Example 7 Find the largest real number $c$ such that for any $n \in \mathbf{N}^{*}$, we have $\{\sqrt{2} n\}>\frac{c}{n}$.


Notice that, for any $n \in \mathbf{N}^{*}$, let $k=[\sqrt{2} n]$, then
$$\begin{aligned}
\{\sqrt{2} n\} & =\sqrt{2} n-[\sqrt{2} n]=\sqrt{2} n-k=\frac{2 n^{2}-k^{2}}{\sqrt{2} n+k} \\
& \geqslant \frac{1}{\sqrt{2} n+k}>\frac{1}{\sqrt{2} n+\sqrt{2} n}=\frac{1}{2 \sqrt{2} n}
\end{aligned}$$

Here we use the fact that $\sqrt{2}$ is irrational, so $\{\sqrt{2} n\}>0$. Thus, $2 n^{2}-k^{2}>0$ (of course, $2 n^{2}-k^{2} \geqslant 1$), and $k<\sqrt{2} n$. Therefore, $c \geqslant \frac{1}{2 \sqrt{2}}$.

On the other hand, for all positive integer solutions $\left(k_{m}, n_{m}\right)$ of the Pell equation $x^{2}-2 y^{2}=-1$, we have
$$n_{m}\left\{\sqrt{2} n_{m}\right\}=\frac{n_{m}}{\sqrt{2} n_{m}+k_{m}}=\frac{n_{m}}{\sqrt{2} n_{m}+\sqrt{1+2 n_{m}^{2}}} \rightarrow \frac{2}{2 \sqrt{2}}$$

Here the fundamental solution is $\left(k_{1}, n_{1}\right)=(1,1)$, and
$$k_{m}+n_{m} \sqrt{2}=(1+\sqrt{2})^{2 m+1}$$

The above limit holds as $m \rightarrow+\infty$, so $c \leqslant \frac{1}{2 \sqrt{2}}$.
In conclusion, the maximum real number $c=\frac{1}{2 \sqrt{2}}$.