# 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

• (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 $x \in K$

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:

$|f(y) - f(x)| < \epsilon$ 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

• (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.

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. $\square$

2 Theorem Let $f$ be a real-valued continuous function on an open interval. Then the following are equivalent.

• (i) f is injective.
• (ii) f is strictly monotonic.
• (iii) f is an open mapping.

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. $\square$

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 $p^n_k(x) = {n \choose k} x^k (1-x)^{n-k}$ (w:Bernstein polynomial).

Then $f_n \to f$ uniformly on $[0, 1]$
Proof [1]: 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 ([2] 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.