Discussion
Let \(V\) be a closed subspace of \(L^{2}[0,1]\) and let \(f, g \in L^{2}[0,1]\) be given by \(f(x) = x\) and \(g(x) = x^{2}\). If \(V^{\perp} = \text{Span}\{f\}\) and \(Pg\) is the orthogonal projection of \(g\) on \(V\), then \((g - Pg)(x),\ x \in [0,1]\), is
Answer: A
\((g - Pg)(x)\) is \(\frac{3}{4}x\).
\(g - Pg\) is the component of \(g\) orthogonal to V, which is the projection of \(g\) onto \(V^{\perp} = \text{Span}\{f\}\):
\[ \begin{aligned} g - Pg &= \frac{\langle g, f \rangle}{\langle f, f \rangle} f \\ \langle g, f \rangle &= \int_{0}^{1} x^{3}\,dx = \tfrac{1}{4} \\ \langle f, f \rangle &= \int_{0}^{1} x^{2}\,dx = \tfrac{1}{3} \end{aligned} \]
So \(g - Pg = \frac{3}{4}x\). \(\frac{1}{4}x\) forgets to divide by \(\langle f, f \rangle\), and the \(x^{2}\) options are multiples of \(g\) rather than of \(f\).
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