The defining relation for $P_{\mathcal U}$ gives
$$
(f-P_{\mathcal U}f,v)=0
\qquad\forall v\in\mathcal U_n.
$$
Part (a) therefore shows that $P_{\mathcal U}f$ is the best approximation. Moreover, $P_{\mathcal U}f$ and $f-P_{\mathcal U}f$ are orthogonal, so
$$
\|f\|^2=\|P_{\mathcal U}f\|^2
+\|f-P_{\mathcal U}f\|^2.
$$
Thus <orthogonal projection> is a contraction:
$$
\boxed{\|P_{\mathcal U}f\|\leq\|f\|}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 318 / 5 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-318/5/1/b/solution