3.4 Riesz’s representer theorem
Definition linear functional
- $K = \mathbb{R}, \mathbb{C}$
- field
- $X$
- linear space over $K$
- $f:X \rightarrow K$
$f$ is said to be linear function if
\[\begin{eqnarray} f(u + v) & = & f(u) + f(v) & & (u, v \in X), \nonumber \\ f(\alpha u) & = & \alpha f(u) & & (\alpha \in K, u \in X) . \nonumber \end{eqnarray}\]Definition 3.7 dual space
- $X$
- topological linear space,
We denote \(X^{*}\) by a set of continuous linear functionals over $X$. \(X^{*}\) is called dual space.
Remark
$X^{*}$ is a linear space over $K$. Indeed, we define operations as follows:
\[\begin{eqnarray} (f + g)(u) & \Leftrightarrow & f(u) + g(u) & & (u \in X) \nonumber \\ (\alpha f)(u) & = & \alpha f(u) & & (u \in X) . \nonumber \end{eqnarray}\]Definition 3.8 boundedness of linear functionals
- $X$,
- norm space
- $f:X \rightarrow K$,
- linear functional
$f$ is said to be bounded if
\[\begin{equation} \exists M \in \mathbb{R}, \ \text{ s.t. } \ \forall u \in X, \ |f(u)| \le M \| u \| \label{chap03_04_03_36_boundedness} . \end{equation}\]Theorem 3.4
- $X$
- norm space
- $f: X \rightarrow K$,
- linear functional
$f$ is continuous if and only if $f$ is bounded.
proof
We first show if part. Suppose that \(\eqref{chap03_04_03_36_boundedness}\) holds. Then, if \(u_{n} \rightarrow u_{0} \ (n \rightarrow \infty)\), we have
\[\begin{eqnarray} |f(u_{n}) - f(u_{0})| & = & |f(u_{n} - u_{0})| & \le & M \|u_{n} - u_{0}\| \rightarrow 0 . \nonumber \end{eqnarray}\]Hence $f$ is continuous.
We next show only if part. Suppose that $f$ is not bounded. There exists \(\{u_{n}\}\) such that
\[\forall n \in \mathbb{N}, \ |f(u_{n})| > n \|u_{n}\| .\]Obviously $u_{n} \neq 0$. Let
\[v_{n} := \frac{ u_{n} }{ n \|u_{n} \| } .\]\(\|v_{n}\| = 1 / n \rightarrow 0\), $v_{n} \rightarrow 0$. Thefore, by continuity of $f$, we have
\[f(v_{n}) \rightarrow 0 .\]However,
\[|f(v_{n})| = \left| f \left( \frac{ u_{n} }{ n \|u_{n}\| } \right) \right| > \frac{ n \|u_{n}\| }{ n \|u_{n}\| } = 1 .\]If we take the limit of both side of above equation, we have a contradiction. Thefore, $f$ is bounded
Definition 3.9 norm in dual space
- $X$,
- norm space
- $f \in X^{*}$,
\(\|f\|_{X^{*}}\) is called norm of $f$.
Proposition
\[\begin{eqnarray} \|f\| & = & \sup_{u \neq 0} \frac{ |f(u)| }{ \|u\| } & = & \sup_{\|v\| = 1} |f(v)| \nonumber \end{eqnarray}\]proof
\[\forall u \neq 0 \in X, \ \frac{ |f(u)| }{ \|u\| } \le \frac{ \|u\| \|f\| }{ \|u\| } = \|f\| .\]Hence
\[\begin{eqnarray} \sup_{u \neq 0} \frac{ |f(u)| }{ \|u\| } \le \|f\| . \end{eqnarray}\]Conversely,
\[\forall u \in X, \ |f(u)| \le |f(u)| \frac{ \|u\| }{ \|u\| } \le \sup_{u^{\prime} \neq 0} \frac{ |f(u^{\prime})| }{ \|u^{\prime}\| } \|u\| .\]Hence
\[\sup_{u \neq 0} \frac{ |f(u)| }{ \|u\| } \in \{ M \in \mathbb{R} \mid |f(u)| \le M \|u\| \quad (u \in X) \} .\]It follows that
\[\begin{eqnarray} \sup_{u \neq 0} \frac{ |f(u)| }{ \|u\| } \ge \|f\| . \end{eqnarray}\]Propostion
\((X^{*}, \|\cdot\|_{X^{*}})\) is norm space.
proof
\[\sup_{\|v\| = 1} |(f + g)(v)| \le \sup_{\|v\| = 1} \left( |f(v)| + |g(v)| \right) \le \|f\| + \|g\| .\] \[\|\alpha f\| = \sup_{\| v\| = 1} |\alpha f(v)| = \sup_{\| v\| = 1} |\alpha| |f(v)| = |\alpha| \|f\| .\]Proposition 3.3
$N_{f}$ is closed subspace in $X$.
proof
Closedness is obvious since $f$ is continuous. Moreover, linearity is also obvious since $f$ is linear.
Theorem Riesz’s representor theorem
- $X$
- Hilbert space
proof
Uniquness.
Suppose that $f(u) = (u, a_{1}) = (u, a_{2})$. It follows that $(u, a_{1} - a_{2}) = 0$. Hence $a_{1} - a_{2} = 0$.
Existence.
Let $N_{f} := \mathrm{Ker}f$. If $N_{f} = X$, then $f \equiv 0$. In this case, $a = 0$ satisfies the statement. Now we assume $N_{f} \neq X$. There exists $c \in X$ such that $c \notin N_{f}$. Let $p \in N_{f}$ be orthogonal projection of $c$ and $b := c - p$. Then we have
\[\begin{eqnarray} f(b) & = & f(c - p) = f(c) \neq 0 \nonumber \\ b & \perp & N_{f} \nonumber \\ b & \neq & 0 \nonumber \end{eqnarray}\]For $u \in X$, let
\[\begin{eqnarray} \alpha(u) & := & \frac{ f(u) }{ f(b) } \nonumber \\ w & := & u - \alpha(u)b . \nonumber \\ f(w) & = & f(u) - \alpha(u)f(b) = 0 . \nonumber \end{eqnarray}\]Hence $w \in N_{f}$.
\[\begin{eqnarray} 0 & = & (w, b) \nonumber \\ & = & (u, b) - \alpha(i)\|b\|^{2} \nonumber \\ & = & (u, b) - \frac{ f(u) }{ f(b) } \|b\|^{2} . \end{eqnarray}\]Therefore, we have
\[f(u) = \left( u, \frac{ \overline{f(u)} }{ \|b\|^{2} } b \right) .\]It follows that $a := \overline{f(b)}b/|b|^{2}$ satisfies the statement.
For $f \in X^{*}$, we denote $Jf$ by $a$ in \(\eqref{chap03_04_03_44_riesz}\) corresponding to $f$.
\[\begin{equation} f(u) = (u, Jf) \quad (u \in X) . \label{chap03_04_03_47} \end{equation}\]Proposition
- (1) $J:X^{*} \rightarrow X$ is isometric,
- (2) $J$ is antilinear linear (or conjugate-linear), that is,
- (3) $J$ is surjection.
proof
(1)
\[\begin{equation} \|Jf\|_{X} = \|f\|_{X^{*}} . \label{chap03_04_03_48} \end{equation}\]Indeed,
\[\begin{equation} \forall u \in X, \ |f(u)| = |(u, Jf)| \le \|u\| \|Jf\| \label{chap03_04_03_49} . \end{equation}\]Moreover,
\[|f(Jf)| = |(Jf, Jf)| = \|Jf\|^{2} .\]It follows that the equality \(\eqref{chap03_04_03_49}\) holds when $u := Jf$. When $Jf \neq 0$,
\[\|f\| = \sup_{u \neq 0} \frac{ |f(u)| }{ \|u\| } = |Jf| .\]In case of $Jf = 0$, we have
\[|f(u)| \le \|Jf\| \|u\| = 0 .\]Since $f \equiv 0$,
\[\|Jf\| = \|f\| = 0 .\](2)
\[\begin{eqnarray} \forall u \in X, \ (f + g)(u) & = & f(u) + g(u) \nonumber \\ & = & (u, Jf) + (u, Jg) \nonumber \\ & = & (u, Jf + Jg) \nonumber \end{eqnarray}\]It follows that $J(f + g) = Jf + Jg$. Similary,
\[\begin{eqnarray} \forall u \in X, \ (\alpha f)(u) & = & \alpha f(u) \nonumber \\ & = & \alpha(u, Jf) \nonumber \\ & = & (u, \bar{\alpha}Jf) \nonumber \end{eqnarray}\]It follows that $J(\alpha f) = \bar{\alpha}Jf$.
(3)
Let $a \in X$ be fixed. Let $f(u) := (u, a)$. Then $f \in X^{*}$ and $Jf = a$.
Theorem 3.6
- $X$,
- Hilbert space
- $J:X^{*} \rightarrow X$,
- mapping defined in Riesz theorem
$J$ is antilinear, isometric and bijection.
proof
Immeadiately consequence of the above proposition.