Show that as a distribution $u$ on $D(R^1)$, if $u'\ge0$, then $u$ is an increasing function indeed.

Question

The question is that: if $u$ is a distribution on real line, and first distributive derivative of it is a positive distribution, is $u$ indeed a locally integrable function? And if the second derivative is positive, is $u$ a convex function on real line?

Answer 1

Yes to both. The proof is in the book of Laurent Schwartz. Theorie des Distributions. Hermann, Paris, 1950–51. The theorems you want are on page 54. The proof is a bit sketchy and it is in French... sorry! For the first, you first prove that if a derivative of a distribution is constant then the distribution is actually a constant function. Then you prove that if a distribution is nonnegative, that is $T(\phi)\ge 0$ for all $\phi\ge 0$, then $T$ is given by $T(\phi)=\int_R\phi \,d\mu$ for some measure $\mu$. For this you use Hahn-Banach and Riesz representation theorem. Next if a distribution $S$ has nonnegative derivative $S'$ then $S'(\phi)=\int_R\phi \,d\mu$. You consider the function $f(x)=\int_0^x\,d\mu$ and prove that its derivative in the sense of distributions is exactly $S'$. Hence $S'-f'=0$ in the sense of distributions and so $S-f$ is a constant. The convex case is similar.

Edit Given $\phi$ by Fubini's theorem you have $$\int_{\mathbb{R}} \phi'(x)f(x)\,dx=\int_{\mathbb{R}} \phi'(x)\int_0^x1\,d\mu(t)\,dx\\ =\int_{\mathbb{R}} \int_t^\infty\phi'(x)\,dx\,d\mu(t) =-\int_{\mathbb{R}} \phi(t)\,d\mu(t)$$