# Discrete versus Continuous Hilbert Transform

Let me define the Fourier transform of a function $u$, say in the Schwartz space $\mathscr S(\mathbb R)$ as $\hat u(\xi)=\int_{\mathbb R} e^{-2iπ x\cdot \xi} u(x) dx.$ The Hilbert transform $\mathscr H$ is the Fourier multiplier $-i \operatorname{sign} \xi$ and can be expressed equivalently as the convolution with $\operatorname{pv}\frac1{π x}.$ We have $$(\mathscr H u)(x)=\lim_{\epsilon\rightarrow 0_+}\int_{\vert x-y\vert\ge \epsilon} \frac{u(y)}{π(x-y)} dy.$$ That operator can be realized as a unitary skew-adjoint operator on $L^2(\mathbb R)$: this is easy when we look at the Fourier multiplier realization $-i\operatorname{sign}(D)$ given above. A related operator is the so-called Hardy $\mathscr H_a$ operator, whose kernel is $$k(x,y)=\frac{H(x)H(y)}{π(x+y)}, \quad H=\mathbf 1_{\mathbb R_+}.$$ It is easy to prove that $\mathscr H_a$ is a selfadjoint operator on $L^2(\mathbb R)$ with operator-norm less than 1 and slightly more difficult to prove that its operator-norm is actually equal to 1. On the other hand, one may look at the discrete Hardy operator $\mathscr H_{ad}$, say on $\ell^2(\mathbb N^*)$ by the kernel $$J(k,l)=\frac{1}{π(k+l)}.$$ To prove that $\mathscr H_{ad}$ is bounded selfadjoint with operator norm less than 1 is not difficult, using the boundedness of the discrete Hilbert transform. It is also possible to prove that the norm of $\mathscr H_{ad}$ is actually 1.

Finally my question: What are the explicit links between the discrete results and the "continuous" results mentioned above? In particular, these not-so-trivial results on operator-norms equal to 1 for Hardy operators are more classical in the discrete case. Is it possible to actually deduce their continuous version from the discrete ones? References are welcome.