[proofplan]
We deduce Hausdorff–Young from the [Riesz–Thorin Interpolation Theorem](/theorems/???) by interpolating between two endpoint bounds for the Fourier transform in the symmetric normalisation: the $L^1 \to L^\infty$ bound with constant $(2\pi)^{-n/2}$, and the $L^2 \to L^2$ Plancherel isometry with constant $1$. Riesz–Thorin yields constant $(2\pi)^{-n(1-\theta)/2}$ on the interpolated line, which simplifies to $(2\pi)^{n(1/p' - 1/2)}$.
[/proofplan]
[step:Record the two endpoint operator norms under the symmetric normalisation]
Throughout this proof we use the symmetric normalisation
\begin{align*}
\mathcal{F}: \mathcal{S}(\mathbb{R}^n) &\to \mathcal{S}(\mathbb{R}^n), \qquad \hat{f}(\xi) := \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n} f(x)\, e^{-ix \cdot \xi}\, d\mathcal{L}^n(x).
\end{align*}
The two endpoint norms are:
\begin{align*}
\mathcal{F}: \quad &\|\hat{f}\|_{L^\infty} \le (2\pi)^{-n/2}\|f\|_{L^1} \quad (M_0 = (2\pi)^{-n/2}), \\
&\|\hat{f}\|_{L^2} = \|f\|_{L^2} \quad (M_1 = 1).
\end{align*}
Riesz–Thorin produces the interpolated constant $M_0^{1-\theta} M_1^\theta = (2\pi)^{-n(1-\theta)/2}$ from these two endpoint norms. We use the Lebesgue measure $\mathcal{L}^n$ on $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$, which is $\sigma$-finite — a hypothesis of the [Riesz–Thorin Interpolation Theorem](/theorems/???).
[/step]
[step:Establish the endpoint bound $\|\hat{f}\|_{L^\infty} \le (2\pi)^{-n/2}\|f\|_{L^1}$]
For $f \in L^1(\mathbb{R}^n)$, the integral defining $\hat{f}(\xi)$ is absolutely convergent for every $\xi \in \mathbb{R}^n$, since $|f(x)\, e^{-ix \cdot \xi}| = |f(x)|$ is integrable. Estimating directly,
\begin{align*}
|\hat{f}(\xi)| = \frac{1}{(2\pi)^{n/2}}\left| \int_{\mathbb{R}^n} f(x)\, e^{-ix \cdot \xi}\, d\mathcal{L}^n(x) \right| \le \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n} |f(x)|\, d\mathcal{L}^n(x) = (2\pi)^{-n/2}\|f\|_{L^1}.
\end{align*}
Taking the supremum over $\xi \in \mathbb{R}^n$, $\|\hat{f}\|_{L^\infty} \le (2\pi)^{-n/2}\|f\|_{L^1}$, so $M_0 = (2\pi)^{-n/2}$.
[/step]
[step:Establish the endpoint bound $\|\hat{f}\|_{L^2} = \|f\|_{L^2}$ via Plancherel]
By the [Plancherel Theorem](/theorems/???) under the symmetric normalisation, the Fourier transform extends from $\mathcal{S}(\mathbb{R}^n)$ to a unitary operator $\mathcal{F}: L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$ satisfying
\begin{align*}
\|\hat{f}\|_{L^2} = \|f\|_{L^2} \qquad \text{for all } f \in L^2(\mathbb{R}^n).
\end{align*}
The operator norm is $M_1 = 1$.
[/step]
[step:Compute the interpolated exponents and verify they cover $1 \le p \le 2$]
The [Riesz–Thorin Interpolation Theorem](/theorems/???) produces, for each $\theta \in (0, 1)$, an exponent pair $(p_\theta, q_\theta)$ defined by
\begin{align*}
\frac{1}{p_\theta} = \frac{1 - \theta}{p_0} + \frac{\theta}{p_1} = \frac{1 - \theta}{1} + \frac{\theta}{2} = 1 - \frac{\theta}{2},
\end{align*}
\begin{align*}
\frac{1}{q_\theta} = \frac{1 - \theta}{q_0} + \frac{\theta}{q_1} = \frac{1 - \theta}{\infty} + \frac{\theta}{2} = \frac{\theta}{2}.
\end{align*}
Adding,
\begin{align*}
\frac{1}{p_\theta} + \frac{1}{q_\theta} = \left( 1 - \frac{\theta}{2} \right) + \frac{\theta}{2} = 1,
\end{align*}
so $q_\theta = p_\theta'$, the Hölder conjugate of $p_\theta$, for every $\theta \in (0, 1)$.
As $\theta$ ranges over $(0, 1)$, $1/p_\theta = 1 - \theta/2$ ranges over $(1/2, 1)$, hence $p_\theta$ ranges over $(1, 2)$. Together with the endpoint cases $p = 1$ and $p = 2$, this exhausts $1 \le p \le 2$.
[/step]
[step:Apply Riesz–Thorin to obtain the interpolated bound]
The [Riesz–Thorin Interpolation Theorem](/theorems/???) with the strong-type bounds $M_0 = (2\pi)^{-n/2}$ at $(p_0, q_0) = (1, \infty)$ and $M_1 = 1$ at $(p_1, q_1) = (2, 2)$ gives, for each $\theta \in (0, 1)$,
\begin{align*}
\|\hat{f}\|_{L^{q_\theta}} \le M_0^{1 - \theta}\, M_1^\theta\, \|f\|_{L^{p_\theta}} = (2\pi)^{-n(1-\theta)/2}\, \|f\|_{L^{p_\theta}}
\end{align*}
for all $f \in L^{p_\theta}(\mathbb{R}^n)$. Since $q_\theta = p_\theta'$ and $1 - \theta = 2/p_\theta - 1 = 2(1 - 1/p_\theta') - 1 = 1 - 2/p_\theta'$, the exponent is
\begin{align*}
-\frac{n(1-\theta)}{2} = -\frac{n(1 - 2/p_\theta')}{2} = \frac{n}{p_\theta'} - \frac{n}{2} = n\!\left(\frac{1}{p'} - \frac{1}{2}\right).
\end{align*}
Thus
\begin{align*}
\|\hat{f}\|_{L^{p'}} \le (2\pi)^{n(1/p' - 1/2)}\, \|f\|_{L^p}
\end{align*}
for $p = p_\theta \in (1, 2)$, and for the endpoints $p = 1$ and $p = 2$ by the direct bounds of steps 2 and 3.
[/step]
[step:Extend to all of $L^p$ via density]
For $f \in L^p(\mathbb{R}^n)$ with $1 \le p \le 2$, write $f = f_1 + f_2$ with
\begin{align*}
f_1 := f \cdot \mathbb{1}_{\{|f| > 1\}}, \qquad f_2 := f \cdot \mathbb{1}_{\{|f| \le 1\}}.
\end{align*}
Then $f_1 \in L^1$ and $f_2 \in L^2$: on $\{|f| > 1\}$, $|f_1| \le |f|^p$ pointwise (since $|f| \ge 1$ implies $|f|^p \ge |f|$ when $p \ge 1$), giving $\|f_1\|_{L^1} \le \|f\|_{L^p}^p$; on $\{|f| \le 1\}$, $|f_2|^2 \le |f|^p$ pointwise (since $|f| \le 1$ and $p \le 2$ imply $|f|^2 \le |f|^p$), giving $\|f_2\|_{L^2}^2 \le \|f\|_{L^p}^p$. Define $\hat{f}(\xi) := \hat{f}_1(\xi) + \hat{f}_2(\xi)$, where $\hat{f}_1$ is given by the absolutely convergent integral and $\hat{f}_2$ is the Plancherel extension. This definition is independent of the decomposition.
Take a sequence $g_k \in \mathcal{S}(\mathbb{R}^n)$ with $g_k \to f$ in $L^p$. The bound from step 5 applied to $g_j - g_k$ gives $\|\hat{g}_j - \hat{g}_k\|_{L^{p'}} \le (2\pi)^{n(1/p' - 1/2)}\|g_j - g_k\|_{L^p} \to 0$, so $(\hat{g}_k)$ is Cauchy in $L^{p'}$. Its limit equals $\hat{f}$. Passing the inequality to the limit,
\begin{align*}
\|\hat{f}\|_{L^{p'}} \le (2\pi)^{n(1/p' - 1/2)}\,\|f\|_{L^p}.
\end{align*}
[/step]
