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Let $f(x)$, $x>0$ be a convex function. Then it's distributional second derivative is defined by the rule $$ \langle f''(x),\varphi(x)\rangle = \langle f(x), \varphi''(x)\rangle $$ for any $\varphi \in \mathcal{D}(0,\infty)$. If function $f(x) \in C^2(0,\infty)$ then $$ \langle f''(x),\varphi(x)\rangle \geqslant 0 $$ for any $\varphi \in \mathcal{D}(0,\infty)$, $\varphi \geqslant 0$. It is true for general convex function $f(x)$? There is a theorem that states that any distribution may be approximated by sequence of smooth functions. I tried to use it, but without success.

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One possible approach would be to try to check if you can modify that theorem: try to prove that if $f$ is convex, you can approximate the distribution defined by $f$ be a sequence of smooth convex functions... – N. S. Dec 1 '12 at 18:20
@N.S. aha! That's what I'm trying to do! – Nimza Dec 1 '12 at 18:24
up vote 4 down vote accepted

Here is another approach using mollifiers, adapted from Evans and Gariepy "Measure theory and fine properties of functions."

Let $\eta_\epsilon$ be a mollifier (smooth approximation to the identity) and set $f^\epsilon=f*\eta_\epsilon$. Then $f_\epsilon$ is smooth and convex (convexity is reasonably trivial to check). Thus for all $\varphi\in\mathcal{D}(0,\infty)$ with $\varphi\geq 0$,

$$ \langle f^{\prime\prime}_\epsilon,\varphi\rangle\geq 0 $$

Integrate by parts:

$$ \langle f^{\prime\prime}_\epsilon,\varphi\rangle=\langle f_\epsilon,\varphi^{\prime\prime}\rangle\geq 0 $$

Now let $\epsilon\searrow 0$; $f_\epsilon\rightarrow f(x)$, and hence $\langle f,\varphi^{\prime\prime}\rangle\geq 0$

Hopefully I didn't overlook any tricky details here.

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