : X Y an isometric embedding and f(x) is dense, then f, gotten from the above part, is an isometry. Proof. Let Cauchy(X) be the set of all Cauchy sequences d(x n, y n ) = lim n (d(x n, y n )). Lemma 5.4. Given a, b, c, d X then d(b, d) d(a, c) d(a, b) + d(c, d). Proof. Triangle Inequality. Lemma 5.5. If x n, y n are Cauchy sequences, then d(x n, y n ) defines a Cauchy sequence. Proof. Given ɛ, take N so that for all m, n N we have x m x n < ɛ/2 and y m y n < ɛ/2. Then, for any n, m > N, we have d(x n, y n ) d(x m, y m ) < ɛ. This follows from the above lemma. Lemma 5.6. The distance function gotten from the above lemma on Cauchy sequences Cauchy(X) satisfies the triangle inequality. Proof. Definition 5.7. A space X is a premetric space if X satisfies the triangle inequality and symmetry, and d(x, x) = 0, but we do not necessarily have x y = d(x, y) 0. Given X a premetric space, define an equivalence relation x 1 x n if d(x 1, x 2 ) = 0. Then define X = Cauchy(X)/. Using the above lemma we can see (on the homework) that this is indeed a metric space, satisfying the properties claimed in the theorem.

11 MATH 55 NOTES 11 Corollary 5.8. For any isometric embedding i 1 : X X 1, with X 1 complete and i 1 has dense image. Then, there exists a commutative diagram X X. Proof. This follows from part c of 5.3 X 1 Corollary 5.9. Let Q denote the metric completion of Q (not the algebraic closure!). Then, Q = R Proof. We have an inclusion Q R satisfying the conditions of Theorem 5.3 Theorem Let f : [0, 1] R. Then there is x [0, 1] so that f(x) f(x ) for all x [0, 1]. Proof. First, observe f([0, 1]) is bounded above. To see this, if it were not bounded above, then there exists a sequence x n with f(x n ) n. By Weierstrass, x n has a convergent subsequence, x in x. Then, f(x in ) f(x), implying f(x) is bigger than every integer, a contradiction. Now, since f([0, 1]) is bounded above, take LUB(f([0, 1])). Observe there is x [0, 1] with f(x) = y. To see this, note there is some x i with f(x i ) im f([0, 1]) with f(x i ) (y 1 n, y], and then there is some convergent subsequence x in x. Since f(x in ) y, f(x) = y. Then, this x satisfies the hypotheses of the theorem. Definition A topological space X is sequentially compact if if every sequence has a convergent subsequence. Proposition For f : X Y, if X is sequentially compact, so is f(x), where we give f(x) the subspace topology. Proof. Take a sequence f(x n ). Say x in is a convergent subsequence of x n, then f(x in ) f(x), and f(x) is sequentially compact Heine-Borel. We next prove a more standard version of Heine Borel. Lemma If Y Y with Y sequentially compact and Y first countable and Hausdorff, then Y is closed in Y. Proof. Because Y is first countable, it is enough to show that if y n and y i Y then y Y (using problem set 1). Let y n be a sequence in Y. We then have some subsequence y in y Y. Because Y is Hausdorff, any sequence has a unique limit, and therefore y = y implying y Y and so Y is closed. Lemma Say Y Y with Y Hausdorff, and Y is closed, Y sequentially compact. Then Y is sequentially compact. Proof. Say y n y. with y Y. Then, Y Y, then y Y, because Y is closed. Therefore, Y is sequentially compact. Theorem (Heine-Borel, standard version) We have A R sequentially compact if and only if A is bounded an closed. Proof. Suppose A is bounded and closed. Then, A [a, b]. We wish to show A is sequentially compact. We know [a, b] is sequentially compact, by Weierstrass. The above lemma implies that A is sequentially compact. Conversely, suppose A is sequentially compact. We know A is closed by a lemma above. Additionally, A is bounded, because if A were unbounded, we can construct a subsequence that has no convergent subsequence (constructed so that each point is distance at least 1 from all the previous points).

12 12 AARON LANDESMAN Lemma If A R is bounded and closed, then LUB(A) A Proof. Done above in the proof of completions. Lemma Heine-Borel implies Theorem 5.10 Proof. We know [0, 1] is bounded and closed, so [0, 1], is sequentially compact, so f([0, 1]) is sequentially compact, hence bounded and closed, by Heine-Borel. Definition (Important Definition!) A topological space is compact if α A U α = X then there is a finite subset A A so that α A U α = X. Proposition If f : X Y is continuous and X compact, then f(x) is compact. Proof. Choose an open cover f(x) = α U α. Then, α f 1 (U α ) = X. Choose a finite subset B A. Then, β f 1 (U β ) = X implies β U β = f(x). Lemma Y Y with Y compact (in the subset topology) and Y Hausdorff, then Y is closed. Proof. It suffices to show U = Y \ Y is open. For this, it suffices to show each x Y, we have some open set V with x V U. To do this, for each y Y, take A y, B y open sets with y A y, x B y and A y B y =, which exist as Y is Hausdorff. Therefore, y Y A y. By compactness of Y, we have a finite subset y 1,..., y n so that Y n i=1 A y i. Therefore, V = n i=1 B y i U, which is a finite intersection of opens, hence open. It follows that x V U is an open set of the desired form. Hence, U is open, and Y is closed. Definition We say X is first countable if every x X has a countable neighborhood base. That is, there exist countably many U 1,..., U n,... so that for every open x U there is some U i with U i U. Proposition Say X is first countable and compact. Then X is sequentially compact. Proof. Say x n is a sequence. Suppose it has no partial limits. For every x X there is U x X with U x contains only finitely may x i in the sequence. Then, x U x = X. By compactness, there exists a finite cover n i=1 U x i = X. But this only contains a finite number of elements of our sequence, implying the entire space has only finitely many elements of the sequence, a contradiction. Definition A neighborhood base for a topological space X is a collection of open sets of X, U i so that any open set U in X can be written as U = i U i. Definition A space X is second countable if if has a countable neighborhood base. That is, a neighborhood base containing only countable many closed subsets. Proposition If X is second countable, and sequentially compact, then X is compact. Proof. 6. 2/12/15 Lemma 6.1. (1) If a i a, b i b then if we define c i = a i + b i then c i a + b. (2) a i b i ab (3) For a 0, we have 1/a i 1/a. Proof. This follows from continuity of addition multiplication, and inversion, which can be checked by simple ɛ, δ arguments. Lemma 6.2. (Sandwich Rule) If a n a n a n and a n a, a n a then a n a. Proof. Use ɛ, δ arguments. Example 6.3. For a < 1 we have a n 0. To see this, write b = 1 a and then bn = (1 + (b 1)) n n(b 1). Then, 0 a n 1 (b 1)n and by the sandwich rule, an 0.

13 MATH 55 NOTES 13 Lemma 6.4. For any a R 0, a 1 n exists. Proof. Intermediate value theorem, applied to f(x) = x n. Example 6.5. We have a 1 n 1 as n. To see this, it suffices to show it for a > 1, using the fact that 1 a i 1 implies a i 1. Then, 1 a 1 a 1 + a 1 n, using the binomial theorem. Example 6.6. We have n 1 n 1. We can again use the sandwich rule, bounding 1 n 1 n 2 n 1, again using the binomial theorem, since (1 + (n 1 n 1)) n n(n 1)(n 1 n 1)/2. Definition 6.7. Given a i a sequence the series a i refers to the partial sums b i = n i=1 a i. Notation 6.8. From now on, we will often use a i to denote a series, and b i to denote the corresponding partials sums b i = i j=1 a j. Definition 6.9. Let a i be a series with n i=1 a i = b i. Then we say the series a i converges if the sequence b i converges. Definition The series a i converges absolutely if the series a i converges. Lemma Let a n = a n for a < 1. This series converges absolutely. Proof. We have i ai = 1 1 a 1 1 a an+1 1 a 1. Lemma If the series a n converges then the sequence a n converges to 0. Proof. Say b n a and b n+1 a then a i = b n+1 b n a a = 0. Lemma The series a n converges if and only if for every ɛ > 0 there exists N so that n 2 i=n 1 a i < ɛ Proof. This condition is equivalent to the sequence b i of partial sums being Cauchy. But, since the reals are complete, Cauchy convergence is equivalent to convergence. Lemma If a i converges absolutely, then it converges. Proof. We have n 2 i=n 1 a i ɛ then certainly n 2 i=n 1 a i ɛ by the triangle inequality, and this implies convergence by the previous lemma. Theorem Let a n converge absolutely. and φ : N N, then a φ(i) converges and the sum equals i a i. Proof. Theorem Let a n be convergent but not absolutely convergent. Then, for any b R, there exists φ : N N with i a φ(i) = b. Proof. Omitted, see Rudin. { 1/ i/2, if n 0 (mod 2) Example A sequence which converges but not absolutely is a i = 1/ (i 1)/2 otherwise Lemma Let b n be a bounded, monotonically increasing sequence. Then b n is convergent. Proof. (Sketch) Let B = lim sup b i, which exists as b i is bounded. We claim b i B. To see this, take any ɛ > 0 and then there exists some N so that for any n > N we have b n > B r. Since B is an upper bound, and b i are arbitrarily close to B, we have b i B. Proposition Let a n have positive terms. Then, a n converges if and only if there exists A > 0 so that i a i A. Proof. If the partial sums converge to b, then b bounds b n for all n. For the converse, use the above lemma.

14 14 AARON LANDESMAN Corollary Say a i, ã i have positive terms. converges then a i converges. Proof. If n i=1 a i n i=1 ãi for all n and ã i Proposition (1) Let a i be so that lim sup n a n < 1 = a n converges absolutely. (2) Let a n be so that lim sup a n 1 n > 1 then a n diverges. Proof. (1) Let a = lim sup a n 1 n. Take a = lim sup a n 1 n and ã i = a i. For all but finitely many i, we have a i 1 n a if and only if a i a i, and we know ã i converges as it is a geometric series. (2) We have the a i do not converge to 0, as there are infinitely many i for which a i 1. Proposition Assume that lim sup ai+1 a i < 1. Then, a i converge absolutely. Proof. Let r = lim sup an+1 a n and for all but finitely many i we have ai+1 a i and so we may take r r < 1 we may assume ai+1 a i < r and so a i+1 r i a 1. Example Let a n = xn an+1 n!. This converges absolutely. Note a n = x/n It satisfies the ratio test, and hence converges. Example Take a n = 1 n a for a > 0. Then, if we use the ration test, we have an+1 a n 1. The root test also is inconclusive, since 1 n a n 1. Lemma Let a n be a series of positive numbers that is monotonically decreasing, and let ã n = 2 n a 2 n. Then a n converges if and only if ã n does. Proof. Homework. Theorem The series 1 n a converges for a > 1 and diverges for a 1. Proof. For our series, using the associated zeta series from the above lemma ã n = 1 1 b, which converges if and only if b > 1, which is equivalent to a > 1. n Definition Let exp(x) = i=0 xi i! 2 (a 1)n = Lemma If a n converge absolutely to A and b n converge absolutely to B and c n = i a ib n i then c n converges absolutely and converges to AB Proof. We want to find and N for which 2N i=0 c i AB) < ɛ. Consider ( n i=0 a n) ( n i=0 b n AB < ɛ 3 for n N. These two sums differ by a sum of a ib j with i + j 2n but i > n or j > n. Then, a i b j < b j a i < B a i < B ɛ/b = ɛ, j n,n i 2n 1 j n n<i 2n where the penultimate step uses cauchyness of b i. Theorem exp(x) exp(y) = exp(x + y). n<i 2n Proof. Note (x+y)n n! = n x i y n i i=0 i!(n i)!, and so we can use the above lemma. Definition Let X be a sequentially compact topological space. Let F unct cont (X, Y ) denote the set of continuous functions f : X Y. Define a metric on F unct cont (X, R) by ρ(f, g) = max x X f(x) g(x). Lemma The function ρ defined above is a metric. Proof. Omitted. Definition Let f : X Y be a map between metric spaces. We say f is uniformly continuous if for all ɛ there exists δ such that ρ(x 1, x 2 ) < δ, then ρ(f(x 1 ), f(x 2 )) < ɛ.

15 MATH 55 NOTES 15 Example If ρ(f(x 1 ), f(x 2 )) C ρ(x 1, x 2 ) then f is uniformly continuous. Such functions are called lipschitz functions. Proposition X f i Y with X X a dense injection, f uniformly continuous, Y complete. Then, there exists f : X Y so that f = f i and f uniformly continuous. Proof. Given x X choose x n x, with x n X. Lemma We have f(x n ) is convergent. Proof. We see x n is Cauchy. By uniform continuity, f(x n ) is Cauchy. As Y is complete, f(x n ) y. Define y = lim n f(x n ), and define f(x) = y. We only need show f is uniformly continuous. Given ɛ, let δ be so that ρ(x 1, x 2 ) < δ = ρ(f(x 1 ), f(x 2 )) < ɛ for x 1, x 2 X. We want the same for x 1, x 2 X. Take x 1, x 2 X let ρ(x 1, x 2 ) = δ < δ. We want to show ρ(f(x 1 ), f(x 2 )) < ɛ. Choose x n 1 x 1, x n 2 x 2. So, for n > N, we have ρ(x n 1, x n 2 ) < δ. So, ρ(f(x 1 ), f(x 2 )) = lim n ρ(f(x n 1 ), f(x n 2 )). Therefore, rho(f(x 1 ), f(x 2 )) ɛ. Uniqueness follows from the fact that Y is Hausdorff, so the sequence converges to a unique limit point. Definition Define F unct cont,pw linear (X, R) to be the set of continuous piecewise linear functions f : X R. Corollary Given X F unct cont ([a, b], R) R F unct cont,pw linear ([a, b], R) we get a unique map : F unct cont ([a, b], R) R. Proof. We need to check uniform continuous of. If max x f(x) g(x) < C then f g (b a)c then f g (b a) maxx ( f(x) g(x) ). If f is piecewise linear, then f < C implies f < (b a)c. Theorem We have F unct cont,pw linear ([a, b], R) is dense in F unct cont ([a, b], R). Proof. Let us prove this, assuming Theorem 6.39 below. But this is immediate, as [a, b] is compact. Given f : [a, b] R, and ɛ we will produce f piecewise linear so that ρ(f(x) f(x) < ɛ. Choose the δ given by uniform continuity. Choose a partition of [a, b] into [x i, x i+1 ] with x i x i+1 < δ. Then, define f(x) = f(x i ), for x i x < x i+1. It is then clear that Theorem Let X be a compact metric space and f : X Y. Then f continuous implies f is uniformly continuous. Proof. Given ɛ we want δ so that ρ(x 1, x 2 ) < δ implies f(x 1, x 2 ) < ɛ. Proof 1. For all x let δ x be so that ρ(x, x) < δ x implies ρ(f(x), f(x )) < ɛ/2. Consider the balls B(x, δ x ). Let X = x B(x, δx/2). We can replace this by a finite subcover, and let δ = min(δ x /2). We claim this δ works, because ρ(x 1, x 2 ) < δimplies there exists x with x 1, x 2 B(x, δ x ) implies ρ(f(x 1 ), f(x)) < ɛ/2, ρ(f(x 2 ), f(x)) < ɛ/2 = ρ(f(x 1 ), f(x 2 )) < ɛ.

16 16 AARON LANDESMAN Proof 2. For all n, there exists x n 1, x n 2 so that ρ(x n 1, x n 2 ) < 1/n but ρ(f(x 1 ), f(x 2 )) ɛ. Choose convergent subsequences of x n i x i, and we obtain x 1 = x 2 = x. This is a contradiction, as for N large enough, we have ρ(f(x) f(x n i )) < ɛ/3 implying ρ(f(x 1, x 2 )) < ɛ. Definition We define a map : F unct cont ([a, b], R), called integration by first defining it on linear functions, then continuous piecewise linear functions, then positive continuous functions, then all continuous functions. (1) If f is linear on [a, b] then [a,b] f = (b a)d + b2 a 2 c. 2 (2) If f is linear of [x 0, x 1 ],..., [x n 1, x n ], with x 0 = a, x n = b. The define [a,b] f = n i=1 [x i 1,x i] (3) For positive functions, define the integral by the extension of piecewise linear integrals, using the theorems above. (4) For general functions f let f + = max(f, 0), f = min(0, f) and define f = f+ + f. f. Lemma (1) f 1 + f 2 = f 1 + f 2. (2) cf = c f. (3) [a,b] [a,c] f + [c,b] f (4) f 0 = f 0. (5) If f 0 and f = 0 then f = 0. Proof. (1) Look at F unct cont,pw linear ([a, b], R) 2 R F unct cont ([a, b], R) 2 and consider the two functions f 1, f 2 f 1 + f 2 and f 1, f 2 f 1 + f 2. (2) Consider F unct cont,pw linear ([a, b], R) R F unct cont ([a, b], R) The two functions f c f, f cf obviously agree of piecewise linear functions, hence everywhere. (3) Similar to the above part. (4) Lemma We have F unct cont,pw linear ([a, b], R 0 ) is dense in F unct cont ([a, b], R).

17 MATH 55 NOTES 17 Proof. We have F unct cont,pw linear ([a, b], R 0 ) R 0 F unct cont ([a, b], R 0 ) We need to just show the image of the left map is dense. Given ɛ if ρ(f, f) < ɛ define f + = max(f, 0) then ρ(f, f + ) < ɛ. Let x [a, b] be so that f(x) with f(x) = r > 0 and choose δ so that on [x δ, x + δ], f(x) r/2. We then have that the integral is positive on [x δ, x + δ] hence on [a, b] as it is positive outside [x δ, x + δ] by the previous part. Corollary If f g then f g. Proof. Use points 3 and 4. Notation Choose [a, b] an interval and a partition [x 0,..., x n ] = p with x i [x i 1, x i ]. Define S p (f) = (x i x i 1 ) f(x i ). Theorem Let p n be a sequence of partitions so that a sequence r n 0 so that for all n, we have r n x i+1 x i is the nth partition. Then S pn (f) f. ɛ b a. Proof. Given ɛ, let δ be so that for all i, we have ρ(y, z) < δ = f(y) f(x) < Then, f = i [x f, and S(f) = i 1,x i] i (x i+1 x i ) f(x i ) and f(x) f(x i ) < ɛ/b a implying [x f i,x i+1] [x f(x i,x i+1] i ) (x i x i+1 ) ɛ/b a, and so the sum of all of these is at most ɛ. 7. 2/19/15 Today, we are simply reviewing problems from problem sets. Exercise 7.1. Show that if y n = x n n is a Cauchy sequence in Cauch(X) then y n have a limit. Proof. First, we can see X is dense in Cauch(X). Using this, we can choose x n X so that ρ(x n, y n ) < 1/n, and the limit of a sequence of these x n agrees with the limit of the y n. Exercise 7.2. If p : X Y X is the projection with Y compact, then p is closed. Proof. Let Z X be closed. Let x X, x / p(z). Take x X, and note x Y is not contained in Z. Then, around each x, y x Y, we can find an open set not intersecting Z. We can take these open subsets of the form U y V y. Then, we can take finitely many such subsets, by compactness of Y. Finally, intersecting the corresponding U y we obtain U y Y is an open set not intersecting Z and so U y is an open set not intersecting p(z). Definition 7.3. Let X be a topological space. X is locally compact if for every x X, x U, there is some open V and compact C with x V C U. Definition 7.4. The one point compactification of X is the set X = X { } so that open sets are of the form U =, U = X, U X is open, or U and the complement of U is compact. Lemma 7.5. Given X, the one point compactification, X, is compact. Proof. Take an open cover. One element U of the open cover contains. Then, we only need to find a finite number covering X \ U, which exists because it is compact. Lemma 7.6. If X is Hausdorff, then X is a topological space.

18 18 AARON LANDESMAN Proof. It is straightforward to check that arbitrary unions of opens and finite intersections are open. Lemma 7.7. We have X is Hausdorff if and only if X is locally compact. Proof. Again, follows from the definitions. Exercise 7.8. For r R the function x r exists. Proof. Say r n r. We want to show x rn is Cauchy. Lemma 7.9. For x > 0, x r is monotonic. Lemma x 1 n 1 as n. Then, x rn x rm = x rn (x rm rn 1) 0. Exercise Show, x r is continuous. Proof. If x i x then x r i xr = x r ((x i /x) r 1). Since ( ) x i r x is monotonic in r so this ratio converges to 0 as x i x. Exercise Show a x is continuous in x. Proof. It suffices to show a ri 1 as r i 0, which follows as a 1 n 1 and the squeeze theorem. Lemma If f : [a, b] R is continuous and strictly increasing on rationals, then it is monotonic on all reals. Proof. For x < y, want f(x) < f(y). Just write both real numbers as a limit of rationals. Lemma The function a x is surjective onto R >0. Proof. This follows from the intermediate value theorem. 8. 2/24/15 Definition 8.1. A function f : X Y is continuous at x if for all f(x) U y there exists x U x with f(u x ) U Y. Definition 8.2. A function ˆf : X x Y, we say the limit of ˆf at x is y if f can be extended to f : X Y so that f(x) = y and f is continuous at x. Lemma 8.3. The limit of f at x is y if for all U y y there exists U x x so that for all x U x x we have f(x ) U y. Proof. Remark 8.4. Let Y = R. If lim x x f i (x ) = y i and lim x x f 1 (x )f 2 (x ) = y 1 y 2, lim x x f 1 (x )+ f 2 (x ) = y 1 + y 2. If lim x x f(x ) = 0 and g is bounded in X x then lim x x f(x )g(x ) = 0. Definition 8.5. Let X be a metric space and f : X x 0 R. Then, f is o(ρ(x, x 0 ) n ), or equivalently f o(ρ(x, x 0 ) n ), viewing o(ρ(x, x 0 ) n f ) as a set of functions, if lim x x0 ρ(x,x 0) = n 0 if and only if for all ɛ the is some δ so that ρ(x, x 0 ) < δ implies f(x) < ɛ ρ(x, x 0 ) n. Remark 8.6. Observe f is o(1) if and only if lim x0 x f(x) = 0. Example 8.7. If X = R and f(x) = x, x 0 = 0. Then, for which n is f in o((x x 0 ) n )? Answer: only n = 0.

19 MATH 55 NOTES 19 Definition 8.8. We say f : [a, b] R is differentiable at x 0 if there exists f (x 0 ) R so that f(x) f(x 0 ) (x x 0 )f (x 0 ) is o(x x 0 ). I.e., if f(x) f(x 0 ) (x x 0 )f (x 0 ) lim = 0. x x 0 x x 0 Equivalently, for every ɛ there exists δ so that x x 0 < δ implies f(x) f(x 0) (x x 0 )f (x 0 ) x x 0 < ɛ. Lemma 8.9. If f (x 0 ) exists, then it is unique. Proof. Take the difference of two such derivatives, which is f 1(x 0 ) f 2(x 0 ), but tends to 0, hence equals 0. Lemma If f is differentiable at x 0 it is continuous at x 0. Proof. We want to see lim x x0 f(x) = f(x 0 ). Observe ( f(x) f(x0 ) f ) (x 0 )(x x 0 ) lim f(x) = lim (x x 0 )+f(x 0 )+f (x 0 )(x x 0 ) = f(x 0 ). x x 0 x x 0 x x 0 Example Say f(x) = x. What is f (x)? We see f (x 0 ) = 1, directly from the definition. Lemma If f, g are differentiable at x 0 then f g is differentiable at x 0 and (f g) (x 0 ) = f (x 0 )g(x 0 ) + f(x 0 )g (x 0 ). Proof. Behold lim f(x)g(x) f(x 0)g(x 0 ) f (x 0 )g(x 0 ) (x x 0 ) g (x 0 )f(x 0 )(x x 0 ) x x 0 = lim f(x)g(x) f(x 0)g(x) (x x 0 )f (x 0 )g(x) + f(x 0)g(x) f(x 0 )g(x) (x x 0 )g (x 0 )f(x 0 ) + x x 0 x x 0 (x x 0 )g(x) g(x 0 )f (x 0 ) x x 0 and all three terms tend to 0. Example We have f(x) = x n then f (x) = nx n 1. Lemma (Chain rule) Suppose f : (a, b) (c, d), g : (c, d) R. If f is differentiable at x 0 and g at y 0 = f(x 0 ) then g f is differentiable at x 0 and (g f) (x 0 ) = g (y 0 ) f (x 0 ). Proof. Behold lim g f(x) g f(x 0) g (y 0 )f (x 0 )(x x 0 ) x x 0 = lim g(f(x)) g(y 0) g (y 0 )f(x) y 0 ) + g (y 0 )(f(x) y 0 g (x 0 )(x x 0 ) x x 0 x x 0 = lim g(f(x)) g(y 0) g (y 0 )f(x) y 0 ) x x 0 Now, we claim this goes to 0. First, for ɛ given, choose µ so that y y 0 < µ implies y y 0 < µ implies g(y) g (y 0) g (y 0)(y y 0) y y 0 < ɛ. Let δ be so that x x 0 < δ = f(x) y 0 < µ, and then g(f(x)) g(y 0 ) g (y 0 )(f(x) y 0 ) ɛ f(x) y 0. x x 0 x x 0 Note that lim f(x) y0 x x 0 = f (x 0 ), is bounded, hence the above expression goes to 0.

20 20 AARON LANDESMAN Theorem Let f : [a, b] R and f continuous on [a, b] is differentiable on (a, b). Then f attains its maximum at x 0 (a, b) implies f (x 0 ) = 0. Proof. Assume f (x 0 ) > 0. Then, f(x) = f(x 0 ) + (x x 0 )(f (x 0 ) + f(x) f(x0) f (x 0)(x x 0) x x 0 ), and so for x > x 0 we have f(x) > f(x 0 ). Theorem Let f : [a, b] R be continuous and differentiable on (a, b) with f(a) = f(b). Then, there exists x (a, b) so that f (x) = 0. Proof. Let x be the maximum point. The only problem is if the maximum occurs at one of the endpoints. Then let x be the minimum point. If that is also at one of the endpoints, the the function is constant. Theorem There exists x so that f(b) f(a) b a = f (x). Consider g(x) = f(x) f(b) f(a) b a (x a). Then there is an x for which g (x) = 0 and so g (x) = f (x) f(b) f(a) b a. Proof. (1) If f is differentiable on (a, b) then f (x) 0 implies f is nondecreasing (2) f (x) 0 implies f is non-increasing (3) f (x) = 0 implies f is constant. (1) Suppose not. Then there exists x, y with x < y but f(x) > f(y). Then f(y) f(x) y x = f (z), a contradiction. (2) Similar to the above (3) If f is not increasing and not decreasing, then f is constant. Definition Let f : (a, b) R. We say that f is differentiable n times at X 0 if f is differentiable n 1 times on (a, b ) x 0 and f (n 1) : (a, b ) R is differentiable at x 0. Theorem Let f be differentiable n times at x 0 so that f (k) (x 0 ) = 0 for k = 0,..., n. Then, f = o((x x 0 ) n ). Proof. If n = 1 we have f(x) x x 0 0 is equivalent to f o((x x 0 ) n ). Next, we will do n = 2 as the general case is similar. Take f(x) = (x x 0 )f (x 1 ), we have f(x)/(x x 0 ) 2 = f (x 1 )/(x x 0 ) f (x 1 )/(x 1 x 0 ) 0. For the n = 3 case, f(x) (x x 0 ) 3 = f (x 1 ) (x x 0 ) 2 f (x 1 ) (x x 0 )(x 1 x 0 ) f (x 0 ) f (x 2 ) 0. x x 0 x 2 x 0 Theorem Let f be differentiable n times at x 0, then n f i (x 0 )(x x 0 ) i f(x) i! is o((x x 0 ) n ). i=0 Proof. Note that this function is n times differentiable at x 0, and so by the previous theorem, it suffices to check this has first n derivatives equal to 0. But we precisely constructed this difference to have first n derivatives 0. Theorem Let f : (a, b) R by f(x) = { 0, if x 0 e 1/x. x 0 Proof. For now, assume e x is differentiable. We ll prove this later. We need to show f(x) = o(x n ) for all n. We want to show e 1/x /x n 0 as x 0. Letting y = 1 x then yn /e y 0 e y /y n.

21 MATH 55 NOTES /26/15 Theorem 9.1. Let f : [a, b] R be continuous. Let F (x) = f Then, F (x) is differentiable and F (x) = f(x), and F is continuous on [a, [a,x] b]. Proof. We want to show lim F (x ) F (x) f(x)(x x) x 0. x f f(x) x x Given ɛ, we want δ so that x x < δ implies < ɛ. Since f is continuous, there is δ so that f(x ) f(x) ɛ. This same δ works to bound F (x). We know F is continuous on (a, b). To see it is continuous at a, we have f C, then f C(x a) < ɛ for x [a,x] close to a. Similarly, it is continuous at b. Lemma 9.2. Let f be a continuous function on [a, b] differentiable on (a, b) so that f = 0. Then, f is constant. Proof. For any d there exists a e for which, f(c) f(d) = (c d)f (e) = 0 implying f(c) = f(d) Corollary 9.3. Let F be continuous on [a, b] and differentiable on (a, b) and F extends to a continuous function on [a, b], then F (b) F (a) = [a,b] F. Proof. Set G(x) = x a F. By the preceding theorem, G is differentiable and note that G F has zero derivative. Hence, it is a constant. Then, G(x) F (x) G(a) F (a) = F (a). Therefore, G(b) F (b) = F (a). This implies [a,b] F = F (b) F (a). Definition 9.4. We say f n f converge uniformly, if they converge as L functions (in the sup metric.) That is, for all ɛ there is N so that f(x) f n (x) ɛ. Theorem 9.5. Let f n be a sequence of continuous functions on [a, b] and differentiable on (a, b). Assume each f n extends to a continuous function on [a, b] and f n g uniformly and f n (x) f(x) uniformly. Then, f differentiable and f = g. Proof. Take f(x) = g + (lim f n (c) g). [a,x] n [a,c] By the fundamental theorem of calculus, f is differentiable and f = g. We need to check f n (x) f uniformly. We have f n (x) = f n + (f n (c) [a,x] [a,c] f n) [a,x] g + (lim f n (c) g). [a,c] We claim all three terms converge uniformly to the corresponding terms. second are by definition. We need to show f n g f n g ɛ(x a). [a,x] [a,x] [a,x] The first and Exercise 9.6. If the f n do not converge uniformly the above theorem may not hold. For example, consider 0, if x < 1/2 1/n g n = 1 if x > 1/2 + 1/n. n/2(x 1/2 + 1/n)otherwise Let f n (x) = x 0 g n. We claim f n f where { 0, if x 1/2 f = x 1 2 otherwise.

22 22 AARON LANDESMAN We want f n (x) f(x). Then, f n (x) = [0,x] g n g 0. However, f is not differentiable at 1 2. Proposition 9.7. Letting exp(x) = n i=0 xn n!, then (exp(x)) = exp(x). Proof. Using the above theorem, we may note that f n = n i=0 xn n! are polynomials. Choose an arbitrary interval [a, b]. Note f n are polynomials, hence continuous on [a, b]. We want f n = f n 1 exp uniformly, which holds as the difference at b bounds the difference everywhere. This shows exp (x) = n i=0 xn 1 (n 1)! = exp(x), as desired. Definition 9.8. Let Γ : (a, b) R n. Then, writing γ = (γ 1,..., γ n ) with γ i : (a, b) R, we say γ is differentiable if for all i, γ i is differentiable. Notation 9.9. Let γ : [a, b] R n be differentiable on (a, b) so that f : (a, b) R n extends to a continuous function on [a, b]. Choose a norm on R n. Define lg(γ) = γ. Notation Let p, be a partition a = c 0 c 1 c m = b of [a, b]. Let m lg pn (γ) = γ(c i ) γ(c i 1 ). i=1 Theorem We have lim n lg pn (γ) = lg(γ) for any sequence of partitions p n so that max m (c i c i 1 ) 0. [a,b] Proof. Letting d i be arbitrary points, we want to show m γ (d i ) γ. i=1 The main step is to show [a,b] f [a,b] f. [a,b] 10. 3/3/15 Definition Let V be a finite dimensional normed vector space. Let U V be an open subset, contained in a compact set. Then f : U R, is differentiable at x if there exists Df(x) V so that In this case, Df is called the differential. Lemma If Df exists, it is unique. f(x + v) f(x) Df(x), v lim = 0. v 0 v Proof. Same as 1 variable case, if you have two derivatives, subtract them, and obtain their difference is 0. Lemma The definition of derivative is independent of the choice of norm on V. Proof. Any two norms differ by a constant and if g is bounded, and h 0 then gh 0, where we take g to be the constant that the norms differ by and h to be f(x + v) f(x) Df(x), v lim = 0 v 0 v with respect to one of the norms. Lemma Choose v V, and let g v (t) = f(x + tv). If f is differentiable at x with differential Df(x) then g v is differentiable at 0, and g v(0) = Df(x), v.