= Solution
We prove the estimate first for $u$ in the <Schwartz space>. By the <Fourier inversion theorem> and the <Cauchy-Schwarz inequality>,
$$
\begin{aligned}
|u(x)|
&\leq C\int_{\mathbb R^n}|\widehat u(\xi)|\,d\xi\\
&\leq C
\left(\int_{\mathbb R^n}(1+|\xi|^2)^{-n}\,d\xi\right)^{1/2}
\lVert u\rVert_{H^n}.
\end{aligned}
$$
The integral is finite because $n>n/2$.
For $0<\alpha\leq1$, the elementary bound
$$
|e^{it}-1|\leq C_\alpha|t|^\alpha
$$
gives
$$
\begin{aligned}
|u(x)-u(y)|
&\leq C|x-y|^\alpha
\int_{\mathbb R^n}|\xi|^\alpha|\widehat u(\xi)|\,d\xi\\
&\leq C|x-y|^\alpha
\left(\int_{\mathbb R^n}
|\xi|^{2\alpha}(1+|\xi|^2)^{-n}\,d\xi\right)^{1/2}
\lVert u\rVert_{H^n}.
\end{aligned}
$$
The last integral is finite near zero for every $\alpha>0$ and at infinity exactly when $2\alpha<n$. Choose, for example, $\alpha=1/4$, which works for every $n\geq1$. We obtain
$$
\lVert u\rVert_{C^{0,1/4}}
\leq C_n\lVert u\rVert_{H^n}.
$$
Density of the Schwartz space in $H^n$ extends the estimate and supplies a unique Hölder-continuous representative. Therefore
$$
H^n(\mathbb R^n)\hookrightarrow C^{0,1/4}(\mathbb R^n)
$$
continuously. This is the <Fourier proof of Hölder regularity from a Sobolev norm>.
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