Proof of Nearest-Point Projection onto a Nonempty Closed Convex Subset of a Real Hilbert Space

We use the notation and claims of Elementary Identities in a Real Inner Product Space and The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity; $d(x,y)=|x-y|$ is the metric of $H$. Real-number facts are taken from Elementary Order Arithmetic in an Ordered Field, Elementary Arithmetic in an Ordered Field and Properties of the Absolute Value in an Ordered Field. The proof uses countable choice in claim 1 to select a minimising sequence.

Claim 1. Fix $x\in H$. The set $S=\{|x-y|:y\in K\}$ of real numbers is nonempty since $K$ is, and bounded below by $0$; let $\delta=\inf S$, which exists by Existence of the Infimum of a Nonempty Subset of $\mathbb{R}$ Bounded Below and satisfies $0\le\delta$ (a lower bound is at most the greatest one).

A minimising sequence. For $m\in\mathbb{N}$ put $\eta_{m}=\min\bigl(1,\ (m(2\delta+1))^{-1}\bigr)$, a positive number (Minimum of Two Elements of a Totally Ordered Set; positivity of $m$ and $2\delta+1$ and claim 7 of Elementary Order Arithmetic in an Ordered Field). By claim 4 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ there is $y_{m}\in K$ with $|x-y_{m}|<\delta+\eta_{m}$; fix one such $y_{m}$ for every $m$ (countable choice). Since $0\le|x-y_{m}|$, claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $|x-y_{m}|^{2}\le(\delta+\eta_{m})^{2}=\delta^{2}+\eta_{m}(2\delta+\eta_{m})\le\delta^{2}+\eta_{m}(2\delta+1)\le\delta^{2}+m^{-1}$, using $\eta_{m}\le 1$ and $\eta_{m}\le(m(2\delta+1))^{-1}$. Also $\delta\le|x-y_{m}|$, so $\delta^{2}\le|x-y_{m}|^{2}$.

The sequence is Cauchy. Let $m,n\in\mathbb{N}$ and put $u=x-y_{m}$, $v=x-y_{n}$. By Elementary Identities in a Real Inner Product Space §parallelogram, $|u-v|^{2}=2|u|^{2}+2|v|^{2}-|u+v|^{2}$. Here $u-v=y_{n}-y_{m}$ and $u+v=2\bigl(x-\tfrac12(y_{m}+y_{n})\bigr)$, where $\tfrac12(y_{m}+y_{n})=\tfrac12 y_{m}+(1-\tfrac12)y_{n}\in K$ by convexity; hence $|u+v|^{2}=4\,|x-\tfrac12(y_{m}+y_{n})|^{2}\ge 4\delta^{2}$ by Elementary Identities in a Real Inner Product Space §homogeneity and the definition of $\delta$ as a lower bound of $S$ (squaring by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field). Therefore

We note that for $m,N\in\mathbb{N}$ with $N\le m$ one has $m^{-1}\le N^{-1}$ in $\mathbb{R}$ (natural numbers being read in $\mathbb{R}$ through the canonical map as in The Real Numbers and Standard Notation): $0<N$ and $0<m$ in $\mathbb{R}$ by claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, and $N\le m$ holds in $\mathbb{R}$ by claim 6 there when $N\ne m$ and trivially when $N=m$; both inverses exist and are positive by claim 7 of Elementary Order Arithmetic in an Ordered Field, and multiplying $N\le m$ by the nonnegative number $m^{-1}N^{-1}$ (claim 5 of Elementary Arithmetic in an Ordered Field) gives $m^{-1}=m^{-1}N^{-1}N\le m^{-1}N^{-1}m=N^{-1}$. Given $\varepsilon>0$, choose by claim 3 of The Archimedean Property of the Real Numbers an $N\in\mathbb{N}$ with $N^{-1}<\varepsilon^{2}/4$; for $m,n\ge N$ we have $m^{-1}\le N^{-1}$ and $n^{-1}\le N^{-1}$, so $|y_{n}-y_{m}|^{2}<\varepsilon^{2}$ and hence $|y_{n}-y_{m}|<\varepsilon$ by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. Thus $(y_{m})$ is a Cauchy sequence in $(H,d)$.

Its limit is a nearest point. Since $(H,d)$ is complete by Real Hilbert Space §hilbert, $(y_{m})$ converges to some $z\in H$, and $z\in K$ because $K$ is closed, by Sequential Characterization of Closed Subsets of a Metric Space. By The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §linear-limits and The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §continuity, the real sequence $(|x-y_{m}|)$ converges to $|x-z|$, hence $(|x-y_{m}|^{2})$ converges to $|x-z|^{2}$ by claim 2 of Arithmetic of Limits of Real Sequences. The real sequence $(m^{-1})$ converges to $0$: given $\varepsilon>0$, claim 3 of The Archimedean Property of the Real Numbers gives $N$ with $0<N^{-1}<\varepsilon$, and $|m^{-1}-0|=m^{-1}\le N^{-1}<\varepsilon$ for $m\ge N$ by the monotonicity noted above. Hence $(\delta^{2}+m^{-1})$ converges to $\delta^{2}$ by claim 1 of Arithmetic of Limits of Real Sequences (the constant sequence $(\delta^{2})$ converging to $\delta^{2}$). Since $\delta^{2}\le|x-y_{m}|^{2}\le\delta^{2}+m^{-1}$ for all $m$, claim 2 of Order Properties of Limits of Real Sequences shows that $(|x-y_{m}|^{2})$ converges to $\delta^{2}$, and by claim 1 of Uniqueness of Limits and Boundedness of Convergent Real Sequences its limit $|x-z|^{2}$ equals $\delta^{2}$; then $|x-z|=\delta$ by claim 3 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, both numbers being nonnegative. As $\delta$ is a lower bound of $S$, $|x-z|\le|x-y|$ for every $y\in K$.

Uniqueness. Let $z,z'\in K$ both satisfy the nearest-point property. Any nearest point $z$ satisfies $|x-z|\le|x-y|$ for all $y\in K$, so $|x-z|$ is a lower bound of $S$ that belongs to $S$; being a lower bound it is at most $\delta$, and being an element of $S$ it is at least $\delta$, so $|x-z|=\delta$, and likewise $|x-z'|=\delta$. The parallelogram law with $u=x-z$, $v=x-z'$ and convexity give, exactly as above, $|z-z'|^{2}=2\delta^{2}+2\delta^{2}-4|x-\tfrac12(z+z')|^{2}\le 4\delta^{2}-4\delta^{2}=0$; together with $0\le|z-z'|^{2}$ this gives $|z-z'|^{2}=0=0^{2}$, hence $|z-z'|=0$ by claim 3 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, and $z=z'$ by Elementary Identities in a Real Inner Product Space §vanishing. We denote the unique nearest point by $P_{K}x$.

Claim 2. Let $x\in H$ and $z\in K$.

Suppose first that $z=P_{K}x$, and let $y\in K$. For real $t$ with $0<t\le 1$ the point $y_{t}=z+t(y-z)=t\,y+(1-t)\,z$ lies in $K$ by convexity, so $|x-z|^{2}\le|x-y_{t}|^{2}$ (squaring the nearest-point inequality). By Elementary Identities in a Real Inner Product Space §expansion, Elementary Identities in a Real Inner Product Space §bilinear and Elementary Identities in a Real Inner Product Space §homogeneity,

so $2t\langle x-z,y-z\rangle\le t^{2}|y-z|^{2}$, and dividing by $2t>0$ (claim 5 of Elementary Arithmetic in an Ordered Field with the positive multiplier $(2t)^{-1}$) yields $\langle x-z,y-z\rangle\le\tfrac{t}{2}|y-z|^{2}$ for every $t$ with $0<t\le 1$. If we had $\langle x-z,y-z\rangle>0$, then choosing such a $t$ with $\tfrac{t}{2}|y-z|^{2}<\langle x-z,y-z\rangle$ (possible by claim 3 of The Archimedean Property of the Real Numbers applied to $\varepsilon=\langle x-z,y-z\rangle/(|y-z|^{2}+1)$ and $t=n^{-1}$, since $\tfrac{t}{2}|y-z|^{2}\le t(|y-z|^{2}+1)$) would contradict this. Hence $\langle x-z,y-z\rangle\le 0$.

Conversely suppose $\langle x-z,y-z\rangle\le 0$ for every $y\in K$. For $y\in K$, by Elementary Identities in a Real Inner Product Space §expansion,

so $|x-z|\le|x-y|$ by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. Thus $z$ has the nearest-point property, and $z=P_{K}x$ by the uniqueness in claim 1.

Claim 3. If $x\in K$, then $|x-x|=0\le|x-y|$ for every $y\in K$, so $x$ has the nearest-point property and $P_{K}x=x$ by uniqueness. Conversely $P_{K}x\in K$ by definition.

Claim 4. Let $x,x'\in H$, $z=P_{K}x$ and $z'=P_{K}x'$. By claim 2 applied to $x$ with $y=z'$ and to $x'$ with $y=z$: $\langle x-z,z'-z\rangle\le 0$ and $\langle x'-z',z-z'\rangle\le 0$. Since $z-z'=-(z'-z)$, the second reads $-\langle x'-z',z'-z\rangle\le 0$ by Elementary Identities in a Real Inner Product Space §bilinear. Adding the two inequalities (compatibility of the order with addition, an axiom of Ordered Field, applied twice, and transitivity) and using bilinearity,

where $|z'-z|=|z-z'|$ by Elementary Identities in a Real Inner Product Space §homogeneity. Hence $|z-z'|^{2}\le-\langle x-x',z'-z\rangle=\langle x-x',z-z'\rangle\le|x-x'|\,|z-z'|$ by claim 3 of Properties of the Absolute Value in an Ordered Field and The Cauchy-Schwarz Inequality in a Real Inner Product Space. If $|z-z'|=0$ the claim holds since $0\le|x-x'|$. Otherwise $|z-z'|>0$ and multiplying by its inverse (claim 5 of Elementary Arithmetic in an Ordered Field) gives $|z-z'|\le|x-x'|$.