$C$ such that $\sum_{k\in \mathbb{Z}^n} k_i^2k_j^2|a_{ij}|^2 \leq C \sum_{k\in \mathbb{Z}^n} \|k\|^4|a_{ij}|^2$

Question

More generally, can we find $C_n>0$ such that

$$\sum_{k\in \mathbb{Z}^n} k_i^2k_j^2|a_{ij}|^2 \leq C \sum_{k\in \mathbb{Z}^n} \|k\|^2|a_{ij}|^4$$

for all $\{a_k\}_{k\in \mathbb{Z}^n} \in \ell^2(\mathbb{Z}^n)$ where the above sums converge?

My ideas so far: $ij \leq i^2 + j^2$ shows that $C=1$ works. Is 1 the sharpest possible bound?

(This comes from showing that for $u,f$ periodic and smooth in $Q=[0,1]^n$ such that $\Delta u=f$, we have that $$\int_Q \left| \frac{\partial^2 u}{\partial x_i \partial x_j} \right|^2 \leq C \int_Q |f|^2,$$ and since we have been considering Fourier series of the form $$\sum_{k\in \mathbb{Z}^n}\hat{u}(k)e^{2\pi i \langle x,k \rangle}$$ in the problem, my strategy was to use that the $\ell^2$ norm is the same as the $L^2$ norm.)

Answer 1

Here is your mistake. When you write out the right hand integral for the case $n=2$, you should get $$ \sum_{i,j} (i^2+j^2)^2 |a_{ij}|^2 .$$ In other words, you knew what you were doing. You just made an easy to make mistake.

Answer 2

Considering $a_{i,j}:=b_ic_j$, the existence of $C$ would imply those of a $C$ such that for each $(c_j)$ for which $\sum j^2c_j^2$ is finite, we have $$\sum_{j\geqslant 1}j^2c_j^2\leqslant C\sum_{j\geqslant 1}c_j^2,$$ which is not possible (take $c_N=1$ for a fixed $N$ and $c_j=0$ if $j\neq N$).