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Given some convex function $f(x)$, $x >0$ we can define a distribution $F \in \mathcal{D}'(0,\infty)$ using Lebesgue-Stieltjes integral $$ \langle F, \varphi \rangle =\int\limits_{0}^{\infty}\varphi(x) \, df'(x), \; \; \; \varphi\in \mathcal{D}(0,\infty) $$ If function $f(x) \in C^{2}(0,\infty)$, then $$ \langle F, \varphi \rangle = \int\limits_{0}^{\infty} \varphi(x)f''(x)\,dx = \int\limits_{0}^{\infty} f(x) \varphi''(x) \, dx, \;\;\; \varphi \in \mathcal{D}(0,\infty) $$ How to show that for general convex function $f(x)$, $x>0$ we also have $$ \int\limits_{0}^{\infty} \varphi(x) \, df'(x) = \int\limits_{0}^{\infty} f(x) \varphi''(x) \, dx $$ for any $\varphi(x) \in \mathcal{D}(0,\infty)$? It is sufficient to show that for any convex function $f(x)$, $x>0$ we can find a sequence of $C^{2}(0,\infty)$ functions $f_{n}(x)$, such that $f_{n}'(x) \to f'(x)$ in points of continuity of $f'(x)$ and at the same time for any $\varphi(x) \in \mathcal{D}(0,\infty)$ $$ \int\limits_{0}^{\infty} f_{n}(x) \varphi(x) \, dx \to \int\limits_{0}^{\infty} f(x) \varphi(x) \, dx. $$ If such sequence $f_{n}(x)$ is found we can pass to the limit in equality $$ \int\limits_{0}^{\infty} \varphi(x) \, df_{n}'(x) = \int\limits_{0}^{\infty} f_{n}(x) \varphi(x) \, dx $$ to obtain the desired result.

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