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:

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


Proof

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

By Kernel of Bounded Linear Transformation is Closed Linear Subspace, $M := \ker L$ is a closed linear subspace of $H$.

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.

$\blacksquare$


Examples

$L^2$ Space

Let $\struct{ X, \Sigma, \mu }$ be a measure space.

Let $\map {L^2} \mu$ be the associated $L^2$ space.

Let $F: \map {L^2} \mu \to \GF$ be a bounded linear functional.


Then there exists a unique $f_0 \in \map {L^2} \mu$ such that:

$\ds \forall f \in \map {L^2} \mu: \map F f = \int f \overline{f_0} \rd \mu$


Space of Square Summable Mappings

Let $\map {\ell^2} \N$ be the space of square summable mappings on $\N$.

Let $N \in \N$.

Let $L_N: \map {\ell^2} \N \to \GF$ be defined by:

$\map {L_N} {\sequence{ a_n } } := a_N$

Let $\delta_N \in \map {\ell^2} \N$ be given by:

$\forall n \in \N: \paren{ \delta_N }_n = \begin{cases}

1 & n = N \\ 0 & n \ne N \end{cases}$


Then for all $a \in \map {\ell^2} \N$:

$\map {L_N} a = \innerprod a {\delta_N}$


Source of Name

This entry was named for Frigyes Riesz.


Sources

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