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Let $f: \mathbb{R} \to \mathbb{R}$ a continuous function so that $\forall x \in \mathbb{R}$

$$f_+'(x)=\lim_{\delta \to 0^+}\frac{f(x+\delta)-f(x)}{\delta}=0.$$ Prove that $f$ is constant.

This is what I've done:

Let's consider $g(x)=f(1-x)$, since $f$ is continuous, $g$ is continuous as well and $g_+'(x)=-f_+'(1-x)=0$. $$g_+'(x)=\lim_{\delta \to 0^+}\frac{g(x+\delta)-g(x)}{\delta}=\lim_{\delta \to 0^+}\frac{f(1-x-\delta)-f(1-x)}{\delta}$$ Now, with the substitutions $\lambda=-\delta$ and $y=1-x$ $$g_+'(x)=\lim_{\lambda \to 0^-}-\frac{f(y+\lambda)-g(y)}{\lambda}=-f_-'(y)$$ $\Rightarrow -f_-'(x)=g_+'(x)=0$

$\Rightarrow f_-'(x)=f_+'(x)=f'(x)=0$, $ \forall x \in \mathbb{R}$.

$f$ is continous and $f'(x)=0 \forall x \in \mathbb{R}$

$\Rightarrow$ $f$ is constant $\forall x \in \mathbb{R}$.

Is this correct? Thank you.

Carloss
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    it's correct as far as $f$ is derivable, because there is no reason that $\lim_{\delta \to 0^+}\frac{f(x-\delta )-f(x)}{\delta }$ exist. – Surb Jan 06 '24 at 10:02
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    You didn't use continuity of $f$, right? Without it, it is false (e.g. step function). – freakish Jan 06 '24 at 10:08
  • why does $g'_+(x)$ exist? – whoisit Jan 06 '24 at 11:29
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    See https://math.stackexchange.com/q/1485828/42969 for a more general statement. – Martin R Jan 06 '24 at 13:00
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    $g_+'(x)=-f_+'(1-x)$ is wrong, it should be $g_-'(x)=-f_+'(1-x)$ (and that does not help). – $g_+'(x)=-f_-'(x)$ is also wrong. – Martin R Jan 06 '24 at 13:02
  • @MartinR thank you, I understood the first error but I don't understand why the second one is also wrong – Carloss Jan 06 '24 at 16:59
  • @Carloss: If $g(x) = f(1-x)$ then the derivatives of $g$ at $x$ are determined by the derivatives of $f$ at $1-x$. They cannot be determined by the derivatives of $f$ at the same point $x$. – Martin R Jan 06 '24 at 19:44
  • @MartinR obvious... stupid me, thanks – Carloss Jan 06 '24 at 20:28

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