Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 22F (Linear Analysis)

Let $d={\inf }_{v\in V}\Vert f-v\Vert$ and choose $v_n\in V$ with $\Vert f-v_n\Vert \to d$. The parallelogram identity givesCompleteness and closedness give $v_n\to v\in V$, attaining the distance. Put $w=f-v$. For $u\in V$ and real $t$, minimality gives $\Vert w-tu{\Vert }^2\ge \Vert w{\Vert }^2$, so $\mathrm{Re}⟨w,u⟩=0$. In a complex Hilbert space, also replace $u$ with $iu$ to get the imaginary part zero. ThereforeThe decomposition is unique since $V\cap V^{\perp }=\{0\}$. This proves the required orthogonal projection result rather than assuming it before the averaging argument.

For the unitary operator,Thus it is an isometry, and every power has norm one. The triangle inequality gives

The set $X$ is the range of the linear operator $I-U$, hence is a linear subspace. If $f=g-Ug$, the sum telescopes:Since $U$ is an isometry, $\Vert A_{n}f\Vert \le 2\Vert g\Vert /n\to 0$. Hence the averages tend to zero on $X$.

If $v\in \overline{X}$, choose $x\in X$ with $\Vert v-x\Vert <\epsilon$. By part (i),The last term tends to zero by part (ii). Taking the limit superior and then $\epsilon \downarrow 0$ proves $A_{n}v\to 0$ for every $v\in \overline{X}$.

If $w\perp X$, then $⟨w,(I-U)g⟩=0$ for all $g$. The adjoint identity gives $(I-U^{\ast })w=0$, so $U^{\ast }w=w$. Multiplying by $U$ givesSince $X\subseteq V=\overline{X}$, $w\perp V$ implies $w\perp X$ and yields the same result. Conversely a fixed vector is perpendicular to $X$ by this calculation, identifying $V^{\perp }$ as the fixed subspace.

Apply the proved projection result to $V=\overline{X}$ and write $f=v+w$ with $w\perp V$. By part (iii), $A_{n}v\to 0$, while part (iv) gives $U^{k}w=w$ and hence $A_{n}w=w$. ThusThe limit is the orthogonal projection onto the fixed vectors of $U$. This is norm convergence, not just weak convergence.