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 (i), $\langle x,z-z'\rangle=0$ 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 (iii), 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 (viii); 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 (viii) together with Elementary Properties of Bounded Linear Maps and Functionals on Real Inner Product Spaces (vi). 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 (ii), 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 (i) 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 (iv), $w_{0}\notin M$, 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 (iii)), $\ell(x')=\ell(x)-\ell(x)\,\ell(w_{0})^{-1}\,\ell(w_{0})=0$, 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 (i),

$$0=\langle x',w_{0}\rangle=\langle x,w_{0}\rangle-\ell(x)\,\ell(w_{0})^{-1}\,\langle w_{0},w_{0}\rangle=\langle x,w_{0}\rangle-\ell(x)\,\ell(w_{0})^{-1}\,|w_{0}|^{2}.$$

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

$$\ell(x)=\ell(w_{0})\,(|w_{0}|^{2})^{-1}\,\langle x,w_{0}\rangle=\langle x,z\rangle,\qquad z=\ell(w_{0})\,(|w_{0}|^{2})^{-1}\,w_{0},$$

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