Equivalent definition for C^1 functions

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Equivalent definition for C^1 functions

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Source: Romania NMO 2025 11.3

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Ciobi_

#1 h Apr 2, 2025, 1:54 PM

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Prove that, for a function $f \colon \mathbb{R} \to \mathbb{R}$ , the following $2$ statements are equivalent:
a) $f$ is differentiable, with continuous first derivative.
b) For any $a\in\mathbb{R}$ and for any two sequences $(x_n)_{n\geq 1},(y_n)_{n\geq 1}$ , convergent to $a$ , such that $x_n \neq y_n$ for any positive integer $n$ , the sequence $\left(\frac{f(x_n)-f(y_n)}{x_n-y_n}\right)_{n\geq 1}$ is convergent.

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KAME06

#2 h Apr 3, 2025, 2:17 AM

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Let $(x_n)_{n\geq 1},(y_n)_{n\geq 1}$ sequences that converge to $a$ and $x_n \neq y_n$ for any $n$ .
First, note that $(x_n)$ or $(y_n)$ mustn't have an $n$ such that $x_n=a$ or $y_n=a$ (because $x_n \neq y_n$ and they are convergent to $a$ ), so WLOG consider $y_n \neq a$ for any $n$ .
(a) $\rightarrow$ (b):
If $f$ is differentiable, with continuous first derivative, then the function $\lim_{b \rightarrow x} \frac{f(x)-f(b)}{x-b}$ is continuous.
That implies that: $\lim_{x \rightarrow a} \frac{f(a)-f(b)}{a-b}$ = $\lim_{n \rightarrow \infty}\left(\lim_{b \rightarrow x_n} \frac{f(x_n)-f(b)}{x_n-b}\right)$ .
As $(x_n)$ and $(y_n)$ converge, for large enough $n$ , $|x_n-a|<\frac{\epsilon}{2}$ , $|y_n-a|<\frac{\epsilon}{2}$ , so $|x_n-y_n|<\epsilon$ .
That implies (and the fact that that $y_n \neq x_n$ for any $n$ ) if $n \rightarrow \infty$ , then $y_n \rightarrow x_n$ , so:
$\lim_{n \rightarrow \infty}\left(\lim_{b \rightarrow x_n} \frac{f(x_n)-f(b)}{x_n-b}\right)=\lim_{n \rightarrow \infty} \frac{f(x_n)-f(y_n)}{x_n-y_n}$ . The limit exists.
We conclude that $\left(\frac{f(x_n)-f(y_n)}{x_n-y_n}\right)_{n\geq 1}$ is convergent.
(b) $\rightarrow$ (a):
As above: $\lim_{n \rightarrow \infty} \frac{f(x_n)-f(y_n)}{x_n-y_n}=\lim_{n \rightarrow \infty}\left(\lim_{b \rightarrow x_n} \frac{f(x_n)-f(b)}{x_n-b}\right)=\lim_{n \rightarrow \infty}\left(\lim_{b \rightarrow a} \frac{f(x_n)-f(b)}{a-b}\right)$ .
$\lim_{n \rightarrow \infty}f(x_n)$ must be defined because the limit exist. If $f(x_n)$ doesn't converge to $f(a)$ , notice that $b \rightarrow a$ and $a-b$ converges to $0$ and it can be as little as we want, so $\lim_{b \rightarrow a} \frac{f(x_n)-f(b)}{a-b}$ is $\infty$ or $-\infty$ , but $\lim_{n \rightarrow \infty} \frac{f(x_n)-f(y_n)}{x_n-y_n}$ converges. Contradiction.
Then $f(x_n)$ converges to $f(a)$ and $b \rightarrow a$ , so $\lim_{n \rightarrow \infty}\left(\lim_{b \rightarrow x_n} \frac{f(x_n)-f(b)}{x_n-b}\right)=\lim_{b \rightarrow a} \frac{f(a)-f(b)}{a-b}$ .
We conclude that $f$ is differentiable and that $\lim_{b \rightarrow x} \frac{f(x)-f(b)}{x-b}$ is continuous.

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