Bound on $\sum_{\mathbb{Z}^d}|\widehat{f}({\boldsymbol{k}})||{\boldsymbol{k}}|^7$.

Question

Let me assume that we have a function $f$ which is in the Sobolev space with smoothness $\alpha$ in the d-dimensional torus:

$f:\|f\|_{H^{\alpha}(\mathbb{T}^d)}=\sum_{{\boldsymbol{k}}\in\mathbb{Z}^d}|\widehat{f}({\boldsymbol{k}})|^2(1+|{\boldsymbol{k}}|^{2\alpha})<\infty.$

Here $|k|$ is the usual Euclidean norm and $\widehat{f}({\boldsymbol{k}})$ is the Fourier coefficient.

Then, is it possible to bound this quantity $\sum_{{\boldsymbol{k}}\in\mathbb{Z}^d}|\widehat{f}({\boldsymbol{k}})||{\boldsymbol{k}}|^7$ finite, by choosing appropriate $\alpha$, but without $\alpha$ depending on the dimensionality $d$?

I know if one uses Cauchy-Schwarz or so, then one can have the quantity finite but $\alpha$ depending $d$, like $\alpha>(d+14)/2$. But is it necessary to have $d$? For me it is a bit counter intuitive.

Thank you for your help!

Answer 1

For any sequence $\{a_{\bf k}\}_{\bf k\in\mathbb{Z}^d}\in l^2(\mathbb{Z}^d)$ there exists $f\in L^2(\mathbb{T}^d)$ with $\hat{f}(\textbf k) = a_{\bf k}$ for all $\bf k\in\mathbb{Z}^d$, so the question boils down to the following question: does there exist $\alpha$ independent of $d$ such that, for all $d$ and all sequences $\{a_{\bf k}\}_{\mathbf k\in\mathbb{Z}^d}$, $$\text{if }\sum_{\bf k\in\mathbb{Z}^d}{|a_{\mathbf k}|^2(1+|\mathbf{k}|^{2\alpha})}<\infty\quad\text{then}\quad\sum_{\mathbf k\in\mathbb{Z}^d}{|a_{\mathbf k}||\mathbf k|^7}<\infty?$$ The answer is no. To see an example of this, observe that in dimension $d$ we have, for any number $s$, $$\sum_{\mathbf{k}\in\mathbb{Z}^d\backslash\{\mathbf{0}\}}{|\mathbf{k}|^{-s}}<\infty\iff s>d.$$ To see this, it suffices to study the convergence of $\sum_{\mathbf{k}\in\mathbb{Z}^d\backslash\{\mathbf{0}\}}{\|\mathbf{k}\|_{l^{\infty}}^{-s}}$ since $\|\mathbf k\|_{l^{\infty}}\le|\mathbf k|\le\sqrt{d}\|\mathbf k\|_{l^{\infty}}$, and since $$|\{\mathbf k\in\mathbb{Z}^d\,|\,\|\mathbf k\|_{l^{\infty}} = k\}| = (2k+1)^d-(2k-1)^d = (d2^d+o(1))k^{d-1}$$ we have $$\sum_{\mathbf{k}\in\mathbb{Z}^d\backslash\{\mathbf{0}\}}{\|\mathbf{k}\|_{l^{\infty}}^{-s}} = \sum_{k=1}^{\infty}{k^{-s}\cdot(d2^d+o(1))k^{d-1}},$$ which converges if and only if $-s+d-1<-1$, i.e. $s>d$.

Thus, given any $\alpha>0$, take $d>2\alpha-14$. Notice this implies that $\frac{d}{2}+\alpha < d+7$, so let $s$ be some number with $\frac{d}{2}+\alpha < s < d+7$ (note that the bounds imply $2s-2\alpha>d$ and $s-7<d$). Let $a_{\mathbf 0} = 0$ and $a_{\mathbf k} = |\mathbf k|^{-s}$ for $\mathbf k\ne 0$. Then $$\sum_{\mathbf k\in\mathbb{Z}^d}{|a_{\mathbf k}|^2(1+|\mathbf{k}|^{2\alpha})} = \sum_{\mathbf k\in\mathbb{Z}^d\backslash\{0\}}{(|\mathbf k|^{-2s}+|\mathbf k|^{-2s+2\alpha})}<\infty,$$ since $2s > 2s-2\alpha > d$. On the other hand, $$\sum_{\mathbf k\in\mathbb{Z}^d}{|a_{\mathbf k}||\mathbf k|^7} = \sum_{\mathbf k\in\mathbb{Z}^d\backslash\{0\}}{|\mathbf k|^{-s+7}},$$ which diverges since $s-7<d$.