# 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:

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

### Corollary

The norm of $L$ satisfies:

- $\norm L = \norm {h_0}$

## 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 $h \in H$:

\(\ds L \paren {z L h - h L z}\) | \(=\) | \(\ds L z L h - L h L z\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 0\) |

So:

- $z L h - h L z \in \ker L = M$

Then:

\(\ds L h\) | \(=\) | \(\ds L h \innerprod z z\) | as $\norm z = 1$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \innerprod {z L h} z\) | linearity in the first argument | |||||||||||

\(\ds \) | \(=\) | \(\ds \innerprod {z L h - h L z + h L z} z\) | adding and subtracting $h L z$ in the first argument | |||||||||||

\(\ds \) | \(=\) | \(\ds \innerprod {z L h - h L z} z + \innerprod {h L z} z\) | linearity in the first argument | |||||||||||

\(\ds \) | \(=\) | \(\ds \innerprod {h L z} z\) | $z L h - h L z \in M, z \in M^\perp$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \innerprod h {z \paren {L z}^*}\) | conjugate symmetry |

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 $h \in H$:

\(\ds \innerprod h {h_0}\) | \(=\) | \(\ds \innerprod h {h_1}\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \innerprod h {h_0} - \innerprod h {h_1}\) | \(=\) | \(\ds 0\) | |||||||||||

\(\ds \) | \(=\) | \(\ds \innerprod h {h_0 - h_1}\) | additivity in the second argument |

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) $I.3.4$