# 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.

## Sources

- 1990: John B. Conway:
*A Course in Functional Analysis*... (previous) ... (next) $I.3.4$