Riesz representation theorem

Let ${\displaystyle M=\{u\in H\ |\ \varphi (u)=0\}}$. Clearly ${\displaystyle M}$ is closed subspace of ${\displaystyle H}$. If ${\displaystyle M=H}$, then we can trivially choose ${\displaystyle f=0}$. Now assume ${\displaystyle M\neq H}$. Then ${\displaystyle M^{\perp }}$ is one-dimensional. Indeed, let ${\displaystyle v_{1},v_{2}}$ be nonzero vectors in ${\displaystyle M^{\perp }}$. Then there is nonzero real number ${\displaystyle \lambda }$, such that ${\displaystyle \lambda \varphi (v_{1})=\varphi (v_{2})}$. Observe that ${\displaystyle \lambda v_{1}-v_{2}\in M^{\perp }}$ and ${\displaystyle \varphi (\lambda v_{1}-v_{2})=0}$, so ${\displaystyle \lambda v_{1}-v_{2}\in M}$. This means that ${\displaystyle \lambda v_{1}-v_{2}=0}$. Now let ${\displaystyle g}$ be unit vector in ${\displaystyle M^{\perp }}$. For arbitrary ${\displaystyle x\in H}$, let ${\displaystyle v}$ be the orthogonal projection of ${\displaystyle x}$ onto ${\displaystyle M^{\perp }}$. Then ${\displaystyle v=\langle g,x\rangle g}$ and ${\displaystyle \langle g,x-v\rangle =0}$ (from the properties of orthogonal projections), so that ${\displaystyle x-v\in M}$ and ${\displaystyle \langle g,x\rangle =\langle g,v\rangle }$. Thus ${\displaystyle \varphi (x)=\varphi (v+x-v)=\varphi (\langle g,x\rangle g)+\varphi (x-v)=\langle g,x\rangle \varphi (g)+0=\langle g,x\rangle \varphi (g)}$. Hence ${\displaystyle f=\varphi (g)g}$. We also see ${\displaystyle \|f\|_{H}=|\varphi (g)|}$. From the Cauchy-Bunyakovsky-Schwartz inequality ${\displaystyle \varphi (x)\leq \|g\|\|x\||\varphi (g)|}$, thus for ${\displaystyle x}$ with unit norm ${\displaystyle \varphi (x)\leq |\varphi (g)|}$. This implies that ${\displaystyle \|\varphi \|_{H*}=|\varphi (g)|}$.

Given any continuous linear functional g in H*, the corresponding element ${\displaystyle x_{g}\in H}$ can be constructed uniquely by ${\displaystyle x_{g}=g(e_{1})e_{1}+g(e_{2})e_{2}+...}$, where ${\displaystyle \{e_{i}\}}$ is an orthonormal basis of H, and the value of ${\displaystyle x_{g}}$ does not vary by choice of basis. Thus, if ${\displaystyle y\in H,y=a_{1}e_{1}+a_{2}e_{2}+...}$, then ${\displaystyle g(y)=a_{1}g(e_{1})+a_{2}g(e_{2})+...=\langle x_{g},y\rangle .}$