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?
Show that as a distribution $u$ on $D(R^1)$, if $u'\ge0$, then $u$ is an increasing function indeed.
Asked
Active
Viewed 160 times
1 Answers
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 $$\int_{\mathbb{R}} \phi'(x)f(x)\,dx=\int_{\mathbb{R}} \phi'(x)\int_0^x1\,d\mu(t)\,dx\\ =\int_{\mathbb{R}} \int_t^\infty\phi'(x)\,dx\,d\mu(t) =-\int_{\mathbb{R}} \phi(t)\,d\mu(t)$$
Gio67
- 21,870
-
actually, i am very confused about the measure. In wiki, it is said a Radon measure which has the property of inner regularity and locally finite property and it is defined on a $\sigma-$algebra of Borel set(which is clearly the sigma-algebra generated by all open sets). So it seems if $T$ is a positive distribution , then the "some measure" is a Radon measure? But I am not sure. – stephenkk Jul 30 '17 at 00:43
-
yes a Radon measure. Since you are on the real line, all you need is to be finite on compact sets. – Gio67 Jul 30 '17 at 01:07
-
sorry，i don‘t know how to show that derivative of f(x) is S'. – stephenkk Jul 30 '17 at 01:18
-
It's not easy to prove. It's integration by parts for Lebesgue-Stiljies (see here https://math.stackexchange.com/questions/221521/integration-by-parts-and-lebesgue-stieltjes-integrals – Gio67 Jul 30 '17 at 01:22
-
could u give me a proof of this thing. I know Lebesgue-Stiljies integral. But I don't see easy ways to pass to the integration by parts, since f(x) is non-decreasing, but it may be not absolutely continuous with respect to lebesgue measure. Even so, I don't see clear ways to apply the integration by parts. Since they are for two increasing functions, while we have only f as a non-decreasing function. – stephenkk Jul 30 '17 at 01:49
-
I just did. I actually used Fubini's and not integration by parts. – Gio67 Jul 30 '17 at 11:32