$\int_0^1 u(t)\phi''(t)dt \geq 0,\ \forall \phi\in C_0^1((0,1)), \ \phi\geq 0$. Is $u$ convex? 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.
 A: 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.
A: 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.
