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The closest point theorem in a Hilbert space says that, for every nonempty closed convex set $C\subset H$ and $x\in H$, there is exactly one $p\in C$ minimizing $\Vert x-p\Vert$.

Put $d={\inf }_{y\in C}\Vert x-y\Vert$ and choose $y_n\in C$ with $\Vert x-y_n\Vert \to d$. The midpoint belongs to $C$ because it is a convex set. The parallelogram law givesThus $(y_n)$ is a Cauchy sequence. Completeness of the Hilbert space and closedness of $C$ give a limit $p\in C$, with $\Vert x-p\Vert =d$. Applying the same identity to two minimizers gives their squared distance at most zero, proving uniqueness.

The resulting projection is characterized byIndeed, differentiate $\Vert x-p-t(z-p){\Vert }^2$ at $t=0^{+}$; the minimum there gives the inequality. Conversely, expanding $\Vert x-z{\Vert }^2$ proves minimality from this inequality. For a closed linear subspace, both signs of each direction are allowed, so $x-p$ is orthogonal to that subspace: this recovers the orthogonal projection.

Use the Hilbert space interpretation of a linear inverse problem, with a bounded linear operator $K:\mathcal{U}\to \mathcal{V}$. A least-squares solution minimizes $\Vert Ku-f{\Vert }_{\mathcal{V}}$ over $u\in \mathcal{U}$. A minimum-norm least-squares solution additionally has the smallest $\mathcal{U}$-norm among all least-squares solutions. When the equation is consistent this is the smallest-norm exact solution.

The least-squares solutions form a nonempty closed affine subspace precisely whenThe closest point theorem in a Hilbert space then gives a unique nearest point to zero, hence a unique minimum-norm least-squares solution. Equivalently, it is the unique least-squares solution in $\mathcal{N}(K{)}^{\perp }$. If $\mathcal{R}(K)$ is closed, this exists for every datum; a nonclosed operator range can leave some data without any least-squares solution.

For example, on the l2 sequence space, take $(Ku{)}_n=u_n/n$ and $f_n=1/n$. The equation would require $u_n=1$ for every $n$, which is not square summable. However, setting the first $N$ coordinates of $u$ to one and the rest to zero givesThus the least-squares infimum is zero but is not attained, and no minimum-norm least-squares solution exists for these data.

Write ${\iota }_C$ for the indicator functional of a constraint set $C$ and $q(u)=\Vert u{\Vert }^2/2$. ThenBoth terms are convex, hence their infimal convolution is the convex squared distance to a convex set. The closest point theorem in a Hilbert space supplies the unique metric projection onto a closed convex set $P_{C}u$ and gives $d_C(u)=\Vert u-P_{C}u{\Vert }^2/2$.

Identify the Hilbert space with its dual space using the Riesz representation theorem. Completing the square yields $q^{\ast }(p)=\Vert p{\Vert }^2/2$, while ${\iota }_C^{\ast }(p)={\sup }_{v\in C}⟨p,v⟩={\sigma }_C(p)$ is the support function. The conjugate of the squared distance to a convex set is thereforeFor the closed unit ball, the metric projection onto a closed convex set is $P_{C}u=u$ if $\Vert u\Vert \le 1$, and $P_{C}u=u/\Vert u\Vert$ otherwise. ThereforeThe last equality uses the Cauchy-Schwarz inequality to compute the unit ball's support function, attained in the direction of $p$ when $p\ne 0$.

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.