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


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

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 non-zero $z \in M^\perp$ with norm $1$. This is possible because, as $M^\perp$ is closed, $\dfrac{z_0}{\Vert z_0 \Vert} \in M^\perp$ for $z_0 \ne 0$.

By linearity of $L$:

So $zLh = hLz$ for any $h \in M \oplus M^\perp$.

Then,

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 $x \in H$:

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