Proof of The Riesz Representation Theorem for a Real Hilbert Space

We use the notation of Elementary Identities in a Real Inner Product Space and write $0_{H}$ for the zero vector of $H$. We prove claims 2 and 3 first; neither uses claim 1.

Claim 2. Suppose $\langle x,z\rangle=\langle x,z'\rangle$ for every $x\in H$. By Elementary Identities in a Real Inner Product Space §bilinear, $\langle x,z-z'\rangle=0$ for every $x\in H$; taking $x=z-z'$ gives $|z-z'|^{2}=\langle z-z',z-z'\rangle=0$, so $z-z'=0_{H}$ by Elementary Identities in a Real Inner Product Space §vanishing, that is, $z=z'$ by claim 2 of Elementary Identities in a Vector Space.

Claim 3. Suppose $\ell(x)=\langle x,z\rangle$ for every $x\in H$. Then $\ell$ is the map $x\mapsto\langle x,z\rangle$, whose norm is $|z|$ by Elementary Properties of Bounded Linear Maps and Functionals on Real Inner Product Spaces §functionals; hence $\lVert\ell\rVert=|z|$.

Claim 1. Let $M=\ker\ell=\{x\in H:\ell(x)=0\}$, a closed linear subspace of $H$ by Elementary Properties of Bounded Linear Maps and Functionals on Real Inner Product Spaces §functionals together with Elementary Properties of Bounded Linear Maps and Functionals on Real Inner Product Spaces §kernel-closed. Let $M^{\perp}$ be its orthogonal complement and $P_{M}$ the projection of Orthogonal Projection onto a Closed Linear Subspace of a Real Hilbert Space.

If $M=H$, then $\ell(x)=0=\langle x,0_{H}\rangle$ for every $x\in H$ by Elementary Identities in a Real Inner Product Space §zero, and $z=0_{H}$ serves.

Otherwise choose $w\in H$ with $w\notin M$ and put $w_{0}=w-P_{M}w$. By Orthogonal Projection onto a Closed Linear Subspace of a Real Hilbert Space §characterisation applied with $x=w$ and $z=P_{M}w$, $w_{0}\in M^{\perp}$. Moreover $w_{0}\ne 0_{H}$: otherwise $w=P_{M}w\in M$. Since $M\cap M^{\perp}=\{0_{H}\}$ by Orthogonal Projection onto a Closed Linear Subspace of a Real Hilbert Space §complement, $w_{0}\notin M$, that is, $\ell(w_{0})\ne 0$, so $\ell(w_{0})$ has a multiplicative inverse.

Now let $x\in H$ and put $x'=x-\ell(x)\,\ell(w_{0})^{-1}\,w_{0}$. By the linearity of $\ell$ (Bounded Linear Maps and Bounded Linear Functionals on Real Inner Product Spaces, and the Operator Norm §functional), $\ell(x')=\ell(x)-\ell(x)\,\ell(w_{0})^{-1}\,\ell(w_{0})=0$, so $x'\in M$ and therefore $\langle x',w_{0}\rangle=0$ because $w_{0}\in M^{\perp}$ (and $\langle x',w_{0}\rangle=\langle w_{0},x'\rangle$ by symmetry). Expanding with Elementary Identities in a Real Inner Product Space §bilinear,

Since $w_{0}\ne 0_{H}$, $|w_{0}|^{2}>0$ by Elementary Identities in a Real Inner Product Space §vanishing and claim 5 of Elementary Order Arithmetic in an Ordered Field, so it is invertible, and solving for $\ell(x)$ gives

the last equality by Elementary Identities in a Real Inner Product Space §bilinear. As $z$ does not depend on $x$, this proves claim 1.