Example 12 Given $a, b \in \mathbf{R}$ and $a \sqrt{2-2 b^{2}}+b \sqrt{1-2 a^{2}}=1$, find the value of $2 a^{2}+b^{2}$.
By Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(2 a^{2}+b^{2}\right) \cdot\left[\left(1-b^{2}\right)+\left(1-2 a^{2}\right)\right] \\
=\left[(\sqrt{2} a)^{2}+b^{2}\right] \cdot\left[\left(\sqrt{1-b^{2}}\right)^{2}+\left(\sqrt{1-2 a^{2}}\right)^{2}\right] \\
\geqslant\left(\sqrt{2} a \cdot \sqrt{1-b^{2}}+b \cdot \sqrt{1-2 a^{2}}\right)^{2} \\
=\left(a \sqrt{2-2 b^{2}}+b \sqrt{1-2 a^{2}}\right)^{2}=1 .
\end{array}$$

Let $2 a^{2}+b^{2}=t$, then $t(2-t) \geqslant 1$,
which means $(t-1)^{2} \leqslant 0$,
so $t=1$, i.e., $2 a^{2}+b^{2}=1$.