Suppose $u,f$ periodic and smooth in $Q=[0,1]^n$ such that $\Delta u=f$. Show that for each $i,j$, $$\int_Q \left| \frac{\partial^2 u}{\partial x_i \, \partial x_j} \right|^2 \leq C \int_Q |f|^2.$$
My attempt so far: In previous parts of the problem, we have been asked to find the Fourier series for $f$ in terms of the Fourier series of $u$, $$u(x)=\sum_{k\in \mathbb{Z}^n}\hat{u}(k)e^{2\pi i \langle x,k\rangle}.$$
I come up with $$f(x) = \sum_{k\in \mathbb{Z}^n \setminus 0}\hat{u}(k)(2\pi i)^2 \|k\|^2 e^{2\pi i \langle x,k\rangle}.$$
Now if I try to write $\frac{\partial^2 u}{\partial x_i \, \partial x_j}$ in a series, I get
$$\sum_{k\in \mathbb{Z}^n}\hat{u}(k)(2\pi i )^2k_ik_je^{2\pi i \langle x,k \rangle},$$
and now my idea has been to compare the $\ell^2$ norms of the sequences of coefficients. In this question I tried to come up with a relevant inequality, but found that what I was looking for was impossible. Any ideas?
The Fourier series approach is exactly on target, and the negative answer to your other question was merely a result of the question being misstated.
For every $x\in\mathbb R^n$ and every pair of indices $i,j$,
$$|x_ix_j|\le \sum_{r=1}^n x_r^2 $$ It is tempting to put $\frac12 $ on the right, but this does not work when $i=j$. Indeed, when the function depends on $x_1$ only, the second derivative in $x_1$ is exactly the same as the Laplacian. So, $C=1$ cannot be improved.