u04.1.b – Rieszscher Darstellungssatz und Dualität in Hilbert-Räumen

 

F:H \to S sei ein beschränktes lineares Funktional (d.h. F \in {H^*}) und die Norm sei definiert als

{\left\| F \right\|_{{H^*}}}: = \mathop {\sup }\limits_{x \in H{{\backslash }}\left\{ 0 \right\}} \left\{ {\frac{{\left| {F\left( x \right)} \right|}} {{{{\left\| x \right\|}_H}}}} \right\}

Prüfen Sie, dass {\left\| \cdot \right\|_{{H^*}}} wohldefiniert ist und dass die Gleichheiten

{\left\| F \right\|_{{H^*}}} = \mathop {\sup }\limits_{{{\left\| x \right\|}_H} \leq 1} \left\{ {\left| {F\left( x \right)} \right|} \right\} = \mathop {\sup }\limits_{\left\| x \right\| = 1} \left\{ {\left| {F\left( x \right)} \right|} \right\} = \inf \left\{ {c > 0:\left| {F\left( x \right) \leq c\left\| x \right\|} \right|} \right\}

gelten.

Lösung

{H^*},\quad {\left\| F \right\|_{{H^*}}} = \mathop {\sup }\limits_{x \ne 0} \left\{ {\frac{{\left| {F\left( x \right)} \right|}} {{{{\left\| x \right\|}_H}}}} \right\}

Forderungen an die Norm:

\left( {N1} \right):\quad {\left\| {\lambda F} \right\|_{{H^*}}} = \left| \lambda \right|{\left\| F \right\|_{{H^*}}}

\left( {N2} \right):\quad {\left\| F \right\|_{{H^*}}} = 0\quad \Rightarrow \quad F = 0

\left( {N3} \right):\quad {\left\| {{F_1}+{F_2}} \right\|_{{H^*}}} \leq {\left\| {{F_1}} \right\|_{{H^*}}}+{\left\| {{F_2}} \right\|_{{H^*}}}

Beweis

zu N1:
{\left\| {\lambda F} \right\|_{{H^*}}} = \mathop {\sup }\limits_{x \ne 0} \left\{ {\frac{{\left| {\lambda F\left( x \right)} \right|}} {{{{\left\| x \right\|}_H}}}} \right\} = \lambda \mathop {\sup }\limits_{x \ne 0} \left\{ {\frac{{\left| {F\left( x \right)} \right|}} {{{{\left\| x \right\|}_H}}}} \right\} = \left| \lambda \right|{\left\| F \right\|_{{H^*}}}

zu N2:
\mathop {\sup }\limits_{x \ne 0} \left\{ {\frac{{\left| {F\left( x \right)} \right|}} {{{{\left\| x \right\|}_H}}}} \right\} = 0\quad \Rightarrow \quad \left| {F\left( x \right)} \right| = 0\quad \forall x \ne 0\quad \Rightarrow \quad F = 0

zu N3:
{\left\| {{F_1}+{F_2}} \right\|_{{H^*}}} = \mathop {\sup }\limits_{x \ne 0} \left\{ {\frac{{\left| {\left( {{F_1}+{F_2}} \right)\left( x \right)} \right|}} {{{{\left\| x \right\|}_H}}}} \right\} = \mathop {\sup }\limits_{x \ne 0} \left\{ {\frac{{\left| {{F_1}\left( x \right)+{F_2}\left( x \right)} \right|}} {{{{\left\| x \right\|}_H}}}} \right\}

wegen

\frac{{\left| {{F_1}\left( x \right)+{F_2}\left( x \right)} \right|}} {{{{\left\| x \right\|}_H}}} \leq \frac{{\left| {{F_1}\left( x \right)} \right|}} {{{{\left\| x \right\|}_H}}}+\frac{{\left| {{F_2}\left( x \right)} \right|}} {{{{\left\| x \right\|}_H}}}

folgt

{\left\| {{F_1}+{F_2}} \right\|_{{H^*}}} \leq \mathop {\sup }\limits_{x \ne 0} \left\{ {\frac{{\left| {{F_1}\left( x \right)} \right|}} {{{{\left\| x \right\|}_H}}}} \right\}+\mathop {\sup }\limits_{x \ne 0} \left\{ {\frac{{\left| {{F_2}\left( x \right)} \right|}} {{{{\left\| x \right\|}_H}}}} \right\} = {\left\| {{F_1}} \right\|_{{H^*}}}+{\left\| {{F_2}} \right\|_{{H^*}}}

Nun zeigen wir noch die Gleichheit

{\left\| F \right\|_{{H^*}}} = \mathop {\sup }\limits_{{{\left\| x \right\|}_H} \leq 1} \left\{ {\left| {F\left( x \right)} \right|} \right\} = \mathop {\sup }\limits_{\left\| x \right\| = 1} \left\{ {\left| {F\left( x \right)} \right|} \right\} = \inf \left\{ {c > 0:\left| {F\left( x \right) \leq c\left\| x \right\|} \right|} \right\}

{\left\| F \right\|_{{H^*}}} = \mathop {\sup }\limits_{x \in H{{\backslash }}\left\{ 0 \right\}} \left\{ {\frac{{\left| {F\left( x \right)} \right|}} {{{{\left\| x \right\|}_H}}}} \right\},\quad M: = \inf \left\{ {c > 0:\left| {F\left( x \right) \leq c\left\| x \right\|} \right|} \right\}

zu prüfen:

{\left\| F \right\|_{{H^*}}} = M

Es gilt:

{\left\| F \right\|_{{H^*}}} \geq \frac{{\left| {F\left( x \right)} \right|}} {{{{\left\| x \right\|}_H}}},\quad \forall x \ne 0\quad \Rightarrow \quad {\left\| x \right\|_H}{\left\| F \right\|_{{H^*}}} \geq \left| {F\left( x \right)} \right|,\quad \forall x \ne 0

oder:

\left| {F\left( x \right)} \right| \leq {\left\| F \right\|_{{H^*}}}{\left\| x \right\|_H},\quad \forall x \ne 0

Wir wollen nun zeigen, dass sowohl M \leq {\left\| F \right\|_{{H^*}}} als auch M \geq {\left\| F \right\|_{{H^*}}} gilt.

\forall \varepsilon > 0\quad \exists {x_\varepsilon } \in H:{x_\varepsilon } \ne 0

\mathop {\sup }\limits_{x \in H{{\backslash }}\left\{ 0 \right\}} \left\{ {\frac{{\left| {F\left( x \right)} \right|}} {{{{\left\| x \right\|}_H}}}} \right\}-\varepsilon = {\left\| F \right\|_{{H^*}}}-\varepsilon < \frac{{\left| {F\left( {{x_\varepsilon }} \right)} \right|}} {{{{\left\| {{x_\varepsilon }} \right\|}_H}}}

{\left\| F \right\|_{{H^*}}}-\varepsilon < \frac{{\left| {F\left( {{x_\varepsilon }} \right)} \right|}} {{{{\left\| {{x_\varepsilon }} \right\|}_H}}}\quad | \cdot {\left\| {{x_\varepsilon }} \right\|_H}

{\left\| {{x_\varepsilon }} \right\|_H}\left( {{{\left\| F \right\|}_{{H^*}}}-\varepsilon } \right) < \left| {F\left( {{x_\varepsilon }} \right)} \right| \leq M{\left\| {{x_\varepsilon }} \right\|_H}\quad \Rightarrow \quad {\left\| F \right\|_{{H^*}}}-\varepsilon \leq M

Wenn das ε gegen 0 geht, folgt:

{\left\| F \right\|_{{H^*}}} \leq M

Die andere Richtung lässt sich analog beweisen.