The Riesz Representation Theorem for Hilbert Spaces

# The Riesz Representation Theorem for Hilbert Spaces

Let $H$ be a Hilbert space. For each $g \in H$ let $f_g : H \to \mathbb{R}$ be defined for all $h \in H$ by:

(1)\begin{align} \quad f_g(h) = \langle h, g \rangle \end{align}

Note that each $f_g$ is linear since if $h_1, h_2 \in H$ we have that:

(2)\begin{align} \quad f_g(h_1 + h_2) = \langle h_1 + h_2, g \rangle = \langle h_1, g \rangle + \langle h_2, g \rangle = f_g(h_1) + f_g(h_2) \end{align}

And for all $h \in H$ and all $a \in \mathbb{R}$ we have that:

(3)\begin{align} \quad f_g(ah) = \langle ah, g \rangle = a \langle h, g \rangle = a f_g(h) \end{align}

Furthermore, observe that $f_g$ is bounded since for all $h \in H$, since by The Cauchy-Schwarz Inequality for Inner Product Spaces:

(4)\begin{align} \quad | f_g(h) | = | \langle h, g \rangle | \leq \| h \| \| g \| = \| g \| \| h \| \end{align}

So $\| f_g \| \leq \| g \|$. Furthermore, we have that:

(5)\begin{align} \quad \| g \|^2 = |\langle g, g \rangle| = |f_g(g)| \leq \| f_g \| \| g \| \end{align}

This shows that $\| g \| \leq \| f_g \|$. So $\| f_g \| = \| g \|$.

The Riesz-Representation theorem tells us that every bounded linear functional on $H$ is of the form $f_g$.

Theorem (The Riesz Representation Theorem for Hilbert Spaces): Let $H$ be a Hilbert space. Then for every $f \in H^*$ there exists an $g \in H$ such that for every $h \in H$, $f(h) = \langle h, g \rangle$. |

**Proof:**Let $f \in H^*$. If $f = 0$ then take $g = 0$. Then $f(h) = \langle h, 0 \rangle = 0$ for all $h \in H$. So assume $f \neq 0$.

- Let $V = \ker f$. Then $V$ is a closed subspace of $H$. Since $f \neq 0$ we have that $V \neq H$. So $V^{\perp} \neq \{ 0 \}$.

- Let $t \in V^{\perp}$ be such that $\| t \| = 1$ and $f(t) \neq 0$. Note this is always doable. Just take any nonzero $t' \in V^{\perp}$ and let $\displaystyle{t = \frac{t'}{\| t' \|}}$. Then $\| t \| = 1$ and since $t' \in V^{\perp}$ we have that $f(t) \neq 0$.

- Let:

\begin{align} \quad g = f(t) t \end{align}

- Then for all $h \in H$ we have that:

\begin{align} \quad f \left ( h - \frac{f(h)}{f(t)}t \right ) = f(h) - \frac{f(h)}{f(t)}f(t) = 0 \end{align}

- So for every $h \in H$ we have that:

\begin{align} \quad h - \frac{f(h)}{f(t)} t \in \ker f = V \end{align}

- Since $g \in V^{\perp}$ we have that then for every $h \in H$:

\begin{align} 0 &= \left \langle g, h - \frac{f(h)}{f(t)} t \right \rangle \\ 0 &=\langle \langle g, h \rangle - \left \langle g, \frac{f(h)}{f(t)} t \right \rangle \\ \langle g, \frac{f(h)}{f(t)} t &= \langle g, h \rangle \\ \langle f(t) t, \frac{f(h)}{f(t)} t \rangle &= \langle g, h \rangle \\ f(h) \langle t, t \rangle &= \langle g, h \rangle \\ f(h) \| t \|^2 &= \langle g, h \rangle \\ f(h) &= \langle g, h \rangle \end{align}

- So for every $h \in H$ we have that $f(h) = \langle h, g \rangle$. $\blacksquare$