[proofplan]
The Hilbert transform on Schwartz functions is the Fourier multiplier with symbol $m(\xi) = -i\,\mathrm{sgn}(\xi)$. We first verify that $|m(\xi)| = 1$ almost everywhere with respect to $\mathcal{L}^1$, so multiplication by $m$ is an isometry on $L^2(\mathbb{R})$. Combined with the fact that the Fourier transform in the symmetric normalisation is a unitary operator on $L^2(\mathbb{R})$ (Plancherel's theorem), this gives the $L^2$ identity on the dense subspace $\mathcal{S}(\mathbb{R})$. We then extend $H$ uniquely from $\mathcal{S}(\mathbb{R})$ to $L^2(\mathbb{R})$ by the [extension of bounded linear operators by density](/theorems/???), preserving the isometry property.
[/proofplan]
[step:Define the Hilbert transform on Schwartz functions via its Fourier symbol]
For $f \in \mathcal{S}(\mathbb{R})$, the Hilbert transform is the unique tempered distribution $Hf$ characterised by the Fourier multiplier identity
\begin{align*}
\widehat{Hf}(\xi) = m(\xi)\,\hat{f}(\xi), \qquad m: \mathbb{R} \setminus \{0\} \to \mathbb{C}, \quad m(\xi) = -i\,\mathrm{sgn}(\xi).
\end{align*}
Since $\hat{f} \in \mathcal{S}(\mathbb{R})$ and $m \in L^\infty(\mathbb{R})$ with $\|m\|_{L^\infty} = 1$, the product $m \hat{f}$ lies in $L^2(\mathbb{R})$, and the inverse Fourier transform $Hf := \mathcal{F}^{-1}(m \hat{f})$ is a well-defined element of $L^2(\mathbb{R})$.
[/step]
[step:Compute $\|Hf\|_{L^2}^2$ on Schwartz functions via Plancherel]
We use the [Plancherel identity](/theorems/???), which under the symmetric normalisation states $\|\hat{g}\|_{L^2} = \|g\|_{L^2}$ for every $g \in L^2(\mathbb{R})$. Applied to $g = Hf \in L^2(\mathbb{R})$,
\begin{align*}
\|Hf\|_{L^2}^2 = \|\widehat{Hf}\|_{L^2}^2 = \|m\,\hat{f}\|_{L^2}^2 = \int_{\mathbb{R}} |m(\xi)|^2\,|\hat{f}(\xi)|^2\,d\mathcal{L}^1(\xi).
\end{align*}
For all $\xi \neq 0$ we have $|m(\xi)| = |-i\,\mathrm{sgn}(\xi)| = 1$, and the singleton $\{0\}$ has $\mathcal{L}^1$-measure zero, so $|m|^2 = 1$ $\mathcal{L}^1$-almost everywhere. Therefore
\begin{align*}
\|Hf\|_{L^2}^2 = \int_{\mathbb{R}} |\hat{f}(\xi)|^2\,d\mathcal{L}^1(\xi) = \|\hat{f}\|_{L^2}^2 = \|f\|_{L^2}^2,
\end{align*}
where the last equality is Plancherel applied to $f$. Taking square roots, $\|Hf\|_{L^2} = \|f\|_{L^2}$ for every $f \in \mathcal{S}(\mathbb{R})$.
[/step]
[step:Extend $H$ from $\mathcal{S}(\mathbb{R})$ to $L^2(\mathbb{R})$ by density]
The map
\begin{align*}
H: \mathcal{S}(\mathbb{R}) &\to L^2(\mathbb{R}) \\
f &\mapsto \mathcal{F}^{-1}(m\,\hat{f})
\end{align*}
is linear, and by Step 2 it satisfies $\|Hf\|_{L^2} = \|f\|_{L^2}$ for every $f \in \mathcal{S}(\mathbb{R})$. In particular it is bounded with operator norm $1$ from $(\mathcal{S}(\mathbb{R}), \|\cdot\|_{L^2})$ to $L^2(\mathbb{R})$.
The [density of Schwartz functions in $L^2$](/theorems/???) gives that $\mathcal{S}(\mathbb{R})$ is dense in $L^2(\mathbb{R})$. By the [bounded linear extension theorem](/theorems/???), every bounded linear map from a dense subspace of a normed space into a Banach space admits a unique bounded linear extension to the whole space, with the same operator norm. We apply this with the dense subspace $\mathcal{S}(\mathbb{R}) \subset L^2(\mathbb{R})$ and the Banach space $L^2(\mathbb{R})$ to obtain a unique bounded linear extension
\begin{align*}
H: L^2(\mathbb{R}) \to L^2(\mathbb{R})
\end{align*}
with $\|H\|_{\mathcal{L}(L^2(\mathbb{R}))} = 1$.
[/step]
[step:Conclude that the extension is an isometry on $L^2(\mathbb{R})$]
Let $f \in L^2(\mathbb{R})$. By [density of Schwartz functions in $L^2$](/theorems/???), there is a sequence $f_n \in \mathcal{S}(\mathbb{R})$ with $f_n \to f$ in $L^2(\mathbb{R})$. By Step 3, $H$ is bounded on $L^2(\mathbb{R})$, so $Hf_n \to Hf$ in $L^2(\mathbb{R})$. The norm $\|\cdot\|_{L^2}$ is continuous on $L^2(\mathbb{R})$. Therefore
\begin{align*}
\|Hf\|_{L^2} = \lim_{n \to \infty} \|Hf_n\|_{L^2} = \lim_{n \to \infty} \|f_n\|_{L^2} = \|f\|_{L^2},
\end{align*}
where the middle equality is the isometry on $\mathcal{S}(\mathbb{R})$ from Step 2. This is the desired identity $\|Hf\|_{L^2} = \|f\|_{L^2}$ for all $f \in L^2(\mathbb{R})$, proving that $H: L^2(\mathbb{R}) \to L^2(\mathbb{R})$ is an isometry.
[/step]
