Real Hardy space H1

Remarks

For $n = 1$ the only Riesz transform is the Hilbert transform $H$, so $H^1(\R) = \set{f \in L^1(\R) : Hf \in L^1(\R)}$. Equivalent definitions use maximal functions or atoms. Fix a Schwartz function $\varphi$ with $\int \varphi = 1$ and let $\varphi_t(x) = t^{-n}\varphi(x/t)$. Then $f \in H^1(\R^n)$ exactly when $f$ is a tempered distribution with $\sup_{t > 0} \abs{f * \varphi_t} \in L^1(\R^n)$, and the two norms are comparable. Every $f \in H^1(\R^n)$ has $\int f = 0$.

Example: $n = 1$ and $a = \mathbf{1}_{[0, 1]} - \mathbf{1}_{[-1, 0]}$. For $\abs{x} > 2$, using $\int a = 0$, $Ha(x) = \frac{1}{\pi} \int a(y) \left( \frac{1}{x - y} - \frac{1}{x} \right) dy$, which is bounded by $C/x^2$. Near the points $-1$, $0$ and $1$, $Ha$ has only logarithmic singularities, which are integrable. So $Ha \in L^1(\R)$ and $a \in H^1(\R)$.

Non-example: $f = \mathbf{1}_{[0, 1]}$ on $\R$ is integrable but not in $H^1(\R)$. Its Hilbert transform is $Hf(x) = \frac{1}{\pi} \log \abs{\frac{x}{x - 1}}$, which behaves like $\frac{1}{\pi x}$ as $\abs{x} \to \infty$, so it is not integrable. This agrees with the fact that $\int f = 1 \ne 0$.