Curious Continuity

Problem 1. Define the function $f : [-1, 1] \to \mathbb R$ by

$f(x) = \begin{cases} x^2, & x \in [-1,1] \cap \mathbb Q, \\ -x^2, & x \in [-1,1] \backslash \mathbb Q. \end{cases}$

Prove that $f$ is continuous at $0$ .

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Solution. We observe that $-x^2 \leq f(x) \leq x^2$ . Since $\pm x^2 \to 0$ as $x \to 0$ , by the squeeze theorem, $f(x) \to 0 = 0^2 = f(0)$ as $x \to 0$ , as required.

Problem 2. For any function $f$ that is continuous at $c$ , prove that if $f(c) > 0$ , then there exists $\delta > 0$ such that for $x \in (c - \delta, c + \delta)$ , $f(x) > 0$ .

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Solution. Suppose $f(c) > 0$ . Fix $\epsilon > 0$ that will be chosen later. Since $f$ is continuous at $c$ , for any $k > 0$ , there exists $\delta > 0$ such that

$|x-c| < \delta \quad \Rightarrow \quad |f(x) - f(c)| < k \cdot \epsilon.$

Expanding the right-hand side, $f(x) > f(c) - k \cdot \epsilon$ . Setting $\epsilon = f(c) > 0$ and $k = 1/2$ ,

$\displaystyle f(x) > f(c) - \frac{1}{2} \cdot f(c) = \frac{f(c)}{2} > 0.$

Problem 3. Let $f : [a, b] \to \mathbb R$ be continuous. Suppose that for any $x \in [a, b]$ , $f(x) \neq 0$ . Prove that there exists $C > 0$ such that either $f < -C$ or $f > C$ . In particular, $|f| > C$ .

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Solution. The function $g := | \cdot | \circ f : [a, b] \to \mathbb R$ is a composition of continuous functions and thus continuous. By the extreme value theorem, there exists $c \in [a, b]$ such that for any $x \in [a, b]$ ,

$|f(x)| = g(x) \geq g(c) = |f(c)|.$

Since $f(c) \neq 0$ , $|f(c)| > 0$ , so that defining $C := |f(c)| / 2 > 0$ yields

$|f(x)| \geq |f(c)| > C.$

Now suppose $f(c_0) > 0$ without loss of generality. In the case $f(c_0) < 0$ , apply the first case to the continuous map $-f$ . We claim that $f > 0$ .

Suppose, for a contradiction, there exists $c_1 \in (a, b)$ such that $f(c_1) < 0$ . Assume $c_1 \in (c_0, b)$ without loss of generality. By the intermediate value theorem, there exists $c_2 \in (c_0, c_1)$ such that $f(c_2) = 0$ , a contradiction. Therefore, $f > 0$ . For any $x \in [a, b]$ , $f(x) = |f(x)| > C$ , yielding $f > C$ , as required.

Problem 4. Let $f : \mathbb R \to \mathbb R$ and suppose that for any $x, y \in \mathbb R$ ,

$f(x + y) = f(x) + f(y).$

Prove that if $f$ is continuous at some $c \in \mathbb R$ , then $f$ is continuous on $\mathbb R$ .

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Fix $x \in \mathbb R$ . Since $f$ is continuous at $c$ ,

$\begin{aligned} f(c) &= \lim_{t \to 0} f(c+t) \\ &= \lim_{t \to 0} (f(c)+f(t)) \\ &= \lim_{t \to 0} f(c)+\lim_{t \to 0} f(t) \\ &= f(c) \cdot 1 + \lim_{t \to 0} f(t), \end{aligned}$

so that $\displaystyle \lim_{t \to 0}f(t) = 0$ . Therefore,

$\begin{aligned} \lim_{t \to 0} f(x+t) &= \lim_{t \to 0} (f(x)+f(t)) \\ &= \lim_{t \to 0} f(x) + \lim_{t \to 0} f(t) \\ &= f(x) \cdot 1 + 0 = f(x). \end{aligned}$

Problem 5. Let $f, g : \mathbb R \to \mathbb R$ be functions. Suppose that $g$ is continuous at $0$ and that there exists some sufficiently small $c > 0$ such that $f(t) = g(t)$ for $t \in (-c,c)\backslash \{0\}$ . Show that

$\displaystyle \lim_{t \to 0} f(t) = g(0).$

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Solution. Assume $g(0) = 0$ without loss of generality. Fix $\epsilon > 0$ . Since $g$ is continuous at $0$ , there exists $\delta_0 > 0$ such that

$\displaystyle |t| < \delta_0 \quad \Rightarrow \quad |g(t)| < \epsilon.$

Set $\delta := \min\{ c, \delta_0 \}$ . Fix $0 < |t| < \delta$ . Since $\delta \leq c$ , $f(t) = g(t)$ . Since $\delta \leq \delta_0$ ,

$|f(t)| = |g(t)| < \epsilon.$

Remark 1. By Problem 5, if there exists a continuous function $g_a$ such that for $t \neq 0$ ,

$\displaystyle \frac{f(a+t) - f(a)}{t} = g_a(t),$

then $f'(a)$ exists, and in which case, $f'(a) = g_a(0)$ . For instance, if $f(x) = x^n$ , then using the binomial theorem,

$\displaystyle \frac{f(a+t) - f(a)}{t} = \underbrace{ na^{n-1} + (\text{finite stuff}) \cdot t }_{ g_a(t) },$

Therefore, $\displaystyle f'(a) = g_a(0) = na^{n-1}$ .

Problem 6. Let $f : [0, 1] \to \mathbb R$ be continuous and $f(0) = f(1)$ . Prove that for any $n \in \mathbb N^+$ , there exists $\xi \in [0,1]$ such that $f(\xi+1/n) = f(\xi)$ .

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Solution. For each $n$ , define the function $g := f(\cdot + 1/n) - f$ . We aim to show that $g$ has at least one root. We claim that there exists $x_0 \in [0,1]$ such that $g(x_0) \geq 0$ . Otherwise, $f < f(\cdot + 1/n)$ . In particular,

$\displaystyle f(0) < f(1/n) < f(2/n) < \cdots < f(1) = f(0),$

a contradiction. Therefore, there exists $x_0 \in [0,1]$ such that $g(x_0) \geq 0$ . A similar argument shows that there exists $x_1 \in [0,1]$ such that $g(x_1) \leq 0$ .

If at least one equality holds, then set $\xi = x_0$ or $\xi = x_1$ . Otherwise, $g(x_0) > 0$ and $g(x_1) < 0$ . Use the intermediate value theorem to find $c$ between $x_0$ and $x_1$ such that $g(c) = 0$ , as required.