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),

$$|\Lambda(v)|=|\Lambda(v)-\Lambda(v')|=|\Lambda(v-v')|\le C\lVert v-v'\rVert_{L^{2}}<C\varepsilon .$$

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

$$\langle z,(p+w)-p\rangle_{L^{2}}\le0\qquad\text{and}\qquad\langle z,(p-w)-p\rangle_{L^{2}}\le0,$$

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

$$\eta=\frac{\Lambda(z)}{\lVert z\rVert_{L^{2}}^{2}}\,z\in H .$$

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,

$$0=\langle z,v\rangle_{L^{2}}-\frac{\Lambda(v)}{\Lambda(z)}\,\langle z,z\rangle_{L^{2}}=\langle z,v\rangle_{L^{2}}-\frac{\Lambda(v)}{\Lambda(z)}\,\lVert z\rVert^{2}_{L^{2}},$$

so that

$$\Lambda(v)=\frac{\Lambda(z)}{\lVert z\rVert^{2}_{L^{2}}}\,\langle z,v\rangle_{L^{2}}=\langle\eta,v\rangle_{L^{2}},$$

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

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

$$\lVert\eta\rVert^{2}_{L^{2}}=\langle\eta,\eta\rangle_{L^{2}}=\Lambda(\eta)\le|\Lambda(\eta)|\le C\,\lVert\eta\rVert_{L^{2}} .$$

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 $\lVert\eta\rVert_{L^{2}}\le C$.