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Can you check my proof? I think it is correct but it seems a bit overcomplicated and the language used is not the best.

If correct, some advice to write it in a more clear way will be highly appreciated, thank you.

I cant use the continuity of $f$ here for this proof.

Let $-\infty<a<b<\infty$ and $f:[a,b]\to\Bbb R$ is convex. Prove or disprove: for each $x\in(a,b)$ the limits $\partial_{\pm}f(x)$ exists and $\partial_-f(x)\le\partial_+f(x)$.

By convexity of $f$, we have that

$$\frac{f(x)-f(x-\epsilon)}{\epsilon}\le\frac{f(x+\epsilon)-f(x-\epsilon)}{\epsilon}\le\frac{f(x+\epsilon)-f(x)}{\epsilon}$$

for $(x-\epsilon,x+\epsilon)\subset[a,b]$. Then taking limits above we get

$$\lim_{\epsilon\to 0^+}\frac{f(x)-f(x-\epsilon)}{\epsilon}=\partial_-f(x)\le\partial_+f(x)=\lim_{\epsilon\to 0^+}\frac{f(x+\epsilon)-f(x)}{\epsilon}\tag{1}$$

Using convexity of $f$ again we have that

$$\frac{f(y)-f(x)}{y-x}\le\frac{f(b)-f(x)}{b-x},\quad\forall y\in(x,b)$$

$$\frac{f(x)-f(a)}{x-a}\le\frac{f(x)-f(y)}{x-y},\quad\forall y\in(a,x)$$

Then taking appropriated limits above we get

$$\lim_{y\to x^+}\frac{f(y)-f(x)}{y-x}=\partial_+f(x)\le\frac{f(b)-f(x)}{b-x}\\\frac{f(x)-f(a)}{x-a}\le\lim_{y\to x^-}\frac{f(x)-f(y)}{x-y}=\partial_-f(x)$$

Then $\partial_+f(x)$ is bounded above and $\partial_-f(x)$ is bounded below. This together with $(1)$ means that, if such limits exists, they are bounded.

Remembering the sequential characterization of functional limits if these limits doesnt converge, and because it limit process is bounded, it means that exists sequences that have more than one cluster point.

Let suppose that the above limits doesnt converge, then exists two cluster points $c_1\neq c_2$ for any of the above limits process, and we can choose two sequences that converge each one to a cluster point, i.e. $(v_n)\to c_1$ and $(w_n)\to c_2$.

Without lose of generality let $c_1<c_2$, then we can choose points $x_1<x_2<x_3$, such that $f(x_1)\in (v_n)$,$f(x_2)\in (w_n)$ and $f(x_3)\in (v_n)$, where $f(x_1)<f(x_2)$ and $f(x_3)<f(x_2)$. Then

$$\frac{f(x_2)-f(x_1)}{x_2-x_1}>\frac{f(x_3)-f(x_2)}{x_3-x_2},\quad x_2\in(x_1,x_3)$$

what contradicts the convexity of $f$, so the limits $\partial_{\pm}f(x)$ exists.$\Box$

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