In mathematics, the Hilbert projection theorem is a famous result of convex analysis that says that for every vector ${\displaystyle x}$ in a Hilbert space ${\displaystyle H}$ and every nonempty closed convex ${\displaystyle C\subseteq H,}$ there exists a unique vector ${\displaystyle m\in C}$ for which ${\displaystyle \|c-x\|}$ is minimized over the vectors ${\displaystyle c\in C}$; that is, such that ${\displaystyle \|m-x\|\leq \|c-x\|}$ for every ${\displaystyle c\in C.}$

Some intuition for the theorem can be obtained by considering the first order condition of the optimization problem.

Consider a finite dimensional real Hilbert space ${\displaystyle H}$ with a subspace ${\displaystyle C}$ and a point ${\displaystyle x.}$ If ${\displaystyle m\in C}$ is a minimizer or minimum point of the function ${\displaystyle N:C\to \mathbb {R} }$ defined by ${\displaystyle N(c):=\|c-x\|}$ (which is the same as the minimum point of ${\displaystyle c\mapsto \|c-x\|^{2}}$), then derivative must be zero at ${\displaystyle m.}$

In matrix derivative notation: $${\displaystyle {\begin{aligned}\partial \lVert x-c\rVert ^{2}&=\partial \langle c-x,c-x\rangle \\&=2\langle c-x,\partial c\rangle \end{aligned}}}$$ Since ${\displaystyle \partial c}$ is a vector in ${\displaystyle C}$ that represents an arbitrary tangent direction, it follows that ${\displaystyle m-x}$ must be orthogonal to every vector in ${\displaystyle C.}$

Hilbert projection theorem—For every vector ${\displaystyle x}$ in a Hilbert space ${\displaystyle H}$ and every nonempty closed convex ${\displaystyle C\subseteq H,}$ there exists a unique vector ${\displaystyle m\in C}$ for which ${\displaystyle \lVert x-m\rVert }$ is equal to ${\displaystyle \delta :=\inf _{c\in C}\|x-c\|.}$

If the closed subset ${\displaystyle C}$ is also a vector subspace of ${\displaystyle H}$ then this minimizer ${\displaystyle m}$ is the unique element in ${\displaystyle C}$ such that ${\displaystyle x-m}$ is orthogonal to ${\displaystyle C.}$

Let ${\displaystyle \delta :=\inf _{c\in C}\|x-c\|}$ be the distance between ${\displaystyle x}$ and ${\displaystyle C,}$ ${\displaystyle \left(c_{n}\right)_{n=1}^{\infty }}$ a sequence in ${\displaystyle C}$ such that the distance squared between ${\displaystyle x}$ and ${\displaystyle c_{n}}$ is less than or equal to ${\displaystyle \delta ^{2}+1/n.}$ Let ${\displaystyle n}$ and ${\displaystyle m}$ be two integers, then the following equalities are true: $${\displaystyle \left\|c_{n}-c_{m}\right\|^{2}=\left\|c_{n}-x\right\|^{2}+\left\|c_{m}-x\right\|^{2}-2\left\langle c_{n}-x\,,\,c_{m}-x\right\rangle }$$ and $${\displaystyle 4\left\|{\frac {c_{n}+c_{m}}{2}}-x\right\|^{2}=\left\|c_{n}-x\right\|^{2}+\left\|c_{m}-x\right\|^{2}+2\left\langle c_{n}-x\,,\,c_{m}-x\right\rangle }$$ Therefore $${\displaystyle \left\|c_{n}-c_{m}\right\|^{2}=2\left\|c_{n}-x\right\|^{2}+2\left\|c_{m}-x\right\|^{2}-4\left\|{\frac {c_{n}+c_{m}}{2}}-x\right\|^{2}}$$ (This equation is the same as the formula ${\displaystyle a^{2}=2b^{2}+2c^{2}-4M_{a}^{2}}$ for the length ${\displaystyle M_{a}}$ of a median in a triangle with sides of length ${\displaystyle a,b,}$ and ${\displaystyle c,}$ where specifically, the triangle's vertices are ${\displaystyle x,c_{m},c_{n}}$).

By giving an upper bound to the first two terms of the equality and by noticing that the midpoint of ${\displaystyle c_{n}}$ and ${\displaystyle c_{m}}$ belong to ${\displaystyle C}$ and has therefore a distance greater than or equal to ${\displaystyle \delta }$ from ${\displaystyle x,}$ it follows that: $${\displaystyle \|c_{n}-c_{m}\|^{2}\;\leq \;2\left(\delta ^{2}+{\frac {1}{n}}\right)+2\left(\delta ^{2}+{\frac {1}{m}}\right)-4\delta ^{2}=2\left({\frac {1}{n}}+{\frac {1}{m}}\right)}$$