Math 118B Solutions. Charles Martin. March 6, d i (x i, y i ) + d i (y i, z i ) = d(x, y) + d(y, z). i=1

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1 Math 8B Solutions Charles Martin March 6, Homework Problems. Let (X i, d i ), i n, be finitely many metric spaces. Construct a metric on the product space X = X X n. Proof. Denote points in X as x = (x, x,..., x n ). Given x, y X define d(x, y) = d (x, y ) + + d(x n, y n ). Then clearly d(x, y) and d(x, y) = d(y, x). Furthermore, if d(x, y) = then each d i (x i, y i ) =. That is, each x i = y i and x = y. Finally, given x, y, z X we have d(x, z) = Hence d is a metric on X. n d i (x i, z i ) i=. Let (X, d) be a metric space. Prove that is also a metric on X. δ(x, y) = n d i (x i, y i ) + d i (y i, z i ) = d(x, y) + d(y, z). i= d(x, y) + d(x, y) ; x, y X Proof. Clearly we have that δ(x, y) = δ(y, x) and δ(x, y). If δ(x, y) =, then d(x, y) = and x = y. Finally, given x, y, z X we have Hence δ is a metric on X. d(x, z) δ(x, z) = + d(x, z) = + d(x, z) + d(x, y) + d(y, z) d(x, y) + d(y, z) = + d(x, y) + d(y, z) d(x, y) = + d(x, y) + d(y, z) + d(y, z) + d(x, y) + d(y, z) d(x, y) d(y, z) + + d(x, y) + d(y, z) = δ(x, y) + δ(y, z). 3. Prove that the space l of bounded sequences is complete with respect to the distance. Let c, c denote the subspaces of l of convergent, respectively, convergent to zero, sequences. Prove that c, c are closed subspaces of l.

2 Proof. This is a bit of a notational challenge. Let x, x, x 3... be a sequence of points within l, each of which is itself a sequence of real numbers. That is, for each i we have that x i = (x i, x i, x i3,...). Suppose that the sequence (x i ) is Cauchy in l. Then for each i, m, n we have x im x in sup x im x in = x m x n i as m, n. That is, each sequence of real numbers x i, x i,... is Cauchy in R, hence convergent. Let y i = lim x in and consider the sequence y = (y i ). Let ɛ >. Find N (depending only upon ɛ) so that for all m, n N we have x n x m < ɛ/. For each k this implies that x nk x mk < ɛ/. If we take m then x mk y k and we have that x nk y k ɛ/ for all k. Taking the supremum over k gives x n y ɛ/ < ɛ. This shows that x n y as n. We should also verify that y l ; if n is such that x n y < then so indeed y l. y = x n + (x n y) < x n + <, Now we turn our attention to c and c ; since convergent sequences are bounded we have that c, c l. It is clear that c, c are both subspaces, so we ll only show that they are both closed. Suppose that (x n ) is a sequence in c which converges to some point y l. We wish to show that y c. Let ɛ > and find n so that x n y < ɛ/3. Now find M so that for all j, k M we have x nj x nk < ɛ/3. For all such j, k we have y j y k y j x nj + x nj x nk + x nk y k y x n + x nj x nk + y x k < ɛ 3 + ɛ 3 + ɛ 3 = ɛ. So y is a Cauchy sequence of real numbers, hence convergent. That is, y c. Now suppose that (x n ) is a sequence in c which converges to some y l. Since c c, the above reasoning shows that y c; we need only show that lim y =. Let ɛ > and take n so that y x n < ɛ/. Then for each k we have y k x nk < ɛ/. Find M so that for all k M we have x nk < ɛ/; for all such k we have y k y k x nk + x nk < ɛ + ɛ = ɛ. So lim y = as desired. 4. Describe the intervals [, a] R such that the function f : [, a] [, a], f(x) = sin x, is contractive. Proof. Let a >. Suppose there exists a c < so that for all x, y [, a], Let x (, a] and note that sin x sin y c x y. sin x sin x But if we take x, we have that c, a contradiction. Upon no such interval is the sine function contractive. < c. 5. Let A be a bounded subset of C[, ]. Prove that the set of functions is equicontinuous in C[, ]. F (x) = x f(t) dt, f A,

3 Proof. For convenience denote { x F = } f(t) dt : f A. Choose M > so that for each f A we have f M. Given ɛ > set δ = ɛ/m. Whenever x, y [, ] are such that x y < δ and f A, it follows that x y x f(t) dt f(t) dt = f(t) dt M x y < Mδ = ɛ, so we conclude that F is equicontinuous. 6. Is the sequence of functions sin(nx), n, equicontinuous in C[, ]? y Proof. Suppose that the sequence is equicontinuous. In particular, there exists a δ so that for any n and x, y R with x y < δ, sin nx sin ny <. Let n be so large that π/n < δ. Then π/n < δ, yet a contradiction. The family is not equicontinuous. 7. Start with P = and define for n Prove that P n (x) x uniformly on [, ]. sin(n π/n) sin(n ) =, P n+ (x) = P n (x) + x Pn(x). Proof. We start by proving that P n (x) x for all x [, ] and n. This is true for P (x) =, so assume it true for some n. Then x Pn(x) and P n (x) implies that P n+ (x). Further, notice that [ x P n+ (x) = [ x P n (x)] x + P ] n(x). () Each factor on the right is positive since P n x and ( x + P n (x))/ x ; thus P n+ (x) x. By induction P n (x) x for all x [, ] and n. FIRST PROOF Define the functions g n (x) = x P n (x); we want to show that g n uniformly. Notice that [ g n+ (x) = g n (x) x + P ] [ n(x) g n (x) x ]. Since g (x) = x we find that ( g n (x) x x ) n. We wish to bound the right side from above uniformly in x. It suffices to consider x ; define f : [, ] R by ( f(x) = x x ) n. Since f is continuous on a compact interval, it attains an absolute maximum. To find candidates for the maximum, we need = d ln f(x) dx = x n x x(n + ) =. x( x)

4 The only relative maximum of f occurs at x = /(n + ). Since f() = and f() = n, the absolute maximum is at x = /(n + ). Therefore ( g n (x) x x ) n ( ) f = ( ) n < n + n + n + n +. Thus g n /(n + ) and g n uniformly, as desired. ALTERNATIVE PROOF Once we know that P n (x) x we see that P n+ (x) = P n (x) + x P n(x) P n (x), so for each x the sequence (P n (x)) is monotone and bounded by x, hence pointwise convergent. This limit function must be x ; if we denote P n (x) L then L = L + x L, so L = x because L. Uniform convergence now follows from Dini s theorem: Theorem (Dini). Let X be a compact metric space and suppose that f f f 3 are continuous real-valued functions which converge pointwise to a continuous function f. Then the convergence is uniform. Proof of Dini s theorem. If we consider f n f, we have a monotone sequence of continuous functions which converge pointwise to (from above); henceforth we ll assume the pointwise limit is. If f n does not converge uniformly to, there exists ɛ > and a subsequence (f nk ) so that f nk ɛ for all k. That is, we can find a sequence of points (x nk ) in X so that f nk (x nk ) ɛ for all k. Since X is compact there is a convergent subsequence of (x nk ); for simplicity, we ll not denote a new sequence and instead assume that x nk x X. Here s where we use monotonicity: fix j and note that for any k > j ɛ f nk (x nk ) f nj (x nk ). If we take k then f nj (x nk ) f nj (x) by continuity; thus f nj (x) ɛ for arbitrary j. Taking j we see that f nj (x), a contradiction. The continuous functions x P n (x) monotonically approach pointwise from above, so the convergence is uniform by Dini s theorem. Addendum Here s an even better proof of Dini s theorem! Proof. Let ɛ > and define for each n the set U n = {x X : f n (x) f(x) < ɛ}. Since each f n f is continuous, each U n is open. The monotonicity of the sequence (f n ) implies that the sets U n are nested: U U U 3. At any x X the sequence (f n (x) f(x)) converges to, so each x is in some U n. That is, the sets U n form an open cover of X; by compactness a finite subcover exists: U U U N = X. But the sets U n are nested, so U N = X and for all n N we have X = U N U n X, whence U n = X as well. Translating out of set language, for all n N we have f n f < ɛ throughout X. 8. (a) Give an example of a continuous function f : [, ) [, ) so that f diverges but converges. (b) Give an example of a continuous function f : [, ) [, ) so that f converges but diverges. Proof. (a) Let f be a sawtooth function with f(x) = x on [ /, /], extended periodically. Then f is the area of an infinite number of triangles, each of area /4. Nevertheless f(n) = for each integer n, whence f converges. f f

5 (b) Let f be defined as n (x n) + if n n x n f(x) = n (n x) + if n x n + n otherwise A picture is more illuminating; the graph of f is a sawtooth which is at each integer, a narrow triangle near each integer, and otherwise. The sum f diverges, yet each triangle has area n, so f(x) dx = n= n =. 9. For which values of p, q are the integrals convergent? sin x dx and xp (sin x) q dx x Proof. Answer: p < and q >. Recall that sin x/x as x ; extend sin x/x to a positive, continuous function f on the interval [, ]. Let m, M be the infimum and supremum, respectively, of f on [, ]. Notice that if p < and that if p. Similarly, notice that sin x x p dx = sin x x p dx = f(x) dx M xp f(x) dx m xp dx x p < dx x p = if q > and that if q. (sin x) q dx = x (sin x) q dx = x x q f q (x) dx M q x q dx < x q f q (x) dx m q x q dx =

6 Midterm Practice. Verify the inclusions l l, l. Are any of these spaces closed in the bigger one? Proof. Let x = (x n ) l. That is, x n <. For each k we have x k x n = x, n= so taking the supremum over k gives x x < and we conclude x l. Since the series x n converges the sequence x n converegs to. Hence for some N and all n N we have x n < x n, so x = x n N x n + and x l as well. In fact, we have l l l. x n N N x n + x <, We will show that l is neither closed in l nor in l. For notational convenience (and intuition!) we define the sequences of compact support consisting of those sequences which, after a finite number of terms, are identically : F = {x l : N so that n N, x n = }. Notice the inclusions F l l c c l. We will compute the closure of F in both the l and l topologies; from there we can make conclusions about l. First we compute the closure of F in l. Let x l be arbitrary and ɛ >. Since x n < we can find an N so that x n < ɛ. We can define y F as so that y n = n=n x y = { x n if n < N if n N x n < ɛ. n=n Since x and ɛ are arbitrary, we conclude that a point in l can be approximated to within any error by an element of F. That is, any point of l is a limit point of F. This shows that F l, but since we know F l we can conclude F = l. Now consider the following topological fact: whenever A B, it follows that A B. Thus we know that F l l, where all closures refer to the l topology. We ve shown that F = l ; furthermore, l is closed in its own metric, so we have l l l, which implies that l = l. Since l l, we see that l is not closed in l. Now we compute the closure of F in l. From a previous homework problem we know that c is closed; since F c we know F c (now we use the closure symbol with respect to the l topology). Let x c be arbitrary and ɛ >. Since x converges to we can find N so that for all n N we have x n < ɛ. We can define y F as { x n if n < N y n = if n N so that x n y n = { if n < N x n if n N For all n we have x n y n < ɛ and hence x y < ɛ. Since x and ɛ are arbitrary, we see that points in c can be approximated to within any error by points in F ; that is, F c. Hence F = c and from F l c we conclude as with l above that l = c. Once again the closure of l is not itself, so l is not closed in l.

7 . Let C(, ) denote the space of continuous functions on the open interval (, ). For f, g C(, ) define U(f, g) = {t (, ) : f(t) g(t)}. By continuity U(f, g) is an open set, hence a disjoint union of intervals. Define d(f, g) = length(u(f, g)). Prove that (C(, ), d) is a metric space. Proof. Note that d by definition and that d(f, g) = d(g, f) trivially. Suppose that d(f, g) =. Then U(f, g) = and f = g. Finally, given f, g, h C[, ] we have that [, ] \ U(f, h) = {t [, ] : f(t) = h(t)} {t [, ] : f(t) = g(t) and h(t) = g(t)} = {t [, ] : f(t) = g(t)} {t [, ] : h(t) = g(t)} = ([, ] \ U(f, g)) ([, ] \ U(g, h)) = [, ] \ (U(f, g) U(g, h)) So that U(f, h) U(f, g) U(g, h). Length is both monotonic and subadditive (a fact from measure theory which is intuitively clear in this context), so that d(f, h) = length(u(f, h)) length(u(f, g) U(g, h)) length(u(f, g)) + length(u(g, h)) = d(f, g) + d(g, h). The triangle inequality holds, so d is in fact a metric for C[, ]. 3. Let f : R R be separately continuous. Prove that if each function x f(x, y ) and y f(x, y) is uniformly continuous with respect to x and y, then the function f is continuous. Proof. For simplicity we show that f is continuous at (, ); the same argument shows that the function is continuous at any point in the plane. Let ɛ > and find δ so that for any x, y with x < δ we have f(x, y) f(, y) < ɛ/. Similarly find δ so that for any x, y with y < δ we have f(x, y) f(x, ) < ɛ/. Set δ = min(δ, δ ). If (x, y) < δ then both x < δ and y < δ so we find Hence f is continuous at (, ). f(x, y) f(, ) f(x, y) f(x, ) + f(x, ) f(, ) < ɛ + ɛ = ɛ. 4. Use the Arzelà-Ascoli theorem to show that {f C[, ] : f } cannot be covered by a sequence of compact sets. Proof. Denote B = {f C[, ] : f } as the unit ball in C[, ]. Suppose that K, K,... are compact sets in B so that B = n K n. Each K n is contained in B, hence uniformly bounded. By the Arzelà-Ascoli theorem, each K n is an equicontinuous family of functions in C[, ]. Then more stuff. Whatever.

8 Midterm. Give an example of an incomplete metric space and compute its completion. Proof. A simple example is Q whose completion is R by definition.. Let f : R R be a uniformly continuous function. Prove that the set of its translates f a (x) = f(x a), a R is equicontinuous. Proof. Let ɛ > and choose δ > so that whenever x, y R with x y < δ it follows that f(x) f(y) < ɛ. Given such x, y and any a R we have (x a) (y a) = x y < δ = f a (x) f a (y) = f(x a) f(y a) < ɛ. Thus the family is equicontinuous. 3. Find a set S l which is closed, bounded, bu not compact. Proof. Define the closed unit ball S = {x l : x }. The ball is clearly closed and bounded, but not compact; to see this consider the sequence (x n ) in S wherein each x n is a sequence of all zeroes, except for a in the n-th position. Whenever n m we have x n x m =, so that no subsequence can be Cauchy, let alone convergent. 4. Prove that d(x, y) = x 3 y 3 is a distance on x, y (, ). Proof. This is straight-forward. Clearly d(x, y) and d(x, y) = d(y, x) for any x, y (, ). Further, if d(x, y) = then x 3 y 3 = and x = y. Finally, given any x, y, z (, ) we have so d is a metric. d(x, z) = x 3 z 3 x 3 y 3 + y 3 z 3 = d(x, y) + d(y, z), 5. Let K be a compact subset of a complete metric space (X, d). Prove that the function is a continuous function on X. x dist(x, K) = inf d(x, y) y K Proof. Let ɛ > and choose δ = ɛ/. Given x, y X with d(x, y) < δ, we can find z K so that Since dist(y, K) d(y, z) we have that d(x, z) < dist(x, K) + ɛ/. dist(y, K) d(y, z) d(x, y) + d(x, z) < δ + dist(x, K) + ɛ/ = dist(x, K) + ɛ. Thus dist(y, K) dist(x, K) < ɛ. Reversing the roles of x, y in the above argument gives dist(x, K) dist(y, K) < ɛ; together this gives dist(y, K) dist(x, K) < ɛ, so the function is continuous.