I have added a complement to my answer above.

@Christian Remling: yes, of course, thanks.

@Robert Israel: yes, but I would like a somewhat more explicit description yielding in particular the Poisson summation formula. The latter formula is not exactly a triviality and to provide an algebraic proof would be interesting.

The question is not Hilbertian, but on $\mathcal S'$: I want also to include the Poisson summation formula.

My (Bazin's) answer and Liviu Nicolaescu's answer are correct and contradict George Lowther's answer. The point is to decide what means $\Delta u=0$. In the sense of distributions, it is clear and both answers mentioned above give a clearcut result. Now if you accept the "counterexample" above which is missing a point, the Laplace equation is not satisfied in the distribution sense.

The equality $\sum_{k\ge 0}\phi_k(\xi)=1$ is equivalent to $\sum_{k\ge 0}\phi_k(D)=Id.$

Assume $d=1$. Formally we have $ \mathcal H(G)=\ker \overline \partial, $ so that $$ \bigl(\mathcal H(G)\bigr)^\perp=\bigl(\ker \overline \partial\bigr)^\perp=\overline{\text{ran }\partial }. $$ As a result the orthogonal of $\mathcal H(G)$ is the set of Radon measures $\mu$ of the form $$ \mu=\frac{\partial \nu}{\partial z}, $$ where $\nu$ is an hyperfunction in $U$ such that $\frac{\partial \nu}{\partial z}$ is a Radon measure.

I do not see the connection with D. Serre's result, which is comparing the numerical radius to the norm. The (discrete) Hardy operator is hard stuff to handle, in particular not trace class. To get its $\ell^2$ boundedness, I used the Hilbert transform, and to get the boundedness of the latter, Fourier transform, which is bounded as a sign function.

Are you saying that you can prove the $L^2$ boundedness of the discrete Hilbert transform (matrix $(i-j)^-1)$) or of the Hardy operator (matrix $(i+j)^-1)$) that way ?