Riesz Representation Theorem (Hilbert Spaces)

Theorem

Let $H$ be a Hilbert space.

Let $L$ be a bounded linear functional on $H$.

Then there is a unique $h_0 \in H$ such that

$\forall h \in H: L h = \left\langle{h, h_0}\right\rangle$

Corollary

The norm of $L$ satisfies:

$\left\Vert{L}\right\Vert = \left\Vert{h_0}\right\Vert$

Proof

If $L \equiv 0$ identically, then $Lh = 0 = \left\langle{h,0}\right \rangle$, and the theorem holds.

Otherwise, set:

$M = \ker{L} = L^{-1}\left(\{{0}\}\right)$.

Then $M$ is a subspace.

Because $L$ is bounded, it is continuous.

Because $\{0\}$ is closed,the continuity of $L$ implies that $M$ is closed.

Then we can decompose $H$ as a direct sum:

$H \cong M \oplus M^\perp$

As $L \not \equiv 0$, $M^\perp \ne \left\{{0}\right\}$.

Choose a $z \in M^\perp$ with norm $1$.

By linearity of $L$, for any $h \in H$,

 $\displaystyle L\left({zLh - hLz)}\right)$ $=$ $\displaystyle LzLh - LhLz$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle 0$ $\quad$ $\quad$

So $zLh - hLz \in \ker L = M$.

Then,

 $\displaystyle Lh$ $=$ $\displaystyle Lh \left \langle { z, z }\right \rangle$ $\quad$ $\Vert z \Vert = 1$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left \langle {zLh,z}\right \rangle$ $\quad$ linearity in the first argument $\quad$ $\displaystyle$ $=$ $\displaystyle \left \langle {zLh - hLz + hLz,z}\right \rangle$ $\quad$ adding and subtracting $hLz$ in the first argument $\quad$ $\displaystyle$ $=$ $\displaystyle \left \langle {zLh - hLz,z}\right \rangle + \left \langle {hLz,z}\right \rangle$ $\quad$ linearity in the first argument $\quad$ $\displaystyle$ $=$ $\displaystyle \left \langle {hLz,z}\right \rangle$ $\quad$ $zLh - hLz \in M, z \in M^\perp$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left \langle {h,z (Lz)^* }\right\rangle$ $\quad$ conjugate symmetry $\quad$

Thus $Lh = \left \langle{ h,h_0 }\right \rangle$ for $h_0 = z (Lz)^*$.

To show uniqueness, assume $h_0$ and $h_1$ both satisfy the above equation for all $h \in H$:

 $\displaystyle \left \langle {h,h_0}\right \rangle$ $=$ $\displaystyle \left \langle {h,h_1}\right \rangle$ $\quad$ $\quad$ $\displaystyle \implies \ \$ $\displaystyle \left \langle {h,h_0}\right \rangle - \left \langle {h,h_1}\right \rangle$ $=$ $\displaystyle 0$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left \langle {h,h_0-h_1}\right \rangle$ $\quad$ additivity in the second argument $\quad$

The result follows from Setting $h = h_0 - h_1$ and invoking the positive definiteness of the inner product.

$\blacksquare$

Source of Name

This entry was named for Frigyes Riesz.