Is the function differentiable

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that for $x_0 \in \mathbb{R}$

$$ \lim_{\mathbb{Q} \ni h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}$$

exists. Is this function differentiable at $x_0$?

Answer 1

Let $$ L=\lim_{\mathbb Q\ni h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$$ Let $(h_n)_{n\in\mathbb N}$ be a sequence with $h_n\to 0$. We need to show that $$\lim_{n\to\infty}\frac{f(x_0+h_n)-f(x_0)}{h_n}=L.$$ Consider $$\begin{align}g\colon\mathbb R\setminus\{0\}\to&\mathbb R\\ h\quad\mapsto &\frac{f(x_0+h)-f(x_0)}{h}.\end{align}$$ Then $g$ is continuous and hence $|g(h_n')-g(h_n)|<\frac1n$ holds for all $h_n'\ne 0$ sufficiently close to $h_n$. Since the rationals are dense in $\mathbb R$, we thus can find $h_n'$ such that $h_n'\in\mathbb Q\setminus\{0\}$ and $|h_n'-h_n|<\frac1n$ and $|g(h_n')-g(h_n)|<\frac1n$. Then $\mathbb Q\ni h_n'\to 0$, hence $g(h_n')\to L$ and $g(h_n)\to L$ as was to be shown.

Answer 2

Suppose that the limit $ \displaystyle L := \lim_{\substack{h \to 0; \\ h \in \mathbb{Q}}} \frac{f(x_{0} + h) - f(x_{0})}{h} $ exists.

Claim: $ f $ is differentiable at $ x_{0} $.

Proof: Let $ (h_{n})_{n \in \mathbb{N}} $ be a sequence in $ \mathbb{R} \setminus \{ 0 \} $ that converges to $ 0 $. It suffices to prove that the limit $$ \lim_{n \to \infty} \frac{f(x_{0} + h_{n}) - f(x_{0})}{h_{n}} $$ exists.

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For each $ n \in \mathbb{N} $, use the continuity of $ f $ and the denseness of $ \mathbb{Q} $ to choose a $ q_{n} \in \mathbb{Q} \setminus \{ 0 \} $ arbitrarily close to $ h_{n} $ so that

• $ \forall n \in \mathbb{N}: \quad |f(x_{0} + h_{n}) - f(x_{0} + q_{n})| < \left| \dfrac{h_{n}}{n} \right| $ and

• $ \displaystyle \lim_{n \to \infty} \frac{q_{n}}{h_{n}} = 1 $, which automatically yields $ \displaystyle \lim_{n \to \infty} q_{n} = 0 $.

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Next, observe that \begin{align} \forall n \in \mathbb{N}: \quad \frac{f(x_{0} + h_{n}) - f(x_{0})}{h_{n}} &= \frac{f(x_{0} + h_{n}) - f(x_{0} + q_{n})}{h_{n}} + \frac{f(x_{0} + q_{n}) - f(x_{0})}{h_{n}} \\ &= \frac{f(x_{0} + h_{n}) - f(x_{0} + q_{n})}{h_{n}} + \frac{f(x_{0} + q_{n}) - f(x_{0})}{q_{n}} \cdot \frac{q_{n}}{h_{n}}. \end{align}

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As $$ \forall n \in \mathbb{N}: \quad \left| \frac{f(x_{0} + h_{n}) - f(x_{0} + q_{n})}{h_{n}} \right| < \left| \frac{1}{h_{n}} \cdot \frac{h_{n}}{n} \right| = \frac{1}{n} \stackrel{n \to \infty}{\longrightarrow} 0, $$ the Squeeze Theorem yields $$ \lim_{n \to \infty} \frac{f(x_{0} + h_{n}) - f(x_{0} + q_{n})}{h_{n}} = 0. $$

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Therefore, \begin{align} f'(x_{0}) = &\lim_{n \to \infty} \frac{f(x_{0} + h_{n}) - f(x_{0})}{h_{n}} \\ = &\lim_{n \to \infty} \left[ \frac{f(x_{0} + h_{n}) - f(x_{0} + q_{n})}{h_{n}} + \frac{f(x_{0} + q_{n}) - f(x_{0})}{q_{n}} \cdot \frac{q_{n}}{h_{n}} \right] \\ = &\lim_{n \to \infty} \frac{f(x_{0} + h_{n}) - f(x_{0} + q_{n})}{h_{n}} + \left[ \lim_{n \to \infty} \frac{f(x_{0} + q_{n}) - f(x_{0})}{q_{n}} \right] \left( \lim_{n \to \infty} \frac{q_{n}}{h_{n}} \right) \\ = &0 + \left[ \lim_{n \to \infty} \frac{f(x_{0} + q_{n}) - f(x_{0})}{q_{n}} \right] \cdot 1 \quad (\text{By the previous paragraphs.}) \\ = &\lim_{n \to \infty} \frac{f(x_{0} + q_{n}) - f(x_{0})}{q_{n}} \\ = &L. \quad (\text{By the initial hypothesis.}) \quad \spadesuit \end{align}

Answer 3

This is Hagen von Eitzen's proof, but without sequences.

We may assume $x_0=f(x_0)=0$. The function $g(h):={f(h)\over h}$ $\ (h\ne0)$ is continuous, and after subtracting a linear function from $f$ our basic assumption is $$\lim_{\Bbb Q\ni h'\to 0}g(h')=0\ .$$

Given an $\epsilon>0$ there is a $\delta>0$ such that $$\left|g(h')\right|<\epsilon\qquad\bigl(|h'|<\delta,\ h'\in\Bbb Q)\ .$$ Assume now that $0<|h|<\delta$. Since $g$ is continuous at $h$ there is a $h'\in\Bbb Q$ with $0<|h'|<\delta$ and $|g(h')-g(h)|<\epsilon$. It follows that $|g(h)|<2\epsilon$. This proves $\lim_{h\to0} g(h)=0$, since $\epsilon>0$ was arbitrary.