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:


 * $\forall h \in H: L h = \innerprod h {h_0}$

Proof
If $L \equiv 0$ identically, then $L h = 0 = \innerprod h 0$, and the theorem holds.

Otherwise, set:


 * $M = \map \ker L = \map {L^{-1} } {\set 0}$

Then $M$ is a subspace.

Because $L$ is bounded, it is continuous.

Because $\set 0$ 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 $L \not \equiv 0$:
 * $M^\perp \ne \set 0$

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

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

So:
 * $z L h - h L z \in \ker L = M$

Then:

Thus $L h = \innerprod h {h_0}$ for $h_0 = z (Lz)^*$.

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

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