User:TakuyaMurata/Continuous functions on a compact space

In this section, we will undergo a throughout study of $$C(K)$$, the space of all real-valued or complex-valued functions on a compact space. The most important result in the line of this study is Ascoli's theorem and Stone-Weierstrass theorem. For simplicity, we assume functions in $$C(K)$$ are real-valued. (A discussion will be given later as to why this does not diminish the generality.)

As usual, we topologizes $$C(K)$$ by the norm $$\| \cdot \| = \sup | \cdot |$$. To say that $$C(K)$$ is complete is precisely:

2. Lemma The limit of a uniformly convergent sequence of continuous functions is continuous.

Proof: Suppose $$f_n \in C(K)$$ is a sequence such that $$\sup_K |f_n - f| \to 0$$ for some function $$f$$ defined on $$K$$. For any $$x \in K$$, by the iterated limit theorem,
 * $$\lim_{y \to x} f(y) = \lim_{n \to 0} \lim_{y \to x} f_n(y) = \lim_{n \to 0} f_n(x) = f(x).$$ $$\square$$

Hence, $$C(K)$$ is a Banach space. (For the definition and basic results of Banach spaces, see Functional Analysis.)

2 Theorem (Ascoli) ''Let $$\Gamma \subset C(K)$$. Then $$ \Gamma $$ is relatively compact if and only if'' Proof: First, assume that $$\Gamma$$ is relatively compact. (ii) is then obvious. For (i), let $$ \epsilon > 0$$ and $$x \in K$$ be given. For each $$f \in \Gamma$$, by continuity, we can find a neighborhood $$G_f$$ of $$x$$ such that:
 * (i) Given an $$\epsilon > 0$$ and $$x \in K$$, we can find a neighborhood $$G$$ of $$x$$ such that
 * $$|f(y) - f(x)| < \epsilon$$ for every $$y \in G$$ and every $$f \in \Gamma$$
 * (ii) $$\sup_\Gamma | \cdot (x)| < \infty$$ for every $$x \in K$$
 * $$|f(y) - f(x)| < \epsilon$$ for every $$y \in G_f$$.

Since $$\Gamma$$ is relatively compact, $$\Gamma$$ contains a finite subset $$\gamma$$ such that $$\Gamma$$ is the union of the sets of the form
 * $$\{ f; f \in \Gamma, \sup_K |f - g| < \epsilon / 3 \}$$

over $$g \in \gamma$$. Let $$G$$ be the intersection of $$G_g$$ over $$g \in \gamma$$. Then for every $$f \in \Gamma$$, there is $$g \in \gamma $$ with $$\sup_K |f - g| < \epsilon / 3$$, and so:
 * $$|f(y) - f(x)| \le |f(y) - g(y)| + |g(y) - g(x)| + |g(x) - f(x)| < \epsilon$$ for any $$y \in G$$.

This proves (i). Next, suppose $$E$$ satisfies (i) and (ii). To show that $$\Gamma$$ is totally bounded, let $$\epsilon > 0$$ be given. For each $$x \in K$$, by (i), we can find a neighborhood $$G_x$$ of $$x$$ such that:
 * $$|f(y) - f(x)| < \epsilon / 3$$ for every $$y \in G_x$$ and $$f \in \Gamma$$.

Since $$K$$ is compact, we can find $$z_1, ... z_n \in K$$ such that $$K$$ is the union of $$G_{z_j}$$ over $$j = 1, 2, ... n$$. Let
 * $$A = \{ (f(z_1), f(z_2), ... f(z_n)); f \in \Gamma \}$$.

By (ii), $$A$$ is a bounded (thus totally bounded) subset of $$\mathbf{R}^n$$. That means that $$\Gamma$$ contains a finite subset $$\gamma$$ such that:
 * $$A \subset \bigcup_{g \in \gamma} \{ (t_1, ... t_n); t_j \in \mathbf{R}, \max_j | t_j - g(z_j) | < \epsilon / 3 \}$$

It now follows: given $$f \in \Gamma$$, we can find $$g \in \gamma$$ such that:
 * $$\max_j | f(z_j) - g(z_j) | < \epsilon / 3$$.

Then, for each $$x \in K$$, since $$x \in G_{z_k}$$ for some $$z_k$$,
 * $$|f(x) - g(x)| \le |f(x) - f(z_k)| + |f(z_k) - g(z_k)| + |g(z_k) - g(x)| < \epsilon$$

In other words, $$\sup_K |f - g| < \epsilon$$. Hence, $$\Gamma$$ is totally bounded, or equivalently, relatively compact. $$\square$$

2 Corollary ''Let $$f_n \in C(K)$$. $$f_n$$ is uniformly convergent if and only if it is pointwise convergent and equicontinuous.''

2 Theorem ''Let $$f_n$$ converge pointwise to $$f \in C(K)$$. If $${f_n}'$$ exists and converges uniformly to $$g$$, then $$f_n$$ converges uniformly to $$f$$. Moreover, $$f$$ is differentiable and its derivative is $$g$$. Proof: Let $$M = \sup \{ |f_n'(x)| | n \ge 1, x \in K \}$$. $$M$$ is finite by uniform convergence. By the mean value theorem,
 * $$|f_n(y) - f_n(x)| \le M|y-x|$$

Thus, $$f_n$$ is equicontinuous and converges uniformly by Ascoli's thoerem (or one of its corollaries.) $$\square$$

2 Theorem ''Let $$\Gamma$$ be an equicontinous set of real-valued functions on $$\mathbf{R}$$. If $$\sup_{f \in \Gamma}|f(0)| = b < \infty$$, then there exist an $$a$$ such that:
 * $$\sup_{f \in \Gamma} |f(x)| \le a|x| + b + 1$$

Proof: Let $$\delta > 0$$ be such that:
 * $$|f(x) - f(y)| < 1$$ for all $$f \in \Gamma$$ and $$|x - y| < 2\delta$$

($$ 2\delta$$ isn't a typo; it is meant to simplify the computation.) Let $$x > 0$$ be fixed. Then, for any $$f \in \Gamma$$,
 * $$|f(x)| \le |f(x) - f(0)| + |f(0)|$$,

and we estimate:
 * $$|f(0) - f(x)| \le \sum_{k=0}^{n-1} |f(k\delta) - f((k+1)\delta)| + |f(n\delta) - f(x)| \le n+1$$

where $$n$$ is such that $$n\delta < x \le (n+1)\delta$$. Thus,
 * $$|f(x)| \le \delta^{-1}|x| + 1 + b$$

Since we can get the same estimate for $$x < 0$$, the proof is complete.$$\square$$

2 Corollary (Dini's theorem) ''Let $$f_n \in \mathcal{C}(K)$$ be a sequence such that $$f_n(x) \to f(x)$$ for every $$x \in K$$. If $$f_n$$ is increasing, then $$f_n \to f$$.''

Proof: Set $$g_n = f - f_n$$. Then $$g_n$$ is decreasing and thus satisfies the hypothesis of Ascoli's theorem. Hence, $$g_n$$ admits a convergent subsequence, which converges to 0 since the subsequence converges pointwise to 0. Since $$g_n$$ is decreasing, $$g_n$$ converges as well. $$\square$$

2 Theorem ''Suppose $$K$$ is a metric space. Then $$\Gamma \subset C(K)$$ is equicontinuous if and only if for every $$\epsilon > 0$$ there exists $$\delta > 0$$ such that
 * $$|f(x) - f(y)| < \epsilon$$

for every $$f \in \Gamma$$ and $$x, y \in K$$ with $$|x - y| < \delta$$.''

Proof: $$(\Leftarrow)$$ holds vacuously. For the converse, let $$\epsilon > 0$$ be given. Then for each $$x \in K$$, we can find $$\delta_x$$ such that
 * $$|x - y| < \delta_x$$ implies $$|f(x) - f(y)| < \epsilon / 2$$ for every $$f \in \Gamma$$

By compactness, we find $$x_1, x_2, ... x_n \in K$$ such that:
 * $$K \subset \bigcup_j B(x_j, \delta_{x_j} / 2)$$

Let $$\delta = \min \{ \delta_{x_1}, \delta_{x_2}, ... \delta_{x_n} \} / 2$$, and then suppose we are given $$x, y \in K$$ with $$|x - y| < \delta$$. It follows: there is a $$j$$ with $$|x - x_j| < \delta_{x_j} / 2$$. Since
 * $$|y - x_j| \le |y - x| + |x - x_j| < \delta + \delta_{x_j} / 2 \le \delta_{x_j}$$,

we have:
 * $$|f(x) - f(y)| \le |f(x) - f(x_j)| + |f(x_j) - f(y)| < \epsilon$$

for every $$f \in \Gamma$$. $$\square$$

The Stone-Weierstrass theorem states that polynomials are dense in C(K, R). It is however not the case that the space of polynomials in z is a dense in C(K, C). If it were, we have the equality in the below
 * $$\overline{P(K)} \subset A(K) \subset C(K)$$

But $$A(K) \ne C(K)$$ if K has nonempty interior.

2 Theorem (intermediate value theorem) ''A function $$f:[a, b] \to \mathbf{R}$$ is continuous if and only if Proof: ($$\Rightarrow$$) Obvious. ($$\Leftarrow$$) Suppose $$f(x) < c$$. Since the complement of $$ f^{-1}(\{c\})$$, which contains x, is open, we have: $$f \ne c$$ in some interval U in $$[a, b]$$ containing x. We actually have: $$f < c$$ in U. In fact, if $$y \in U$$ and $$c < f(y)$$, then $$f(x) < c < f(y)$$, which implies U contains a point z such that $$f(z) = c$$, a contradiction. Hence, f is upper semicontinous at x. The same argument applied to $$-f$$ shows that f is also lower semicontinous at x. $$\square$$
 * (i) If $$f(a) < c < f(b)$$, then c is in $$f((a, b))$$.
 * (ii) If $$f^{-1}(\{c\})$$ is closed for every real c.

2 Theorem ''Let $$f$$ be a real-valued continuous function on an open interval. Then the following are equivalent.'' Proof: (ii) $$\Rightarrow$$ (i) is obvious. (iii) $$\Rightarrow$$ (ii): If (ii) is false, then we can assume there exists $$a < c < b$$ such that $$f(a) < f(c)$$ and $$f(c) > f(b)$$. By continuity and compactness, f attains a maximum in some point x in $$[a, b]$$ but by hypothesis $$x \in (a, b)$$ and so $$f(x)$$ is a non-interior point of $$f((a, b))$$, falsifying (iii). If (iii) is false, then $$(a, b)$$ contains a x such that $$f(x)$$ is not an interior point of $$f( (a, b) )$$. Since $$f( (a, b) )$$ is an interval, we may assume that $$\sup_{ (a, b) } f = f(x)$$. It then follows from the intermediate value theorem that f is not injective. $$\square$$
 * (i) f is injective.
 * (ii) f is strictly monotonic.
 * (iii) f is an open mapping.

2 Theorem (mean value theorem) ''Suppose $$f \in C([a, b], \mathbf{R})$$ is differentiable on the open interval $$(a, b)$$. Then''
 * $$f(b) - f(a) = f'(c)(b - a)$$

for some $$c \in (a, b)$$

Proof: FIrst assume $$0 = f(a) = f(b)$$. By the theorem preceding this one, $$f$$ attains a maximum or minimum at $$x \in (a, b)$$; say, a maximum. By definition, we can write:
 * $$f(y) = f(x) + f'(x)(y - x) + o(|y-x|)$$ as $$y \to x$$

Then since x is a maximum,
 * $$0 \ge f(y) - f(x) = f'(x)(y - x) + o(|y-x|)$$.

If $$y > x$$,
 * $$0 \ge f'(x) + O(|y-x|)$$

and letting $$x \to 0$$ gives that $$f'(x) \le 0$$. If $$y < x$$, then by the same argument, we find that $$f'(x) \ge 0$$. Thus, $$f'(x) = 0$$. For the general case, let
 * $$g(x) = f(x) - rx$$ where $$r = {f(b) - f(a) \over b - a}$$.

Then $$g(a) = g(b) = 0$$. Hence, applying the first part of the proof gives: $$g'(c) = 0$$ for some c. Since $$0 = g'(c) = f'(c) - r$$, c is a solution of the equation. $$\square$$

2 Corollary ''Let $$f:\mathbf{R} \to \mathbf{R}$$ be a differentiable function. If $$f'(a) < c < f'(b)$$, then c is in $$f'((a, b))$$.''

Proof:Let $$g(x) = f(x) - cx$$. Then $$g'(a) < 0$$ and $$g'(b) < 0$$. In other words, $$g$$ is increasing at a and decreasing at b. By the theorem above, g is not injective on $$(a, b)$$; i.e., $$g(x) = g(x')$$ for some $$x \ne x'$$ in $$(a, b)$$. It follows:
 * $$0 = g(x) - g(x') = g'(y)(x - x')$$ for some $$y \in (a, b)$$.

and $$g'(y) = 0$$. $$\square$$

2 Corollary A strictly monotonic continuous function with closed range is a homeomorphism.

A real-valued function $$f$$ on $$\mathbf{R}^n$$ is said to be homogeneous of degree $$k$$ ($$k$$ could be any real number) if
 * $$f(tx) = t^k f(x)$$

for all $$x \in \mathbf{R}^n$$ and all $$t > 0$$.

2 Theorem (Euler's relation) ''Let $$f: \mathbf{R}^n \to \mathbf{R} $$ be a function differentiable on $$\mathbf{R}^n \backslash \{0\}$$. Then $$f$$ is homogeneous of degree $$k$$ if and only if:
 * $$\left( \sum_j x_j \partial_j - k \right) f(x) = 0$$

for all $$x \in \mathbf{R}^n$$.

Proof: ($$\Rightarrow$$) Differentiate with respect to $$t$$ both sides of $$f(tx) = t^k f(x)$$ and then put $$t = 1$$. ($$\Leftarrow$$) Note
 * $$(f(tx))' = \sum_j x_j {\partial_j f}(tx) = {k \over t} f(tx)$$.

Thus, if we let
 * $$g(t) = \log \left| {f(tx) \over t^k f(x)} \right| = \log |f(tx)| - k\log t - \log|f(x)|$$,

then the derivative of $$g$$ vanishes identically. Since $$g(1) = 0$$, $$g$$ is identically zero. $$\square$$

The theorem permits a generalization. By definition, any (distribution) solution of the equation
 * $$\left( \sum_j x_j \partial_j - k \right) f = 0$$

is said to be homogeneous of degree $$k$$.

A homogeneous distribution is tempered and its Fourier transform is homogeneous.

2 Theorem (l'Hôpital's rule) ''Let $$0 \le a \le \infty$$. If $$f(x), g(x) \to 0$$ as $$x \to a$$ or if $$g(x) \to \infty$$ as $$x \to a$$, then
 * $$\lim_{x \to a} {f(x) \over g(x)} = \lim_{x \to a} {f'(x) \over g'(x)}$$

provided the limit in the right-hand side exists.

Proof: First assume $$a$$ is finite. We may redefine $$f(a) = g(a) = 0$$ (since the values of functions at $$a$$ are immaterial when we compute the limit.) Fix $$x$$, and define
 * $$h(y) = f(y)g(x) - f(x)g(y)$$

Since $$h(x) = 0 = h(a)$$, by the mean value theorem, we can find a $$c$$ between $$x$$ and $$a$$ such that
 * $$0 = h'(c) = f'(c)g(x) - f(x)g'(c)$$

Since when $$x$$ is close to $$a$$ we may assume $$g'$$ never vanishes,
 * $${ f(x) \over g(x) } = { f'(c) \over g'(c) }$$

Since $$c \to a$$ as $$x \to a$$, the proof of this case is complete. (TODO: handle other cases.) $$\square$$

The next theorem can be skipped without the loss of continuity, for more general results will later be obtained.

2 Theorem (The Weierstrass approximation theorem) ''Let $$f \in C([0, 1])$$, and define
 * $$f_n(x) = \sum_{k=0}^n f(k/n)p^n_k(x)$$ with $$p^n_k(x) = {n \choose k} x^k (1-x)^{n-k}$$ (Bernstein polynomial).

Then $$f_n \to f$$ uniformly on $$[0, 1]$$

Proof : First note that
 * $$1 = \sum_{k=0}^n p^n_k(x)$$

is a partition of unity, by the binomial theorem applied to $$(x + 1 - x)^n$$. Moreover, a simple computation gives the identity:
 * $$\sum (nx - k)^2 p^n_k(x) = nx(x-1)$$

It thus follows: for any $$\delta > 0$$
 * $$|f(x) - f_n(x)| \le \sum_{k : |nx - k| < n\delta} |f(x) - f(k / n)| + {nx(x-1) \over n^2 \delta^2} $$

Since $$f$$ is uniformly continuous on $$[0, 1]$$ by compactness, the theorem now follows. $$\square$$

2 Corollary ''Any continuous function vanishing at infinity is uniformly approximated by Hermite function. (TODO: made the statement more precise and give a proof.)

Example: Let $$f \in L^2([0, 1])$$. Let $$M$$ be the linear span of polynomials. Then $$M$$ is a dense subspace of $$L^2([0, 1])$$ by the above theorem since
 * $$\int_0^1 |f - p_n|^2 dx \le \sup|f - p_n|^2 \to 0$$

2 Theorem If $$f \in C(\mathbf{R})$$ is uniformly continuous and integrable, then $$f \in C_0(\mathbf{R})$$.

Proof: Define $$F(x) = \int_{-\infty}^x f(x)dx$$. That $$f$$ is integrable means that $$M = \lim_{x \to \infty} F(x)$$ exists and is finite. Let $$\epsilon > 0$$ be given. By uniform continuity, there is a $$\delta > 0$$ such that
 * $$|f(y) - f(x)| < \epsilon$$ whenever $$|y-x| < \delta$$.

Then there is an $$R > 0$$ such that
 * $$|F(x + \delta) - F(x)| < \delta\epsilon$$ whenever $$x > R$$

Now, let $$x > R$$ be given. By the mean value theorem, we find $$y$$ such that $$\delta f(y) = F(x + \delta) - F(x)$$ and $$|x - y| < \delta$$. Thus,
 * $$|f(x)| \le |f(x) - f(y)| + \delta^{-1}|F(x + \delta) - M | + \delta^{-1}|M - F(x)| < 3\epsilon$$

$$\square$$

Example ''No function is continuous only on rational points. To see this, let $$f: \mathbf{R} \to \mathbf{R}$$ be a function, and let $$E$$ be the set of all points at which $$f$$ is continuous. It follows immediately from the definition of continuity that $$E$$ is $$G_\delta$$; i.e., it is an intersection of countably many open sets. On the other hand, $$\mathbf{Q}$$ is not $$G_\delta$$.''

2 Theorem ''Let $$U \subset \mathbf{R}$$ be a nonempty open subset.
 * $$\sup_U |f| \le 2 \sqrt{\sup_U |f'| \sup_U |f''|}$$

Proof: Let $$A = \sup_U |f'|$$ and $$B = \sup_U |f'|$$. We assume $$A$$ and $$B$$ are finite; otherwise the inequality is trivial. Given $$x \in U$$, we can find $$h > 0$$ so that an interval $$[x, x + h] \subset U$$. By Taylor's formula,
 * $$f(x + h) = f(x) + f'(x)h + f''(x + \theta h)h^2$$ (where $$0 < \theta < 1$$)

and so:
 * $$|f'(x)| \le {A \over h} + Bh$$

Now, take $$h = {\sqrt{A} \over \sqrt{B}}$$. $$\square$$

2 Theorem (Whitney extension theorem) Every real-valued L-Lipschitz function on a subset of $$\mathbf{R}^n$$ is the restriction of a L-Lipschitz function on $$\mathbf{R}^n$$.

Proof ( pg. 5): Let $$f$$ be a L-Lipschitz function on a subset $$A$$. Define
 * $$F(x) = \inf_{y \in A} (f(y) + L|x - y|) \qquad (x \in \mathbf{R}^n)$$

It is clear that $$F$$ is L-Lipschitz continuous. $$\square$$

Exercise: Every closed set is a zero set of a $$C^\infty$$ function.