Section 7.2 The Riesz representation theorem
In the case of Hilbert space, we recover the characterization of linear functionals as inner products against a representative vector. This powerful idea is one of the central tools of Hilbert space theory.
Theorem 7.2.1. (Riesz-Fréchet).
Let \(\hilbert\) be a Hilbert space and let \(F\) be a continuous linear functional on \(\hilbert\text{.}\) There exists a unique vector \(y \in \hilbert\) so that
\begin{equation*}
F(x) = \ip{x}{y}
\end{equation*}
for all \(x \in \hilbert\text{.}\) Moreover, \(\norm{y} = \norm{F}\text{.}\) (\(y\) is sometimes called the representative vector for \(F\text{.}\))
Proof.
We should first observe that if we can find such a
\(y\text{,}\) Theorem 2.1.4 guarantees that it will be unique, as for all
\(x\in \hilbert\text{,}\) we would have
\begin{equation*}
\ip{x}{y} = F(x) = \ip{x}{y'}.
\end{equation*}
If
\(F\) is the functional that sends every vector to 0, then we can take
\(y = 0\text{.}\) So suppose that
\(F\) is not the zero functional. In that case, the kernel of
\(F\text{,}\) which we denote by
\(M\) is a proper closed subspace of
\(\hilbert\text{.}\) Then by
Theorem 5.5.5 we can decompose
\(\hilbert\) as
\(\hilbert = M \oplus M^\perp\) with
\(M^\perp \neq \{0\}\text{.}\) Now choose a non-zero element
\(z \in M^\perp\text{.}\) We can assume that
\(F(z) = 1\) by normalizing by a scalar. Now, for any
\(x \in \hilbert\text{,}\) decompose
\(x\) into
\begin{equation*}
x = (x - F(x)z) + F(x)z),
\end{equation*}
where \(F(x)z \in M^\perp\) and so \(x - F(x)z \in M\text{.}\) Now, take the inner product of these vectors with \(z\text{.}\) The result is
\begin{align*}
\ip{x}{z} \amp= \ip{(x - F(x)z) + F(x)z)}{z}\\
\amp= \ip{x - F(x)z}{z} + \ip{F(x)z}{z}\\
\amp= \ip{F(x)z}{z}
\end{align*}
since \(z \perp M\text{.}\) Then we get, for all \(x \in \hilbert\text{,}\) that
\begin{equation*}
\ip{x}{z} = F(x)\norm{z}^2.
\end{equation*}
Now, let \(y = z/\norm{z}^2\text{.}\) Taking the inner product with \(x\text{,}\) we compute
\begin{equation*}
\ip{x}{y} = \ip{x}{\frac{z}{\norm{z}^2}} = \frac{1}{\norm{z}^2} F(x)\norm{z}^2 = F(x).
\end{equation*}
This shows that the representing vector for \(F\) exists.
Now, we want to show that
\(\norm{y} = \norm{F}\text{.}\) For any
\(\norm{x} \leq 1\text{,}\) the Cauchy-Schwarz inequality (
Theorem 2.2.4) gives
\begin{equation*}
\abs{F(x)} = \abs{\ip{x}{y}} \leq \norm{x}\norm{y} \leq \norm{y}.
\end{equation*}
On the other hand, choosing the specific unit vector \(x = \frac{y}{\norm{y}}\) gives
\begin{equation*}
\norm{F} \geq \abs{F(x)} = \frac{\abs{F(y)}}{\norm{y}} = \abs{\ip{y}{y}}/\norm{y} = \norm{y};
\end{equation*}
that is, \(\norm{F} \geq \norm{y}\text{.}\) We conclude that \(\norm{F} = \norm{y}\text{,}\) which completes the proof.
It is worth pointing out that the Riesz representation theorem gives an explicit map \(T: \hilbert \to \hilbert\ad\) via
\begin{equation*}
Ty = \ip{\cdot}{y}
\end{equation*}
that is both surjective and isometric. \(T\) is also antilinear or conjugate linear in the sense that \(T(ax + by) = \cc{a}x + \cc{b}{y}\text{.}\) Given the existence of \(T\text{,}\) Hilbert spaces are in some sense their own dual spaces, in the same way that \(\R^n\) is self-dual. (That is to say, Hilbert spaces are in some strong sense the correct notion of “\(\R^\infty\)”).
We’re arrived at one of the major differences between inner product spaces and normed spaces. While the Riesz representation theorem tells us essentially everything about the continuous linear functionals on a Hilbert space, questions turn out to be significantly more complicated in the realm of Banach space. N. Young points out the following, seemingly easy question - given
\(x, y\) distinct elements in a normed space
\(E\text{,}\) does there exist a linear functional that “separates points”; that is, can we find a functional
\(F\) on
\(E\) so that
\(F(x) \neq F(y)\text{?}\) In Hilbert space, the answer is immediate: yes, because
\(F = \ip{\cdot}{x - y}\) will do the job. In a general normed space, the answer is much more subtle and requires the central theorem of functional analysis, the
Hahn-Banach theorem, which is well beyond the scope of these notes and typically introduced in a first course in functional analysis (see e.g.
Conway 2007.).
