Riesz Representation Theorem (Hilbert Spaces)
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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 $h \in H$:
| \(\displaystyle L \paren {z L h - h L z}\) | \(=\) | \(\displaystyle L z L h - L h L z\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle 0\) |
So:
- $z L h - h L z \in \ker L = M$
Then:
| \(\displaystyle L h\) | \(=\) | \(\displaystyle L h \innerprod z z\) | as $\norm z = 1$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \innerprod {z L h} z\) | linearity in the first argument | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \innerprod {z L h - h L z + h L z} z\) | adding and subtracting $h L z$ in the first argument | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \innerprod {z L h - h L z} z + \innerprod {h L z} z\) | linearity in the first argument | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \innerprod {h L z} z\) | $z L h - h L z \in M, z \in M^\perp$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \innerprod h {z \paren {L z}^*}\) | 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 $h \in H$:
| \(\displaystyle \innerprod h {h_0}\) | \(=\) | \(\displaystyle \innerprod h {h_1}\) | |||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle \innerprod h {h_0} - \innerprod h {h_1}\) | \(=\) | \(\displaystyle 0\) | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \innerprod h {h_0 - h_1}\) | 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.
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- 1990: John B. Conway: A Course in Functional Analysis ... (previous) ... (next) $I.3.4$
