Proof of Nearest-Point Projection onto a Nonempty Closed Convex Subset of the Lebesgue Space of Square-Integrable Vector-Valued Functions

Throughout, $\lVert\cdot\rVert$ abbreviates $\lVert\cdot\rVert_{L^{2}}$ and $\langle\cdot,\cdot\rangle$ abbreviates $\langle\cdot,\cdot\rangle_{L^{2}}$, and we use claims 4, 5 and 6 of the inner-product lemma: the pairing is symmetric and linear in each argument, $\langle w,w\rangle=\lVert w\rVert^{2}$, the norm vanishes only at the zero element, $\lVert cw\rVert=|c|\lVert w\rVert$ for real $c$, the Cauchy-Schwarz inequality $|\langle w,w'\rangle|\le\lVert w\rVert\lVert w'\rVert$ and the triangle inequality hold, and $d_{L^{2}}$ is a metric. Expanding by bilinearity, for all $a,b\in H$,

$$\lVert a-b\rVert^{2}+\lVert a+b\rVert^{2}=2\lVert a\rVert^{2}+2\lVert b\rVert^{2}. \tag{P}$$

Claim 1. Existence. Fix $u\in H$ and let $S=\{\lVert u-v\rVert:v\in C\}$, a nonempty set of real numbers bounded below by $0$. Let $\delta$ be its greatest lower bound, which exists because the real numbers are a Dedekind complete ordered field; note $\delta\ge0$.

For each $n\in\mathbb{N}$ the number $\delta+1/n$ is not a lower bound of $S$, so there is $v_{n}\in C$ with $\lVert u-v_{n}\rVert<\delta+1/n$, whence $\lVert u-v_{n}\rVert^{2}<\delta^{2}+2\delta/n+1/n^{2}$. Let $n,m\in\mathbb{N}$. Applying (P) with $a=u-v_{n}$ and $b=u-v_{m}$, and noting $a-b=v_{m}-v_{n}$ and $a+b=2\bigl(u-\tfrac{1}{2}v_{n}-\tfrac{1}{2}v_{m}\bigr)$, so that $\lVert a+b\rVert^{2}=4\lVert u-\tfrac{1}{2}v_{n}-\tfrac{1}{2}v_{m}\rVert^{2}$ by absolute homogeneity, we get

$$\lVert v_{n}-v_{m}\rVert^{2}=2\lVert u-v_{n}\rVert^{2}+2\lVert u-v_{m}\rVert^{2}-4\bigl\lVert u-\tfrac{1}{2}v_{n}-\tfrac{1}{2}v_{m}\bigr\rVert^{2}.$$

Since $C$ is convex, the element $\tfrac{1}{2}v_{n}+\tfrac{1}{2}v_{m}$ lies in $C$, so the subtracted norm is at least $\delta$ and

$$\lVert v_{n}-v_{m}\rVert^{2}<2\bigl(\delta^{2}+\tfrac{2\delta}{n}+\tfrac{1}{n^{2}}\bigr)+2\bigl(\delta^{2}+\tfrac{2\delta}{m}+\tfrac{1}{m^{2}}\bigr)-4\delta^{2}=\tfrac{4\delta}{n}+\tfrac{2}{n^{2}}+\tfrac{4\delta}{m}+\tfrac{2}{m^{2}} .$$

Let $\varepsilon>0$ be real. Since the real numbers are Archimedean, choose $N\in\mathbb{N}$ with $N>(16\delta+8)/\varepsilon^{2}$. For $n\ge N$ we have $1/n^{2}\le1/n\le1/N$, hence

$$\frac{4\delta}{n}+\frac{2}{n^{2}}\le\frac{4\delta+2}{N}<\frac{\varepsilon^{2}}{4},$$

and likewise for $m\ge N$. So for all $n,m\ge N$ the right-hand side of the previous display is less than $\varepsilon^{2}/2$, giving $\lVert v_{n}-v_{m}\rVert^{2}<\varepsilon^{2}$ and therefore $\lVert v_{n}-v_{m}\rVert<\varepsilon$, since both quantities are nonnegative and squaring is strictly increasing on the nonnegative reals. Thus $(v_{n})_{n}$ is a Cauchy sequence in $(H,d_{L^{2}})$. By completeness of $L^{2}([0,T];\mathbb{R}^{d})$ it converges to some $p\in H$, and $p\in C$ because $C$ is closed. Indeed, suppose $p\notin C$. Then $p$ lies in the complement $H\setminus C$, which is open, so there is a real $\rho>0$ with $\{\zeta\in H:d_{L^{2}}(\zeta,p)<\rho\}\subseteq H\setminus C$. Since $(v_{n})_{n}$ converges to $p$, there is $n$ with $d_{L^{2}}(v_{n},p)<\rho$, so that $v_{n}\in H\setminus C$; but $v_{n}\in C$, a contradiction.

Moreover $\lVert u-p\rVert=\delta$: from the triangle inequality, $\bigl|\lVert u-v_{n}\rVert-\lVert u-p\rVert\bigr|\le\lVert v_{n}-p\rVert$, which has limit $0$, so the real sequence $\bigl(\lVert u-v_{n}\rVert\bigr)_{n}$ has limit $\lVert u-p\rVert$; since $\delta\le\lVert u-v_{n}\rVert<\delta+1/n$ for every $n$, that same sequence also has limit $\delta$: given a real $\eta>0$, the Archimedean property supplies $N\in\mathbb{N}$ with $1/N<\eta$, and then $\bigl|\lVert u-v_{n}\rVert-\delta\bigr|<1/n\le1/N<\eta$ for every $n\ge N$. Limits of real sequences are unique by uniqueness of limits, so $\lVert u-p\rVert=\delta$. As $\delta$ is a lower bound of $S$, $p$ satisfies $\lVert u-p\rVert\le\lVert u-v\rVert$ for every $v\in C$.

Uniqueness. Suppose $p,p'\in C$ both satisfy the minimising inequality; then each of $\lVert u-p\rVert$ and $\lVert u-p'\rVert$ lies in $S$ and is a lower bound of $S$. Such a number equals $\delta$: it is at least $\delta$ because $\delta$ is a lower bound of $S$ and the number lies in $S$, and it is at most $\delta$ because it is a lower bound of $S$ while $\delta$ is the greatest lower bound. Hence $\lVert u-p\rVert=\lVert u-p'\rVert=\delta$. Applying (P) with $a=u-p$, $b=u-p'$ and using $\tfrac{1}{2}p+\tfrac{1}{2}p'\in C$ exactly as above,

$$\lVert p-p'\rVert^{2}=2\delta^{2}+2\delta^{2}-4\bigl\lVert u-\tfrac{1}{2}p-\tfrac{1}{2}p'\bigr\rVert^{2}\le4\delta^{2}-4\delta^{2}=0,$$

so $p=p'$. We write $\pi_{C}(u)$ for this unique element.

Claim 2. Suppose first that $\langle u-p,v-p\rangle\le0$ for every $v\in C$. For $v\in C$, bilinearity gives

$$\lVert u-v\rVert^{2}=\lVert(u-p)-(v-p)\rVert^{2}=\lVert u-p\rVert^{2}-2\langle u-p,v-p\rangle+\lVert v-p\rVert^{2}\ge\lVert u-p\rVert^{2},$$

since the middle term is nonnegative and the last is nonnegative. Taking nonnegative square roots, $\lVert u-p\rVert\le\lVert u-v\rVert$ for every $v\in C$, so $p=\pi_{C}(u)$ by the uniqueness in claim 1.

Conversely suppose $p=\pi_{C}(u)$ and let $v\in C$. For a real $s$ with $0<s\le1$, convexity gives $(1-s)p+sv=p+s(v-p)\in C$, so

$$\lVert u-p\rVert^{2}\le\lVert u-p-s(v-p)\rVert^{2}=\lVert u-p\rVert^{2}-2s\langle u-p,v-p\rangle+s^{2}\lVert v-p\rVert^{2}.$$

Hence $2s\langle u-p,v-p\rangle\le s^{2}\lVert v-p\rVert^{2}$, and dividing by $2s>0$,

$$\langle u-p,v-p\rangle\le\tfrac{s}{2}\lVert v-p\rVert^{2}\qquad\text{for every real }s\text{ with }0<s\le1 .$$

If $\langle u-p,v-p\rangle$ were a positive number $\eta$, choosing $s$ with $0<s\le1$ and $s<2\eta/(\lVert v-p\rVert^{2}+1)$ would give $\tfrac{s}{2}\lVert v-p\rVert^{2}<\eta$, a contradiction. Therefore $\langle u-p,v-p\rangle\le0$.

Claim 3. If $u\in C$ then $\lVert u-u\rVert=0\le\lVert u-v\rVert$ for every $v\in C$, so $u$ satisfies the minimising inequality and $\pi_{C}(u)=u$ by uniqueness. Since $\pi_{C}$ takes values in $C$ by construction, it maps $H$ onto $C$.

Claim 4. Let $u,u'\in H$ and put $p=\pi_{C}(u)$, $p'=\pi_{C}(u')$. Applying claim 2 to $u$ with the test element $p'\in C$, and to $u'$ with the test element $p\in C$,

$$\langle u-p,\;p'-p\rangle\le0,\qquad\langle u'-p',\;p-p'\rangle\le0 .$$

Adding the first to the second after replacing $p-p'$ by $-(p'-p)$ in the second, that is adding $\langle u-p,p'-p\rangle\le0$ and $-\langle u'-p',p'-p\rangle\le0$, gives by bilinearity

$$\bigl\langle (u-u')-(p-p'),\;p'-p\bigr\rangle\le0 .$$

Since $\langle p-p',p'-p\rangle=-\lVert p-p'\rVert^{2}$, this reads $\langle u-u',p'-p\rangle+\lVert p-p'\rVert^{2}\le0$, that is

$$\lVert p-p'\rVert^{2}\le\langle u-u',\;p-p'\rangle\le\lVert u-u'\rVert\,\lVert p-p'\rVert$$

by the Cauchy-Schwarz inequality. If $\lVert p-p'\rVert=0$ the asserted inequality is immediate; otherwise dividing by the positive number $\lVert p-p'\rVert$ gives $\lVert p-p'\rVert\le\lVert u-u'\rVert$. In either case $d_{L^{2}}\bigl(\pi_{C}(u),\pi_{C}(u')\bigr)\le d_{L^{2}}(u,u')$, which is exactly the assertion that $\pi_{C}$ is Lipschitz with constant $1$.