 Solution

The closest point theorem in a Hilbert space states that every nonempty closed convex subset $C$ of a Hilbert space has a unique point nearest to any given $x$.

Let $d={\inf }_{y\in C}\Vert x-y\Vert$ and choose $y_n\in C$ with $\Vert x-y_n\Vert \to d$. The parallelogram identity and convexity giveso $(y_n)$ is Cauchy. Completeness and closedness give a limit $y\in C$ with $\Vert x-y\Vert =d$. If $y,z$ were both minimizers, the same identity with their midpoint would force $\Vert y-z\Vert =0$.

Apply the theorem to a closed subspace $F$. For $x\in H$, let $y\in F$ be closest and put $z=x-y$. For every $u\in F$, minimality of $\Vert z-tu{\Vert }^2$ at $t=0$ for real and imaginary $t$ gives $⟨z,u⟩=0$. HenceThe intersection is zero, so $H=F\oplus F^{\perp }$.

If $T$ is a shift with orthonormal basis $(e_n)$, it is an isometry,Conversely, suppose these three properties hold. Choose a unit vector $e_1$ spanning $(\mathrm{Im}T{)}^{\perp }$ and set $e_n=T^{n-1}{e}_1$. Isometry makes this sequence orthonormal. Iterating the orthogonal decompositiongivesA vector orthogonal to every $e_n$ lies in every $\mathrm{Im}(T^n)$ and is therefore zero. Thus $(e_n)$ is an orthonormal basis and $T{e}_n=e_{n+1}$, so $T$ is a shift. This is the wandering-vector characterization of a unilateral shift.