[proofplan]
We specialise the [Fefferman--Stein Riesz Transform Characterisation](/theorems/???) to dimension $n = 1$. In one dimension there is a single Riesz transform, which up to a fixed multiplicative constant equals the Hilbert transform $H$. We make this identification explicit at the level of Fourier multipliers, transfer the equivalence of norms through the multiplier identity, and confirm that the weak-type extension of $H$ from $L^1$ used in the statement coincides with the $H^1 \to L^1$ extension produced by the Fefferman--Stein theorem. The implication (1) $\Rightarrow$ (2) follows because $H$ is a Calderón--Zygmund operator and $H^1 \hookrightarrow L^1$. The converse follows from the multiplier identification together with the Fefferman--Stein characterisation, applied to the single Riesz transform $R_1$.
[/proofplan]
[step:Identify the one-dimensional Riesz transform with the Hilbert transform]
In dimension $n = 1$, the Riesz transform $R_1: L^2(\mathbb{R}) \to L^2(\mathbb{R})$ is the Fourier multiplier with symbol
\begin{align*}
m_{R_1}: \mathbb{R} \setminus \{0\} &\to \mathbb{C} \\
\xi &\mapsto -i\,\frac{\xi}{|\xi|} = -i\operatorname{sgn}(\xi).
\end{align*}
That is, $\widehat{R_1 f}(\xi) = -i\operatorname{sgn}(\xi)\,\hat{f}(\xi)$ for $f \in L^2(\mathbb{R})$.
The Hilbert transform $H: L^2(\mathbb{R}) \to L^2(\mathbb{R})$ is the Fourier multiplier with symbol
\begin{align*}
m_H: \mathbb{R} \setminus \{0\} &\to \mathbb{C} \\
\xi &\mapsto -i\operatorname{sgn}(\xi).
\end{align*}
The two symbols agree pointwise on $\mathbb{R} \setminus \{0\}$, so $R_1 f = H f$ for every $f \in L^2(\mathbb{R})$. By density of $L^1 \cap L^2$ in $L^1$ and the fact that both operators extend to $L^1 \to L^{1,\infty}$ as Calderón--Zygmund operators (see the [weak (1,1) bound for Calderón--Zygmund operators](/theorems/???)), the identification $R_1 = H$ persists at the level of weak-type extensions on $L^1(\mathbb{R})$.
[/step]
[step:Derive (1) $\Rightarrow$ (2) from the $H^1 \to L^1$ boundedness of the Hilbert transform]
Suppose $f \in H^1(\mathbb{R})$. By the [embedding $H^1(\mathbb{R}^n) \hookrightarrow L^1(\mathbb{R}^n)$](/theorems/???), $f \in L^1(\mathbb{R})$ with $\|f\|_{L^1} \le \|f\|_{H^1}$.
The Hilbert transform $H$ is a Calderón--Zygmund operator on $L^2(\mathbb{R})$: it is $L^2$-bounded by Plancherel (its multiplier $-i\operatorname{sgn}(\xi)$ has modulus $1$), and its kernel $K(x,y) = \frac{1}{\pi(x-y)}$ satisfies the size condition $|K(x,y)| \le \frac{1}{\pi}|x-y|^{-1}$ and the Hörmander smoothness condition. Applying the [$H^1 \to L^1$ boundedness of Calderón--Zygmund operators](/theorems/3178) gives
\begin{align*}
\|Hf\|_{L^1(\mathbb{R})} \le C\,\|f\|_{H^1(\mathbb{R})}
\end{align*}
for an absolute constant $C > 0$. Combining the two bounds,
\begin{align*}
\|f\|_{L^1(\mathbb{R})} + \|Hf\|_{L^1(\mathbb{R})} \le (1 + C)\,\|f\|_{H^1(\mathbb{R})}.
\end{align*}
[/step]
[step:Derive (2) $\Rightarrow$ (1) from the Fefferman--Stein Riesz characterisation specialised to $n = 1$]
Suppose $f \in L^1(\mathbb{R})$ and $Hf \in L^1(\mathbb{R})$. By the identification from the first step, $R_1 f = Hf$, so $R_1 f \in L^1(\mathbb{R})$.
The [Fefferman--Stein Riesz Transform Characterisation](/theorems/???) states that for $f \in L^1(\mathbb{R}^n)$,
\begin{align*}
f \in H^1(\mathbb{R}^n) \iff R_j f \in L^1(\mathbb{R}^n) \text{ for all } j = 1, \ldots, n,
\end{align*}
with $\|f\|_{H^1} \asymp \|f\|_{L^1} + \sum_{j=1}^n \|R_j f\|_{L^1}$. Setting $n = 1$, the right-hand side is the single condition $R_1 f \in L^1$, and the norm equivalence becomes
\begin{align*}
\|f\|_{H^1(\mathbb{R})} \asymp \|f\|_{L^1(\mathbb{R})} + \|R_1 f\|_{L^1(\mathbb{R})} = \|f\|_{L^1(\mathbb{R})} + \|Hf\|_{L^1(\mathbb{R})}.
\end{align*}
The hypothesis $f \in L^1$ and $R_1 f = Hf \in L^1$ therefore implies $f \in H^1(\mathbb{R})$, completing the equivalence. The two-sided estimate $\|f\|_{H^1(\mathbb{R})} \asymp \|f\|_{L^1(\mathbb{R})} + \|Hf\|_{L^1(\mathbb{R})}$ holds with absolute implicit constants.
[/step]
[step:Combine the two directions and state the norm equivalence]
The first direction ((1) $\Rightarrow$ (2)) gives, for any $f \in H^1(\mathbb{R})$,
\begin{align*}
\|f\|_{L^1(\mathbb{R})} + \|Hf\|_{L^1(\mathbb{R})} \le (1 + C)\,\|f\|_{H^1(\mathbb{R})}.
\end{align*}
The second direction ((2) $\Rightarrow$ (1)) gives, for any $f \in L^1(\mathbb{R})$ with $Hf \in L^1(\mathbb{R})$, that $f \in H^1(\mathbb{R})$ and
\begin{align*}
\|f\|_{H^1(\mathbb{R})} \le C'\,(\|f\|_{L^1(\mathbb{R})} + \|Hf\|_{L^1(\mathbb{R})}),
\end{align*}
where $C' > 0$ is the implicit upper constant from the Fefferman--Stein theorem in dimension one. Together,
\begin{align*}
\|f\|_{H^1(\mathbb{R})} \asymp \|f\|_{L^1(\mathbb{R})} + \|Hf\|_{L^1(\mathbb{R})},
\end{align*}
which is the claimed two-sided equivalence with absolute constants. This completes the proof.
[/step]
