g 2 (x) (1/3)M 1 = (1/3)(2/3)M.

COMPACTNESS If C R n is closed and bounded, then by B-W it is sequentially compact: any sequence of points in C has a subsequence converging to a point in C Conversely, any sequentially compact C R n is closed and bounded. This is true very generally (metric spaces, for instance.) But closed bounded sets are sequentially compact only in finite dimensions. If F : C R n is continuous (where C is compact: closed and bounded), then f(c) is compact in R n. As a corollary, continuous functions on compact sets are bounded, and achieve their max and min values. Another corollary is that a continuous map on a compact set K R n is closed: maps closed subsets of K to closed subsets of R n. In particular, a bijective continuous map f : K f(k) R m is automatically a homeomorphism. The same is not true for preimages: we say a continuous map f : R n R m is proper if f 1 (C) is compact, whenever C R m is compact. This is equivalent to: f(x n ), for any sequence x n. A continuous map F : C R m defined in a compact subset C R n is uniformly continuous on C. This is easily proved by contradiction, using sequential compactness. A decreasing sequence K n K n+1 of non-empty compact sets (in R n ) has non-empty intersection (Cantor s theorem). This follows from sequential compactness: take x n K n and consider a convergent subsequence. Application. A monotone sequence f n : K R of continuous functions on a compact set (f n f n+1 on K, say) which converges pointwise to f : K R, converges uniformly to f, provided f is continuous. (Dini s theorem.) For the proof, given ɛ > 0, consider K n = {x K; f n (x) f(x) ɛ}. Since the sequence (f n ) is monotone, this is a decreasing sequence of compact sets (K n K n+1 ), yet their intersection is empty (by pointwise convergence). Thus one of the K n is empty, which is uniform convergence. Heine-Borel theorem. A set A R n is compact if, and only if, any open cover {U λ } λ Λ of C admits a finite subcovering. Proof. Let A R n satisfy this condition. Considering the cover {B 1 (x)} x A of A (by open balls of radius one) shows A is bounded. If A is not closed, let (x n ) be a sequence in A, converging to a point x 0 not in 1

A. Then the open sets B 1/n (x 0 ) c (complements of the closed balls at x 0 with radius 1/n) cover A. Extracting a finite subcover, we find N so that x x 0 > 1/N for all x A, contradiction. Conversely, assume A is closed and bounded, and let (U λ ) λ Λ be an open covering of A. By Lindelöf s theorem (see below) we find a countable open subcovering (U λi ) i 1 of A. Let K n = A (U λ1... U λn ) c. Then (K n ) is a decreasing sequence of compact sets, with empty intersection. By Cantor s theorem, some K N must be empty. Then (U λn ) N n=1 is a finite subcovering of A. Lindelöf s theorem. Let A R n. Any open covering of A admits a countable subcovering. Proof. Let (U λ ) λ Λ be an open covering of A. Let E = (x i ) i 1 a countable dense subset of A. Consider the countable set B of open balls B ri (x i ) with center at a point x i E, rational radius r i > 0, and contained in some U λ. We claim the sets (B ri (x i )) i 1 in the collection B cover A. Let x A. Then x U λ for some λ, and since U λ is open we may find r > 0 so that B 2r (x) U λ. Since E is dense in A, we find x i E and a rational r i > 0 so that r i < r and x x i < r i, and it is easy to see that B ri (x i ) B 2r (x) U λ, so B ri (x i ) occurs in B and x B ri (x i ). Now take an enumeration (B i ) i N of B and for each i 1 choose λ i with B i U λi to conclude. Application. f : A R n is locally Lipschitz if for each x A we may find an open ball B rx (x), so that f is Lipschitz in this ball (with some Lipschitz constant M x ): y, z B rx (x) f(y) f(z) M x y z. Exercise. If f : I R is differentiable on the interval I R, with f continuous on I, then f is locally Lipshitz on I. (This follows from the Mean Value Theorem.) Proposition. If K R n is compact and f : K R m is locally Lipschitz on K, then f is Lipschitz on K. The proof of this fact requires the notion of Lebesgue number of a pair A, U, where U = {U λ } λ Λ is an open covering of a subset A R n : a positive number > 0 so that any x, y A with x y < are contained in a single set U λ of the cover. Not every open cover of a set has a Lebesgue number: for instance the cover U 1 = (, 0), U 2 = (0, ) of A = R \ {0} doesn t have one (prove it.) 2

Prop. Every open cover of a compact set A R n has a Lebesgue number. (Otherwise we may find sequences (x k ), (y k ) with x k y k 1/k, but no single U λ contains both x k and y k (for any k); taking subsequential limits leads to a contradiction.) Given this fact, the proof that locally Lipschitz implies Lipschitz on each compact set follows easily. EXTENSION of continuous functions. Exercise: Let A, B R n be closed and disjoint. Define f : R n [0, 1] via: d(x, A) f(x) = d(x, A) + d(x, B). Then f is continuous, takes the value 0 exactly on A, and the value 1 exactly on B. (In Topology a function with these properties would be called a strict Urysohn function for the pair A, B.) Tietze s extension theorem for R n : Let f : A R be continuous and bounded, where A R n is closed: f(x) M in A. Then there exists a continuous extension f : R n R of f, with f M on R n. Proof Consider the sets: B = {x A; f(x) M/3}; C = {x A; f(x) M/3}, both closed in R n and disjoint. Assume first they are both non-empty. Let g 1 : R n [ M/3, M/3] be equal to M/3 in B, and to M/3 in C (Urysohn s lemma.) Then f(x) g 1 (x) 2M/3 on A. (If either B or C is empty, we may find a constant g 1 so that this inequality holds in A.) Let f 1 = f g 1. Repeating the argument with f 1 replacing f and M 1 = 2M/3 replacing M, we find g 2 : R n [ M 1 /3, M 1 /3] continuous, satisfying on A: f(x) (g 1 (x) + g 2 (x)) (2/3) 2 M, g 2 (x) (1/3)M 1 = (1/3)(2/3)M. Inductively we find a sequence g n (x) C b (R n ) satisfying on A: f(x) n g i (x) (2/3) n M, g n (x) (1/3)(2/3) n 1 M( in R n ). i=1 Thus f(x) = i=1 g i(x) converges to f pointwise in A. By the Weierstrass M-test and the second inequality, convergence is uniform in A, and we have 3

in R n : f(x) (M/3) i 1(2/3) i 1 = M. Remark: It is easy to see the theorem fails if we don t assume A is closed, even if f is bounded on A (example?) On the other hand, the result does extend to the case the function is unbounded (if A is closed). Proposition. Let A R n be closed, f : A R continuous. Then f admits a continuous extension f : R n R. Proof. Since R is homeomorphic to ( 1, 1), we may assume f takes values in ( 1, 1) (for example, consider f φ 1, where φ(x) = tan( 2x π ) is a homeomorphism from ( 1, 1) to R.) We know there is continuous extension g : R n [ 1, 1], g = f on A. Consider the closed set: B = g 1 ({ 1, 1}), and let h : R n [0, 1] (continuous) be a Urysohn function: h 1 on A, h 0 on B. Consider the continuous function f : R n [ 1, 1]: f(x) = h(x)g(x). Note f = f on A. On the other hand, f = ±1 at a point only if h = 1 and g = ±1 at that point; this never happens, since h = 0 at the points where g = ±1. Hence in fact f takes values in the open interval ( 1, 1), and extends f continuously to all of R n. Corollary. If A R n is any non-empty set and f : A R is uniformly continuous, then f admits a continuous extension to all of R n. (Exercise.) CONVEX FUNCTIONS Let K R n be a convex set. A function f : K R is convex if: x, y K, t [0, 1] : f(tx + (1 t)y) tf(x) + (1 t)f(y). (Strictly convex if the inequality is strict for t (0, 1).) The epigraph of f is the subset of R n+1 : E f = {(x, y) K R; y f(x)}. Exercise: (i) For any subset K R n and any function f : K R, E f is closed in R n+1 if and only if f is lower semicontinuous on K. This means: ( x 0 K)f(x 0 ) lim inf n f(x n ) if x n x 0, x n K. (ii) f is convex on K if and only if E f is a convex subset of R n+1. 4

Given c R, the sublevel set K c of f defined by c is: K c = {x K; f(x) c}. Exercise. (i) In general (K not nec. convex) K c is closed (or empty) for each c R if, and only if, f is lower semicontinuous. (ii) If K is convex and f is convex on K, each nonempty sub level set of f is convex. The converse is not true: for instance, if f : R R is an increasing function, each sublevel set is either empty, a half-infinite interval or the whole line (hence convex); but f itself need not be convex (say, f(x) = x 3.) Exercise. Three-slopes lemma for convex functions on the real line. m f (a, b) = f(b) f(a) b a. a < b < c m f (a, b) m f (a, c) m f (b, c). Exercise. Convex functions defined on an open interval are continuous (but this is false on other types of intervals.) Hint: Let x 0 I (open interval) and choose > 0 so that [x 0, x 0 +] I. Let M = max{f(x 0 ), f(x 0 + )}. Assume first x 0 < x < x 0 + and write: x = (1 t)x 0 + tx, x 0 = 1 1 + t x + t 1 + t (x 0 ), where t = x x 0. Then, by convexity: f(x) (1 t)f(x 0 ) + tf(x 0 + ), f(x 0 ) 1 1 + t f(x) + t 1 + t f(x 0 ). These inequalities lead respectively to: f(x) f(x 0 ) x x 0 (M f(x 0 )), x x 0 (f(x 0 ) M) f(x) f(x 0 ), and we introduce absolute values to combine them into one estimate: f(x) f(x 0 ) x x 0 M f(x 0 ). This clearly implies continuity at x 0. The case x 0 < x < x 0 is analogous. Exercise. If K R n is convex open, and f : K R is convex, then f is continuous on K. 5

Hint: Given x 0 K, choose > 0 so that the closed n-cube Q = Q (x 0 ) of side length 2 and center x 0 is contained in K (using the fact K is open). Let M be the maximum value of f on the finite set V of vertices of Q. Since Q is the convex hull of V, the sublevel set K M is convex and V K M, it follows that Q K M, or f(x) M for x Q. The closed ball B = {x; x x 0 } is contained in Q. For u on the unit sphere, consider the closed line segment in B: σ u = {x B; x = x 0 + tu; t }. Applying the one-dimensional estimate to this segment we find, for x σ u : f(x) f(x 0 ) x x 0 M f(x 0 ). Therefore this holds for all x B, implying continuity at x 0. (Reference: [Fleming], p.110.) Convexity and differentiability in one dimension. We consider f : I R convex, where I R is an open interval. For h > 0, denote by m f (x, x + h) = f(x+h) f(x) h the slope of the (oriented) secant to the graph of f from x to x + h. The three-slopes lemma implies this is monotone increasing in h; thus its limit as h 0 + exists, the right-derivative of f at x: f +(x) = lim h 0 + m f (x, x + h) = inf h>0 m f (x, x + h). Similarly, the left-derivative exists for each x I: f (x) = lim m f (x k, x) = sup m f (x k, x). k 0 + And since m f (x k, x) m f (x, x + h) for each h, k > 0, it follows that f (x) f +(x), for each x I. Also, if x < y are points in I and h > 0, k > 0 are chosen so that x + h = y k, we have: k>0 f (x) f +(x) m f (x, x + h) m f (y k, y) f (y) f +(y). 6

We see that f, f + are both monotone increasing (nondecreasing) in I, with f +(x) f (y) if x < y. And now it is easy to see that, if a I is a point of continuity of f, then f (a) = f +(a): f (a) f +(a) lim y a + f (y) = f (a). Thus f is differentiable at a. Being monotone, f is continuous at all but a countable set of points in I. We conclude: f convex in I R f differentiable in the complement of a countable set D I. 7