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I have been searching for an answer to this question without much success. I know that any convex function in $\mathbb{R}$ is differentiable almost everywhere (since it is locally Lipschitz, and locally Lipschitz functions are differentiable a.e.). Also, if a convex function $f$ is differentiable at a point $x$, then for all $y$:

$$ f(y) \geq f(x) + f'(x)(y-x) $$

This is also a characterization of convexity if $f$ is $C^1$.

My question is: if we have a function that is differentiable almost everywhere, and for which the inequality above holds, can we also conclude convexity? I have tried to adapt this answer, but I think it breaks because of having differentiability only a.e.

Thanks in advance.

Ismael Medina
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1 Answers1

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If $f$ is not continuous, then the conclusion is clearly false. So, let's assume $f$ is continuous and differentiable a.e.

Given $x$, $y$ and $0<\theta<1$ define $z = \theta x + (1-\theta)y$. We need to show that $$ f(z) \le \theta f(x) + (1-\theta) f(y). $$ If $f$ happens to be differentiable at $z$, we can proceed (as usual) by observing that $$ f(x) \ge f(z) + f'(z)(x-z)\qquad\text{and}\qquad f(y) \ge f(z) + f'(z)(y-z) $$ and then multiplying the first inequality by $\theta$ and the second by $1-\theta$ to obtain the desired conclusion.

If $f$ is not differentiable at this particular $z$, we can tweak $\theta$ by $\epsilon$, little enough as to move $z$ to a point $z'$ where $f$ is differentiable. Proceeding as above, we would get $$ f(z') \le (\theta + \epsilon)f(x) + (1-\theta-\epsilon)f(y). $$ By continuity, we would get the desired inequality for $f(z)$ only letting $\epsilon\to0$.

Leandro Caniglia
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