Let $ u \colon \mathbb{R}^{n} \to \mathbb{R} \in C^{4}(\mathbb{R}^{n}) $ be a convex function such that $\Delta u = 1$. Show that $u$ is a quadratic polynomial.
What I have done:
Every harmonic convex harmonic function is an afim function. The eigenvalues of $D^{2}u$ are positives and the sum of it is 0. Then $D^2 u$ is 0 in every $\mathbb{R}^{n}$. Then using that harmonic function is analytical we conclude that $u$ is linear.
When $\Delta u = 1$, I am able to conclude that $D^{2} u$ is bounded. Is there like a Liouville's Theorem to conclude that $D^2 u$ is constant? If it is true we finish our question.
PS: I still did not use that $u$ is $C^{4}$.
Because $u$ is convex, eache eigenvalue of its hessian matrix is non-negative and since $\Delta u = 1$, eache eigenvalue is at most 1. Then, defining $v \colon \mathbb{R}^{n} \to \mathbb{R}$ such that $v(x) = \frac{1}{2n}\langle x,x \rangle$ we have that $w \colon = u - v$ is an harmonic function and the eigenvalues of $w$ are bounded and so the second derivate is also bounded. Using Liouville's Theorem we get that $D^{2}u(x) = H$ for some matrix $H$ and every $x \in \mathbb{R}^{n}$. Then we can write:
$$w(x) = w(0) + \langle D^1w(0),x \rangle + \frac{\langle Hx , x\rangle }{2},$$
and then
$$u(x) = u(0) + \langle D^{1}w(0) ,x \rangle + \frac{\langle Hx , x\rangle }{2} + \frac{1}{2n}\langle x,x \rangle.$$