Riesz representation theory

Linear Functional

A linear functional $\ell$ is a linear transformation from a Hilbert space to the underlying field of scalars.

Riesz Representation Theorem

Let $\ell$ be a bounded linear functional on a Hilbert space \H. Then, there exists a unique g\in\H, such that $\ell (f)=⟨f,g⟩$ for all f\in \H. Moreover, $\Vert \ell \Vert =\Vert g\Vert$.

Proof Consider the kernel of $\ell$, say $\S :=\ker \ell =\{f\in \stackrel{˝}{:}\ell (f)=0\}$, which is a closed subspace of \H, so $\stackrel{˝}{=}\S \oplus {\S }^{\perp }$. If ${\S }^{\perp }=\varnothing$, then $\ell =0$. Otherwise, pick some $h\in {\S }^{\perp }$ with $\Vert h\Vert =1$, and let $g:=\overline{\ell (h)}h$. Then if we let $u:=\ell (f)h-\ell (h)f$, observe that $u\in \S$ because $\ell (u)=\ell (f)\ell (h)-\ell (h)\ell (h)=0$, therefore, $0=⟨u,h⟩=\ell (f)⟨h,h⟩-⟨f,\overline{\ell (h)}h⟩=\ell (f)⟨h,h⟩-⟨f,g⟩,$which shows that $\ell (f)=⟨f,g⟩$. Moreover, if $g^{\prime }\in {\S }^{\perp }$ is another element such that $\ell (f)=⟨f,g^{\prime }⟩$, then we have $\ell (f)=⟨f,g⟩=⟨f,g^{\prime }⟩,$it follows that $⟨f,g-g^{\prime }⟩=0$ for all f\in\H, which implies that $g=g^{\prime }$. $\square$