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Suppose that $u\in C([0,1])\cap C^1((0,1))$ satisfies for all $\phi\in C_0^2((0,1))$, $\phi\geq 0$ $$\int_0^1 u(t)\phi''(t)dt \geq 0$$

Can we conclude that $u$ is convex?

Note: $C_0^2((0,1))$ is the space of all $C^2$ functions that are zero in an neighbourhood of $\{0\}$ and $\{1\}$.

Update: Maybe this post may help.

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Tomás
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    Try integration by parts and see what you get. Note that, if $u''(t)>0$ then $u$ is convex. – Mhenni Benghorbal Mar 08 '13 at 15:35
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    I know (but don't recall seeing the proof) that $u$ is convex iff its second derivative in the distributional sense is non-negative – Albert Mar 08 '13 at 16:26
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    Maybe it's easier also show, that the first derivative is increasing (one integration by parts is allowed due to the compact support of $\phi$. One could then assume that $u'(t)<0$ for some t (then it has the same sign in a neighborhood of $t$) and choose $\phi$ such that $\phi'$ is mostly negative there (this could be the hard part since $u'$ could be non-positive everywhere and $\phi'$ necessarily has to be positive somewhere if it's negative somewhere else) – Quickbeam2k1 Mar 08 '13 at 17:02
  • @Quickbeam2k1, increasing is too strong. Non-decreasing is better. – Tomás Mar 08 '13 at 18:04
  • You're right. I'm sorry for being imprecise. It might also be worth looking on the fundamental theorem of calculus of variations. The proof should be similar to that – Quickbeam2k1 Mar 08 '13 at 18:13
  • Good idea. I will take a look in it. THank you @Quickbeam2k1 – Tomás Mar 08 '13 at 18:14

2 Answers2

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Let $\psi(x)$ be a nonnegative $C_c^{\infty}$ function with $\int_0^1 \psi(x) = 1$. Then given $0 < a < b < 1$ and a large $N$ look at $$\int_0^1 u'(t)[N\psi(N(x - b)) - N\psi(N(x-a))]\,dx $$ Since $u'(t)$ is continuous and $\int_0^1 \psi(x) = 1$, as $N$ goes to infinity this goes to $u'(b) - u'(a)$. On the other hand, since $\int_0^1 [N\psi(N(x - b)) - N\psi(N(x-a))]\,dx = 0$, there is a $C_c^{\infty}$ function $\alpha_{a,b,N}(x)$ such that $\alpha_{a,b,N}'(x) = N\psi(N(x - b)) - N\psi(N(x-a))$, and if $N$ is large enough its support will be a compact subset of $(0,1)$. In this case, integrating by parts gives $$\int_0^1 u'(t)[N\psi(N(x - b)) - N\psi(N(x-a))]\,dx = -\int_0^1 u(t)\alpha_{a,b,N}''(x)\,dx $$ The construction is such that each $\alpha_{a,b,N}(x) = N \int_0^x [\psi(N(y - b)) - \psi(N(y-a))]\,dy$ is nonpositive for all $x$, so by the assumptions given the right-side of the above equation is nonnegative. Thus taking limits as $N$ goes infinity we see $u'(b) - u'(a) \geq 0$ for all $a$ and $b$. This implies $u$ is convex as requested.

Zarrax
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  • +1, Really nice. Do you mind to explain me what is the intuition behind your argument? – Tomás Mar 11 '13 at 14:44
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    @Tomás Basically, I'm letting $-\alpha_{a,b,N}(x)$ be the characteristic function of $(a,b)$ and $\alpha_{a,b,N}'(x) = \delta(x - b) - \delta(x-a)$. Since delta functions aren't allowed I approximate by $C_c^{\infty}$ functions. – Zarrax Mar 11 '13 at 14:51
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I think I have an answer to this question, please verify if it is correct.

Consider this post. By using the same argument as him (the argument is in where he proves that $\Delta u_\epsilon(x) =0$), we can conclude that $u''_\epsilon (x)\geq 0$, $\forall x\in (0,1)$, or equivalently, $u_\epsilon$ is convex. Now, because $u_\epsilon(x)\rightarrow u(x)$, we can conclude that $u$ is convex.

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