Uniform Continuity

Let $f$ be a function on a metric space $X$. We say that $f$ is uniformly continuous if for every $\varepsilon > 0$ there is a $\delta > 0$ such that if $|x - y| < \delta$ then $|f (x) - f (y) | < \varepsilon$. This is stronger than continuity, since one $\delta$ works for all $x$ and $y$.

Theorem 1: If $X$ is compact then a continuous function on $X$ is uniformly continuous.

This is called the Heine-Cantor theorem according to Wikipedia. It is proved in the book, but the following proof is slightly different. Yet another proof uses the notion of sequential compactness.

Proof. (Click to Expand/Collapse)

Let $\varepsilon > 0$ be given. For every $x \in X$, since $f$ is continuous at $x$ there exists a $\delta_x > 0$ such that if $|y - x| < \delta_x$ then $|f (y) - f (x) | < \frac{\varepsilon}{2}$.

Now \[ X = \bigcup_{x \in X} B_{\delta_x / 2} (x), \] since every $x \in X$ is contained on one of the balls $B_{\delta_x / 2} (x)$, so their union covers $X$. Because $X$ is compact, this cover has a finite subcover, so there exist $x_1, \cdots, x_N$ such that (writing $\delta_i = \delta_{x_i}$) we have
\[ X = B_{\delta_1 / 2} (x_1) \cup \cdots \cup B_{\delta_N / 2} (x_N) . \] (1)

Now let \[ \delta = \min \left( \frac{\delta_1}{2}, \cdots, \frac{\delta_N}{2} \right) . \] We will show that if $|x - y| < \delta$, then $|f (x) - f (y) | < \varepsilon$. This will prove uniform continuity of $f$.

Suppose that $|x - y| < \delta \le \delta_i / 2$. By (1) we have $x \in B_{\delta_i / 2} (x_i)$ for some $i$. Thus $|x - x_i | < \delta_i / 2$. Furthermore \[ |y - x_i | = |y - x + x - x_i | \le |y - x| + |x - x_i | \le \frac{\delta_i}{2} + \frac{\delta_i}{2} = \delta . \] Thus both $x$ and $y$ lie in $B_{\delta_i} (x_i)$, and so $|f (x) - f (x_i) | < \frac{\varepsilon}{2}$, $|f (y) - f (x_i) | < \frac{\varepsilon}{2}$. By the triangle inequality, we therefore have \[ |f (x) - f (y) | \le |f (x) - f (x_i) | + |f (y) - f (x_i) | < \varepsilon . \]

Here is an extremely useful application of uniform continuity.

Proposition 1: Let $f (x, t)$ be a continuous function defined on $[a, b] \times U$ where $U$ is an open interval in $\mathbb{R}$. Assume that the partial derivative $\partial f / \partial t$ is also continuous on $[a, b] \times U$. Then \[ \frac{d}{d t} \int_a^b f (x, t) \, d x = \int_a^b \frac{\partial f}{\partial t} (x, t) \, d x. \]

Proof. (Click to Expand/Collapse)

Let \[ F (t) = \int_a^b f (x, t) \, d x, \] and let $f_2 (x, t) = \frac{\partial f}{\partial t} (x, t)$ be the second partial derivative. We must prove \[ F' (t) = \int_a^b f_2 (x, t) \, d x. \] We have \[ F' (t) = \lim_{u \longrightarrow t} \frac{F (u) - F (t)}{u - t} = \lim_{u \longrightarrow t} \int_a^b \frac{f (x, u) - f (x, t)}{u - t} \, d x \] and so what we must show is that \[ \lim_{u \longrightarrow t} \int_a^b \left[ \frac{f (x, u) - f (x, t)}{u - t} - f_2 (x, t) \right] \, d x = 0. \] Let $\varepsilon > 0$ be given.

Another application.

Theorem 2: Let $F (x, y)$ be a function that is continuous on the rectangle $[a, b] \times [c, d]$. Then
\[ \int_a^b F (x, y) \, d x \] (2)

is a continuous function of $y$, hence integrable. We have
\[ \int_c^d \int_a^b F (x, y) \, d x \; d y = \int_a^b \int_c^d F (x, y) \, d y \, d x. \] (3)

Proof. (Click to Expand/Collapse)

Let $\varepsilon > 0$ be given, and $y \in [c, d]$. Since $R = [a, b] \times [c, d]$ is compact, $F$ is uniformly continuous and so there exists a $\delta$ such that if $\sqrt{(t - v)^2 + (u - w)^2} < \delta$ with $(t, u)$ and $(v, w)$ in $R$, then $|F (t, u) - F (v, w) | < \frac{\varepsilon}{b - a}$. So if $|y_1 - y_2 | < \delta$ we have \[ \left| \int_a^b F (x, y_1) \, d x - \int_a^b F (x, y_2) \, d x \right| \le \int_a^b |F (x, y_1) - F (x, y_2) | \, d x \le \int_a^b \frac{\varepsilon}{b - a} \, d x = \varepsilon . \] This proves the continuity of (2).

Now consider \[ \phi (y, z) = \int_a^z F (x, y) \, d x. \] The partial derivative $\phi_2 (y, z) = \frac{\partial \phi}{\partial z} (y, z)$ equals $F (x, z)$, hence is continuous. Therefore we may differentiate under the integral sign and obtain \[ \frac{\partial}{\partial z} \int_c^d \int_a^z F (x, y) \, d x \; d y = \int_c^d F (z, y) \, d y = \frac{\partial}{\partial z} \int_a^z \int_c^d F (x, y) \, d y \, d x. \] The function \[ \psi (z) = \int_c^d \int_a^z F (x, y) \, d x \; d y - \int_a^z \int_c^d F (x, y) \, d y \, d x \] thus satisfies $\psi' (z) = 0$ for all $z \in [a, b]$ and $\psi (z) = 0$. By the mean value theorem, \[ \psi (b) = \psi (b) - \psi (a) = (b - a) \psi' (z) \] for some $z$, which proves (3).