= Solution
{c}
For any <orthonormal basis> $f,g$ of $U$, <bilinearity> gives
$$
\begin{aligned}
\langle f,h\rangle
&=\langle f,x^2\rangle-\langle f,x^2\rangle\langle f,f\rangle-\langle g,x^2\rangle\langle f,g\rangle=0,\\
\langle g,h\rangle
&=\langle g,x^2\rangle-\langle f,x^2\rangle\langle g,f\rangle-\langle g,x^2\rangle\langle g,g\rangle=0.
\end{aligned}
$$
Together with $\langle f,g\rangle=0$, these identities prove \b[pairwise <orthogonality>]. The construction subtracts the <orthogonal projection onto a finite-dimensional subspace> $U$ from $x^2$, leaving the <orthogonal projection residual>.
Back to article page
