Several Continuity Exercises

Problem 1. Let $f : [a, b] \to \mathbb R$ be continuous and injective. Prove that $f(a) < f(b)$ implies that $f$ is strictly increasing.

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Solution. Suppose $f(a) < f(b)$ . Suppose for a contradiction that $f$ is not strictly increasing. Then there exist $c_1, c_2 \in (a, b)$ such that $f(c_1) \geq f(c_2)$ . If $f(c_1) = f(c_2)$ , then $f$ is not injective, a contradiction. Thus, suppose that $f(c_1) > f(c_2)$ .

If $f(a) < f(c_1)$ , then either $f(a) < f(c_2)$ or $f(a) > f(c_2)$ .

In the former case, we have $f(a) < f(c_2) < f(c_1)$ , and so use the intermediate value theorem to find $c_3 \in (a, c_1)$ such that $f(c_3) = f(c_2)$ , a contradiction.
In the latter case, we have $f(c_2) < f(a) < f(b)$ and hence find $c_4 \in (c_2, b)$ such that $f(c_4) = f(a)$ , a contradiction.

Therefore, suppose $f(a) > f(c_1)$ . Since $f(c_1) < f(a) < f(b)$ , find $c_5 \in (c_1, b)$ such that $f(c_5) = f(a)$ , a contradiction.

Thus, no matter what, if $f$ is not strictly increasing, we obtain a contradiction. Therefore, $f$ must be strictly increasing.

Problem 2. Let $f : [0, 1] \to [0, 1]$ be continuous. Prove that there exists $c \in [0, 1]$ such that $f(c) = c$ .

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Solution. Define $g : [0, 1] \to \mathbb R$ by $g(x) = f(x) - x$ . It suffices to find $c \in [0, 1]$ such that $g(c) = 0$ . We observe that

$\begin{aligned} g(0) &= f(0) - 0 = f(0) \geq 0, \\ g(1) &= f(1) - 1 \leq 1 - 1 = 0.\end{aligned}$

If either $g(0) = 0$ or $g(1) = 0$ , then we are done. Suppose neither, so that $g(0) > 0$ and $g(1) < 0$ . By the intermediate value theorem, there exists $c \in (0, 1)$ such that $g(c) = 0$ , as required.

Problem 3. Let $f : [1, \infty) \to \mathbb R$ be uniformly continuous. Prove that there exists $M > 0$ such that for any $x \geq 1$ , $|f(x)| \leq Mx$ .

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Solution. Fix $\epsilon > 0$ to be tuned. Since $f$ is uniformly continuous, for any $k > 0$ , there exists $\delta > 0$ such that

$|u-v| < \delta \quad \Rightarrow \quad |f(u) - f(v)| < k \cdot \epsilon.$

We will use a discretising technique. Fix $x > 1$ . By the Archimedean property, find the largest $N \in \mathbb N_0$ such that $1+N \cdot \delta/2 < x$ , which implies that $1+(N+1) \cdot \delta/2 \geq x$ . For each $i = 0, 1, \dots, N$ , define $x_i = 1+ i \cdot (x-1)/(N+1)$ so that

$\displaystyle x_{i+1} - x_i = \frac{ x-1 }{ N+1 } \leq \frac{\delta}{2} < \delta.$

Then by the triangle inequality,

$\begin{aligned} |f(x) - f(1)| &= |f(x_{N+1}) - f(x_0)| \\ &= \left| \sum_{i=0}^N (f(x_{i+1}) - f(x_i)) \right| \\ &\leq \sum_{i=0}^N | f(x_{i+1}) - f(x_i) | \\ &\leq \sum_{i=0}^N (k \cdot \epsilon) = k \cdot (N+1) \cdot \epsilon \\ &< k \cdot \left( \frac{x-1}{ \delta/2 } + 1\right) \cdot \epsilon \\ &< k \cdot \left( \frac{2x}{\delta } + x \right) \cdot \epsilon \\ &= k \cdot \left( \frac{2}{\delta } + 1 \right) \cdot \epsilon \cdot x. \end{aligned}$

Setting $\epsilon = k = 1$ ,

$\displaystyle |f(x) - f(1)| < \left( 1 + \frac{2}{\delta} \right) \cdot x.$

By the triangle inequality again, using $x > 1$ ,

$\displaystyle \begin{aligned} |f(x)| &= |f(1) + f(x) - f(1)|\\ &\leq |f(1)| + |f(x) - f(1)| \\ &< |f(1)| \cdot x + \left( 1 + \frac{2}{\delta} \right) \cdot x \\ &< \left(|f(1)| + 1 + \frac{2}{\delta} \right) \cdot x.\end{aligned}$

Hence, setting $M := |f(1)| + 1 + 2/\delta$ yields the desired result, which works since $M$ does not depend on $x$ . Finally, this choice of $M$ works to establish $|f(1)| \leq M \cdot 1$ .