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 = in f_{y \in C} ∥ x - y ∥

and choose

y_{n} \in C

with

∥ x - y_{n} ∥ \to d

. The parallelogram identity and convexity give

∥ y_{n} - y_{m} ∥^{2} = 2∥ x - y_{n} ∥^{2} + 2∥ x - y_{m} ∥^{2} - 4 x - \frac{y _{n} + y _{m}}{2}^{2} \leq 2∥ x - y_{n} ∥^{2} + 2∥ x - y_{m} ∥^{2} - 4 d^{2},

(y_{n})

is Cauchy. Completeness and closedness give a limit

y \in C

with

∥ x - y ∥ = d

. If

y, z

were both minimizers, the same identity with their midpoint would force

∥ y - z ∥ = 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

∥ z - t u ∥^{2}

t = 0

for real and imaginary

t

gives

⟨ z, u ⟩ = 0

. Hence

x = y + z \in F + F^{⊥} .

The intersection is zero, so

H = F \oplus F^{⊥}

T

is a shift with orthonormal basis

(e_{n})

, it is an isometry,

(Im T)^{⊥} = span {e_{1}}, ⋂_{n \geq 1} Im (T^{n}) = {0} .

Conversely, suppose these three properties hold. Choose a unit vector

e_{1}

spanning

(Im T)^{⊥}

and set

e_{n} = T^{n - 1} e_{1}

. Isometry makes this sequence orthonormal. Iterating the orthogonal decomposition

H = span {e_{1}} \oplus Im T

gives

H = span {e_{1}, \dots, e_{n}} \oplus Im (T^{n}) .

A vector orthogonal to every

e_{n}

lies in every

Im (T^{n})

and is therefore zero. Thus

(e_{n})

is an orthonormal basis and

T e_{n} = e_{n + 1}

, so

T

Solved by gpt-5.6-sol high.

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