Proof of Riesz-Frechet Representation of Bounded Linear Functionals on the Lebesgue Space of Square-Integrable Vector-Valued Functions

Throughout, claim numbers for $H$ refer to The Lebesgue Space of Square-Integrable Vector-Valued Functions is a Real Inner Product Space: by claim 4 the pairing is symmetric and linear in each argument, $\langle v,v\rangle_{L^{2}}=\lVert v\rVert_{L^{2}}^{2}$, and $\lVert v\rVert_{L^{2}}=0$ holds only for $v=0$.

Uniqueness. Suppose $\eta$ and $\eta'$ both represent $\Lambda$. Then $\langle\eta-\eta',v\rangle_{L^{2}}=\Lambda(v)-\Lambda(v)=0$ for every $v\in H$ by linearity of the pairing. Taking $v=\eta-\eta'$ gives $\lVert\eta-\eta'\rVert^{2}_{L^{2}}=0$, hence $\eta=\eta'$.

The trivial case. If $\Lambda(v)=0$ for every $v\in H$, then $\eta=0$ represents $\Lambda$, and $\lVert 0\rVert_{L^{2}}=0\le C$.

The kernel. Assume from now on that $\Lambda(u_{0})\ne0$ for some $u_{0}\in H$, and put $N=\{v\in H:\Lambda(v)=0\}$.

$N$ is nonempty and closed under linear combinations. Hypothesis (i) with $s=s'=0$ gives $\Lambda(0)=0$, so $0\in N$. If $v,v'\in N$ and $s,s'$ are real, then $\Lambda(sv+s'v')=s\Lambda(v)+s'\Lambda(v')=0$, so $sv+s'v'\in N$. In particular $N$ is convex, taking $s'=1-s$ with $0\le s\le1$.

$N$ is closed. The metric open subsets determined by $d_{L^{2}}$ form a topology by Metric Open Sets Form a Topology. Let $v$ belong to the closure of $N$ and let $\varepsilon>0$ be real. By Characterization of the Closure in a Metric Space by Open Balls there is $v'\in N$ with $d_{L^{2}}(v,v')<\varepsilon$, so by (i) and (ii),

If $C=0$ then $\Lambda(v)=0$ by (ii). If $C>0$ and $|\Lambda(v)|>0$, then choosing $\varepsilon=|\Lambda(v)|C^{-1}>0$ gives $|\Lambda(v)|<|\Lambda(v)|$, which is false; so $\Lambda(v)=0$ in either case, and $v\in N$. Thus the closure of $N$ is contained in $N$; as it also contains $N$ and is closed, the set $N$ equals its closure and is closed.

A vector orthogonal to the kernel. By claim 1 of Nearest-Point Projection onto a Nonempty Closed Convex Subset of the Lebesgue Space of Square-Integrable Vector-Valued Functions, applied to the nonempty closed convex set $N$, the projection $p=\pi_{N}(u_{0})$ exists; put $z=u_{0}-p$. If $z=0$ then $u_{0}=p\in N$, contradicting $\Lambda(u_{0})\ne0$; so $z\ne0$ and $\lVert z\rVert_{L^{2}}>0$.

Let $w\in N$. Both $p+w$ and $p-w$ lie in $N$, so claim 2 of that theorem gives

that is $\langle z,w\rangle_{L^{2}}\le0$ and $-\langle z,w\rangle_{L^{2}}\le0$, using linearity of the pairing. Hence $\langle z,w\rangle_{L^{2}}=0$ for every $w\in N$.

Moreover $\Lambda(z)\ne0$: otherwise $z\in N$, whence $\lVert z\rVert^{2}_{L^{2}}=\langle z,z\rangle_{L^{2}}=0$ and $z=0$.

Representation. Put

Let $v\in H$ and set $w=v-\dfrac{\Lambda(v)}{\Lambda(z)}\,z$. By (i), $\Lambda(w)=\Lambda(v)-\dfrac{\Lambda(v)}{\Lambda(z)}\Lambda(z)=0$, so $w\in N$ and therefore $\langle z,w\rangle_{L^{2}}=0$. Expanding by linearity of the pairing in the second argument,

so that

the last step by linearity of the pairing in the first argument.

The norm bound. Using the representation with $v=\eta$ and hypothesis (ii),

If $\eta=0$ then $\lVert\eta\rVert_{L^{2}}=0\le C$. Otherwise $\lVert\eta\rVert_{L^{2}}>0$, and multiplying the inequality by $\lVert\eta\rVert_{L^{2}}^{-1}>0$ gives $\lVert\eta\rVert_{L^{2}}\le C$.