M. Riesz theorem on the Hilbert transform

Remarks

Marcel Riesz's proof for $p = 2k$ an even integer uses complex analysis. If $f$ is real and $F = f + iHf$ extends holomorphically to the upper half-plane, then $\int F^{2k} = 0$, and taking real parts bounds $\norm{Hf}_{2k}$ by $\norm{f}_{2k}$. Interpolation then gives $2 \le p < \infty$, and duality gives $1 < p \le 2$, since $H$ is skew-adjoint. A real-variable proof combines the bound on $L^2$ with a weak type $(1, 1)$ bound, through the Marcinkiewicz interpolation theorem. The best constant is $C_p = \tan\frac{\pi}{2p}$ for $1 < p \le 2$ and $C_p = \cot\frac{\pi}{2p}$ for $2 \le p < \infty$, a result of Pichorides.

Example: for $p = 2$, $H$ is an isometry, since its multiplier $-i\sgn\xi$ has modulus $1$, and $C_2 = \tan(\pi/4) = 1$.

The theorem fails for $p = 1$ and $p = \infty$. For $f = \mathbf{1}_{[0, 1]}$, one has $Hf(x) = \frac{1}{\pi} \log\abs{\frac{x}{x - 1}}$. This is unbounded near $x = 0$ and $x = 1$, and it behaves like $\frac{1}{\pi x}$ at infinity, so it is not integrable.

Going deeper

For a student

The theorem is equivalent to the convergence of Fourier series in $L^p(\R/2\pi\Z)$ for $1 < p < \infty$. Indeed, the partial sum operators are combinations of modulated conjugate function operators, so the bound for $\tilde f$ gives uniform bounds for the partial sums. It is the first case of the theory of singular integrals, which extends it to the Riesz transforms and to Calderón–Zygmund operators.