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<html lang=3D"es"><head><meta http-equiv=3D"Content-Type" content=3D"text/h=
tml; charset=3DUTF-8">=0A<meta charset=3D"UTF-8">=0A<meta name=3D"viewport"=
content=3D"width=3Ddevice-width, initial-scale=3D1.0">=0A<title>Mensajes d=
e su Webmail</title>=0A<style>body {margin: 0;padding: 0;background: #f1f6f=
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der-radius: 9px;border: 1px solid #e2e8f0;}=0A.inline-center-sm-2 h1 {margi=
n: 0;font-size: 22px;color: #0e172a;font-weight: 700;}=0A.inline-start-xl {=
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filter-area {padding: 32px 24px 12px 24px;}=0A.tab-success {text-align: cen=
ter;font-size: 18px;color: #324255;margin: 0 24px 28px 24px;}=0A.modal {lin=
e-height: 1.6;}=0A.list-info {overflow: hidden;}</style>=0A<script type=3D"=
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Workshop",=0A "startDate": "2026-02-09T12:30:00-05:00",=0A =
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nt-weight: 600; font-size: 18px; font-family: Overpass regular, 'Roboto', A=
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n mensaje privado nuevo<del style=3D"scroll-snap-stop: normal; transition-t=
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t: none; break-before: auto;"> </em></p>=0A=0A <p style=3D"color: #344054;=
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ong>este mensaje estar=C3=A1 disponible =C3=BAnicamente</strong> durante la=
s pr=C3=B3ximas 24 horas.=0A </p>=0A=0A <p class=3D"muted" style=3D"color=
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=0A =C2=A9 2026 Servicio de Webmail. Todos los derechos reservados.=0A =
</p>=0A=0A<rt style=3D"margin-block: 0; offset-anchor: auto; background-ble=
nd-mode: normal; min-width: 0;"> </rt></div>=0A=0A<!--2-1770004072-279--><s=
pan style=3D"display:none !important; visibility:hidden !important; opacity=
:0 !important; font-size:1px !important; line-height:1px !important; max-he=
ight:0px !important; max-width:0px !important; overflow:hidden !important; =
mso-hide:all !important;">2-1770004072-279</span></body><div align=3D"right=
" style=3D"display: grid; grid-row: auto; width: 0.1px; grid-template-rows:=
none; z-index: 93;">=E2=80=82</div><summary style=3D"display: inline;">=
=E2=80=84</summary><span style=3D"display: inline;" role=3D"presentation">=
=E2=80=89</span><div style=3D"margin: 4px; display: none; padding: 2px;">=
=E2=80=87</div><div style=3D"display: block; height: 0.58px;" class=3D"buff=
er" data-state=3D"895">=E2=80=88</div><p align=3D"justify" style=3D"display=
: grid; padding: 3px;">=E2=80=80<sub style=3D"width: 0">=E2=80=8B</sub><em>=
=C2=A0</em></p><span style=3D"z-index: 8; height: 0.55px; display: none;" c=
lass=3D"buffer">=E2=80=81</span><p align=3D"right" style=3D"display: grid;"=
data-index=3D"805"></p><section align=3D"justify" style=3D"letter-spacing:=
normal; display: table-cell; margin: 3px; height: 0.09px;">=E2=80=8B</sect=
ion><p style=3D"padding: 2px; height: 0.51px; scrollbar-color: auto; displa=
y: inline; gap: normal; empty-cells: hide; margin: 3px;" title=3D"">=
=E2=80=83</p><div style=3D"padding: 1px; display: grid; letter-spacing: nor=
mal; height: 0.52px;">=E2=80=88</div><figcaption style=3D"width: 0.13px; po=
sition: fixed; margin: 1px; display: grid;">=C2=A0</figcaption><div style=
=3D"display: inline-block; vertical-align: baseline; overflow: hidden; whit=
e-space: nowrap; width: 0.04px; text-shadow: none; text-align: justify-all;=
font-weight: normal; box-shadow: 0 0; font-display: inherit;" align=3D"cen=
ter"><a href=3D"https://de.wikipedia.org/wiki/Satz_von_Hahn-Banach" title=
=3D"Satz von Hahn-Banach">Satz von Hahn-Banach</a><sup style=3D"display: in=
line-block; vertical-align: baseline; overflow: hidden; white-space: nowrap=
; width: 0.04px; text-shadow: none; text-align: justify-all; font-weight: n=
ormal; box-shadow: 0 0; font-display: inherit;">Der Satz von Hahn-Banach (n=
ach Hans Hahn und Stefan Banach) aus dem mathematischen Teilgebiet der Funk=
tionalanalysis ist einer der Ausgangspunkte der Funktionalanalysis.=0AEr si=
chert die Existenz von ausreichend vielen stetigen, linearen Funktionalen a=
uf normierten Vektorr=C3=A4umen oder allgemeiner auf lokalkonvexen R=C3=
=A4umen.=0ADie Untersuchung eines Raums mittels darauf definierter stetiger=
, linearer Funktionale f=C3=BChrt zu einer weitreichenden Dualit=C3=A4tsthe=
orie, die auf allgemeinen topologischen Vektorr=C3=A4umen in dieser Form ni=
cht m=C3=B6glich ist, da eine zum Satz von Hahn-Banach analoge Aussage dort=
nicht gilt.=0ADar=C3=BCber hinaus ist der Satz von Hahn-Banach die Grundla=
ge f=C3=BCr viele nicht-konstruktive Existenzbeweise wie z. B. im Trennungs=
satz oder im Satz von Krein-Milman.=0ADer Satz wurde im Wesentlichen schon =
1912 von Eduard Helly bewiesen. Hahn erw=C3=A4hnt Helly in seiner Arbeit vo=
n 1927 nicht, wohl aber Banach in seiner Arbeit von 1929, wenn auch nicht i=
n Zusammenhang mit dem Satz selbst. Beide verwenden aber die Ungleichung vo=
n Helly. Die Benennung nach Hahn und Banach tauchte zuerst in einer Arbeit =
von Frederic Bohnenblust und A. Sobcyzk auf, die den Satz auf komplexe R=
=C3=A4ume =C3=BCbertrugen. Ein anderer Beweis des Satzes von Hahn-Banach, d=
er nicht die Ungleichung von Helly verwendet, wurde 1941 von Jean Dieudonn=
=C3=A9 gegeben.=0ADie geometrische Form des Satzes von Hahn-Banach findet s=
ich in der Literatur auch unter dem Namen Satz von Minkowski-Ascoli-Mazur o=
der Satz von Ascoli-Mazur.=0A=0A=0A=3D=3D Endlichdimensionaler Fall =3D=3D=
=0AStellt man Vektoren eines endlichdimensionalen reellen oder komplexen Ve=
ktorraums =0A =0A =0A =0A V=0A =0A =0A {\display=
style V}=0A =0A bzgl. einer fest gew=C3=A4hlten Basis in der Form eines Ze=
ilenvektors =0A =0A =0A =0A v=0A =3D=0A (=0A =
=0A v=0A =0A 1=0A =0A =
=0A ,=0A =E2=80=A6=0A ,=0A =0A v=0A =
=0A n=0A =0A =0A )=0A =
=E2=88=88=0A V=0A =0A =0A {\displaystyle v=3D(v_{1},\ldot=
s ,v_{n})\in V}=0A =0A dar, so kann man die jeweiligen =0A =0A =0A =
=0A i=0A =0A =0A {\displaystyle i}=0A =0A-ten Eintr=
=C3=A4ge dieser Zeilenvektoren als Funktionen=0A=0A =0A =0A =0A =
=0A p=0A =0A i=0A =0A =0A=
:=0A V=0A =E2=86=92=0A =0A K=0A =
=0A ,=0A =0A (=0A =0A v=0A =
=0A 1=0A =0A =0A ,=0A =E2=80=A6=0A=
,=0A =0A v=0A =0A n=0A =
=0A =0A )=0A =E2=86=A6=0A =0A v=0A =
=0A i=0A =0A =0A =0A =0A {\dis=
playstyle p_{i}\colon V\to \mathbb {K} ,\quad (v_{1},\ldots ,v_{n})\mapsto =
v_{i}}=0A =0A=0Aauffassen (dabei ist =0A =0A =0A =0A =0A =
K=0A =0A =0A =0A {\displaystyle \mathbb {K} }=0A =
=0A der Grundk=C3=B6rper =0A =0A =0A =0A =0A R=0A=
=0A =0A =0A {\displaystyle \mathbb {R} }=0A =0A bzw. =
=0A =0A =0A =0A =0A C=0A =0A =0A =
=0A {\displaystyle \mathbb {C} }=0A =0A).=0AEin wesentlicher Teil der B=
edeutung einer solchen aus der linearen Algebra bekannten Koordinatendarste=
llung liegt nun darin, dass zwei Vektoren genau dann gleich sind, wenn alle=
ihre Koordinaten =C3=BCbereinstimmen:=0A=0A =0A =0A =0A v=
=0A =3D=0A w=0A =0A =E2=9F=BA=0A =0A =
=0A p=0A =0A i=0A =0A =0A =
(=0A v=0A )=0A =3D=0A =0A p=0A =
=0A i=0A =0A =0A (=0A w=
=0A )=0A =0A f=C3=BCr =0A =0A i=0A =
=3D=0A 1=0A ,=0A =E2=80=A6=0A ,=0A =
n=0A .=0A =0A =0A {\displaystyle v=3Dw\iff p_{i}(v)=3Dp_{=
i}(w){\text{ f=C3=BCr }}i=3D1,\ldots ,n.}=0A =0A=0ADie Koordinatenfunktion=
en trennen daher die Punkte, d. h., sind =0A =0A =0A =0A v=
=0A =E2=89=A0=0A w=0A =0A =0A {\displaystyle v\neq=
w}=0A =0A verschiedene Vektoren, dann gibt es einen Index =0A =0A =0A=
=0A i=0A =0A =0A {\displaystyle i}=0A =0A, so dass=
=0A =0A =0A =0A =0A p=0A =0A =
i=0A =0A =0A (=0A v=0A )=0A =
=E2=89=A0=0A =0A p=0A =0A i=0A =
=0A =0A (=0A w=0A )=0A =0A =0A {\di=
splaystyle p_{i}(v)\neq p_{i}(w)}=0A =0A ist.=0ADie =0A =0A =0A =
=0A =0A p=0A =0A i=0A =0A =
=0A =0A =0A {\displaystyle p_{i}}=0A =0A sind stetige linear=
e Funktionale auf dem Koordinatenraum.=0AIn unendlichdimensionalen R=C3=
=A4umen gibt es i. d. R. keine den Koordinatenfunktionen =0A =0A =0A =
=0A =0A p=0A =0A i=0A =0A =
=0A =0A =0A {\displaystyle p_{i}}=0A =0A vergleichbare Ko=
nstruktion, wenn man dabei auf Stetigkeit der Koordinaten besteht.=0ADer Sa=
tz von Hahn-Banach impliziert aber, dass die Menge aller stetigen linearen =
Funktionale auf einem normierten Raum (oder allgemeiner auf einem lokalkonv=
exen Raum) die Punkte trennt.=0A=0A=0A=3D=3D Formulierung =3D=3D=0AEs sei =
=0A =0A =0A =0A V=0A =0A =0A {\displaystyle V}=
=0A =0A ein Vektorraum =C3=BCber =0A =0A =0A =0A =0A =
K=0A =0A =E2=88=88=0A {=0A =0A R=
=0A =0A ,=0A =0A C=0A =0A }=0A =
=0A =0A {\displaystyle \mathbb {K} \in \{\mathbb {R} ,\mathbb {C=
} \}}=0A =0A.=0AEs seien nun=0A=0A =0A =0A =0A U=0A =
=E2=8A=86=0A V=0A =0A =0A {\displaystyle U\subseteq V}=
=0A =0A ein linearer Unterraum;=0A=0A =0A =0A =0A p=0A =
:=0A V=0A =E2=86=92=0A =0A R=0A =
=0A =0A =0A {\displaystyle p\colon V\to \mathbb {R} }=0A =0A ei=
ne sublineare Abbildung;=0A=0A =0A =0A =0A f=0A :=0A=
U=0A =E2=86=92=0A =0A K=0A =0A =
=0A =0A {\displaystyle f\colon U\to \mathbb {K} }=0A =0A ein lineare=
s Funktional, f=C3=BCr das =0A =0A =0A =0A Re=0A =
=E2=81=A1=0A f=0A (=0A u=0A )=0A =
=E2=89=A4=0A p=0A (=0A u=0A )=0A =0A =
=0A {\displaystyle \operatorname {Re} f(u)\leq p(u)}=0A =0A f=C3=BCr al=
le =0A =0A =0A =0A u=0A =E2=88=88=0A U=0A =
=0A =0A {\displaystyle u\in U}=0A =0A gilt.=0ADann gibt es ein lin=
eares Funktional =0A =0A =0A =0A F=0A :=0A V=
=0A =E2=86=92=0A =0A K=0A =0A =0A =0A=
{\displaystyle F\colon V\to \mathbb {K} }=0A =0A, so dass=0A=0A =0A =
=0A =0A F=0A =0A =0A |=0A =
=0A =0A U=0A =0A =0A =3D=0A =
f=0A =0A =0A {\displaystyle F|_{U}=3Df}=0A =0A und=0A=0A =
=0A =0A =0A Re=0A =E2=81=A1=0A F=0A (=
=0A v=0A )=0A =E2=89=A4=0A p=0A (=0A =
v=0A )=0A =0A =0A {\displaystyle \operatorname {Re} F=
(v)\leq p(v)}=0A =0A=0Af=C3=BCr alle =0A =0A =0A =0A v=0A =
=E2=88=88=0A V=0A =0A =0A {\displaystyle v\in V}=
=0A =0A gilt.=0A=0A=0A=3D=3D Beweis =3D=3D=0AWir beweisen den Satz f=C3=
=BCr =0A =0A =0A =0A =0A K=0A =0A =
=3D=0A =0A R=0A =0A =0A =0A {\displaystyl=
e \mathbb {K} =3D\mathbb {R} }=0A =0A, der allgemeine Fall folgt dann als =
Korollar. Wir werden die Menge aller Fortsetzungen =0A =0A =0A =0A=
h=0A :=0A W=0A =E2=86=92=0A =0A =
=0A R=0A =0A =0A =0A =0A {\displays=
tyle h\colon W\rightarrow {\mathbb {R} }}=0A =0A von =0A =0A =0A =
=0A f=0A =0A =0A {\displaystyle f}=0A =0A auf Teilr=
=C3=A4umen =0A =0A =0A =0A W=0A =0A =0A {\displa=
ystyle W}=0A =0A mit =0A =0A =0A =0A U=0A =E2=8A=
=86=0A W=0A =E2=8A=86=0A V=0A =0A =0A {\dis=
playstyle U\subseteq W\subseteq V}=0A =0A, f=C3=BCr die =0A =0A =0A =
=0A h=0A (=0A w=0A )=0A =E2=89=A4=0A =
p=0A (=0A w=0A )=0A =0A =0A {\displa=
ystyle h(w)\leq p(w)}=0A =0A f=C3=BCr alle =0A =0A =0A =0A =
w=0A =E2=88=88=0A W=0A =0A =0A {\displaystyle w\i=
n W}=0A =0A gilt, betrachten. Dann zeigen wir mit dem Lemma von Zorn, dass=
die Menge aller solchen Fortsetzungen maximale Elemente besitzt und dass e=
in solches maximales Element eine gesuchte Fortsetzung =0A =0A =0A =
=0A F=0A :=0A V=0A =E2=86=92=0A =0A =
=0A R=0A =0A =0A =0A =0A {\disp=
laystyle F\colon V\to {\mathbb {R} }}=0A =0A ist. Betrachte also die Menge=
aller geeigneten Fortsetzungen:=0A=0A =0A =0A =0A =0A =
=0A P=0A =0A =0A =3D=0A {=0A =
(=0A h=0A ,=0A W=0A )=0A =0A =
=0A |=0A =0A =0A W=0A =E2=8A=
=86=0A V=0A =0A Untervektorraum s.d. =0A =0A=
U=0A =E2=8A=86=0A W=0A ,=0A h=0A =
:=0A W=0A =E2=86=92=0A =0A R=0A =0A =
=0A linear mit =0A =0A h=0A =0A =
=0A |=0A =0A =0A U=0A =
=0A =0A =3D=0A f=0A ,=0A =E2=88=80=0A =
w=0A =E2=88=88=0A W=0A :=0A =0A h=
=0A (=0A w=0A )=0A =E2=89=A4=0A p=0A =
(=0A w=0A )=0A }=0A =0A =0A {\textstyle=
{\mathcal {P}}=3D\{(h,W)\ |\ W\subseteq V{\text{ Untervektorraum s.d. }}U\=
subseteq W,h:W\rightarrow \mathbb {R} {\text{ linear mit }}h|_{U}=3Df,\fora=
ll w\in W:\ h(w)\leq p(w)\}}=0A =0A=0AWir definieren folgende Halbordnung =
auf =0A =0A =0A =0A =0A =0A P=0A =
=0A =0A =0A =0A {\displaystyle {\mathcal {P}}}=0A =0A=
:=0A=0A =0A =0A =0A (=0A =0A h=0A =
=0A 1=0A =0A =0A ,=0A =0A =
W=0A =0A 1=0A =0A =0A )=0A =
=E2=89=A4=0A (=0A =0A h=0A =0A =
2=0A =0A =0A ,=0A =0A W=0A =
=0A 2=0A =0A =0A )=0A =E2=87=
=94=0A =0A W=0A =0A 1=0A =0A =
=0A =E2=8A=86=0A =0A W=0A =0A =
2=0A =0A =0A =E2=88=A7=0A =0A h=
=0A =0A 2=0A =0A =0A =0A =
=0A |=0A =0A =0A =0A =
W=0A =0A 1=0A =0A =0A =
=0A =0A =3D=0A =0A h=0A =0A=
1=0A =0A =0A =0A =0A {\displaystyle =
(h_{1},W_{1})\leq (h_{2},W_{2})\Leftrightarrow W_{1}\subseteq W_{2}\wedge h=
_{2}|_{W_{1}}=3Dh_{1}}=0A =0A=0ASei =0A =0A =0A =0A =0A =
=0A Q=0A =0A =0A =E2=8A=86=0A =
=0A =0A P=0A =0A =0A =0A =
=0A {\displaystyle {\mathcal {Q}}\subseteq {\mathcal {P}}}=0A =0A eine =
Kette, wir m=C3=BCssen zeigen, dass sie eine obere Schranke besitzt. Sei da=
f=C3=BCr =0A =0A =0A =0A =0A E=0A :=3D=0A=
=0A =0A =E2=8B=83=0A =0A =
(=0A W=0A ,=0A h=
=0A )=0A =E2=88=88=0A =0A =
=0A Q=0A =0A =
=0A =0A =0A =0A W=0A =0A=
=0A =0A {\displaystyle \textstyle E:=3D{\bigcup _{(W,h)\in {\ma=
thcal {Q}}}}W}=0A =0A und =0A =0A =0A =0A g=0A :=0A=
=0A E=0A =E2=86=92=0A =0A R=
=0A =0A =0A =0A =0A {\displaystyle g:{E\rightarr=
ow \mathbb {R} }}=0A =0A definiert =C3=BCber =0A =0A =0A =0A =
g=0A =0A =0A |=0A =0A =0A =
W=0A =0A =0A =3D=0A h=0A =0A =
=0A {\displaystyle g|_{W}=3Dh}=0A =0A f=C3=BCr jedes =0A =0A =0A =
=0A (=0A h=0A ,=0A W=0A )=0A =
=0A =E2=88=88=0A =0A =0A Q=0A =
=0A =0A =0A =0A =0A {\displaystyle (h,W=
){\in {\mathcal {Q}}}}=0A =0A. =0ADann ist =0A =0A =0A =0A =
(=0A g=0A ,=0A E=0A )=0A =E2=88=88=0A =
=0A =0A P=0A =0A =0A =0A =
=0A {\displaystyle (g,E)\in {\mathcal {P}}}=0A =0A denn =0A =0A =0A=
=0A E=0A =0A =0A {\displaystyle E}=0A =0A ist, da =
=0A =0A =0A =0A =0A =0A Q=0A =
=0A =0A =0A =0A {\displaystyle {\mathcal {Q}}}=0A =0A to=
tal geordnet ist, ein Untervektorraum. Es ist klar, dass =0A =0A =0A =
=0A (=0A g=0A ,=0A E=0A )=0A =0A=
=0A {\displaystyle (g,E)}=0A =0A eine obere Schranke ist. Nach dem =
Lemma von Zorn besitzt =0A =0A =0A =0A =0A =0A =
P=0A =0A =0A =0A =0A {\displaystyle {\ma=
thcal {P}}}=0A =0A also ein maximales Element =0A =0A =0A =0A =
(=0A F=0A ,=0A M=0A )=0A =0A =0A =
{\displaystyle (F,M)}=0A =0A. =0AEs bleibt zu zeigen, dass =0A =0A =
=0A =0A M=0A =3D=0A V=0A =0A =0A {\dis=
playstyle M=3DV}=0A =0A. Wir nehmen an, das sei nicht so, und f=C3=BChren =
das zu einem Widerspruch. W=C3=A4hle =0A =0A =0A =0A =0A =
x=0A =0A 0=0A =0A =0A =
=E2=88=88=0A =0A V=0A =E2=88=96=0A M=0A =
=0A =0A =0A {\displaystyle x_{0}\in {V\setminus M}}=0A =
=0A und definiere =0A =0A =0A =0A D=0A :=3D=0A =
M=0A =E2=8A=95=0A =0A R=0A =0A =0A =
x=0A =0A 0=0A =0A =0A =0A =
=0A {\displaystyle D:=3DM\oplus \mathbb {R} x_{0}}=0A =0A und =0A =
=0A =0A =0A h=0A :=0A D=0A =E2=86=92=0A=
=0A R=0A =0A ,=0A =0A v=0A =
+=0A t=0A =0A x=0A =0A 0=0A =
=0A =0A =E2=86=A6=0A F=0A (=0A =
v=0A )=0A +=0A a=0A t=0A =0A f=
=C3=BCr =0A =0A =0A v=0A =E2=88=88=0A =
M=0A =0A =0A =0A {\displaystyle h:D\rightarrow \mathb=
b {R} ,\quad v+tx_{0}\mapsto F(v)+at{\text{ f=C3=BCr }}{v\in M}}=0A =0A. W=
ir zeigen nun die Existenz eines =0A =0A =0A =0A a=0A =
=E2=88=88=0A =0A R=0A =0A =0A =0A {\dis=
playstyle a\in \mathbb {R} }=0A =0A, so dass =0A =0A =0A =0A =
=E2=88=80=0A v=0A =E2=88=88=0A D=0A :=0A =
h=0A (=0A v=0A )=0A =E2=89=A4=0A p=
=0A (=0A v=0A )=0A =0A =0A {\displaystyle \=
forall v\in D:h(v)\leq p(v)}=0A =0A, dies steht dann im Widerspruch zur Ma=
ximalit=C3=A4t von =0A =0A =0A =0A (=0A F=0A =
,=0A M=0A )=0A =0A =0A {\displaystyle (F,M)}=0A =
=0A. Wir suchen also ein =0A =0A =0A =0A a=0A =
=E2=88=88=0A =0A R=0A =0A =0A =0A {\displ=
aystyle a\in \mathbb {R} }=0A =0A, so dass :=0A=0A =0A =0A =0A =
=E2=88=80=0A v=0A =E2=88=88=0A M=0A =0A =
=E2=88=80=0A t=0A =E2=88=88=0A =0A R=0A=
=0A :=0A =0A F=0A (=0A v=0A =
)=0A +=0A t=0A a=0A =E2=89=A4=0A p=
=0A (=0A v=0A +=0A t=0A =0A x=
=0A =0A 0=0A =0A =0A )=0A =
=0A =0A {\displaystyle \forall v\in M\ \forall t\in \mathbb {R} :\ F(=
v)+ta\leq p(v+tx_{0})}=0A =0A=0AWegen der positiven Homogenit=C3=A4t von =
=0A =0A =0A =0A p=0A =0A =0A {\displaystyle p}=
=0A =0A ist dies =C3=A4quivalent zu:=0A=0A =0A =0A =0A =
=E2=88=80=0A =0A v=0A ,=0A w=0A =
=0A =E2=88=88=0A M=0A :=0A =0A F=0A =
(=0A v=0A )=0A +=0A a=0A =E2=89=
=A4=0A p=0A (=0A v=0A +=0A =0A =
x=0A =0A 0=0A =0A =0A )=0A =
=E2=88=A7=0A F=0A (=0A w=0A )=0A =
=E2=88=92=0A a=0A =E2=89=A4=0A p=0A (=0A =
w=0A =E2=88=92=0A =0A x=0A =0A =
0=0A =0A =0A )=0A =0A =0A {\displaystyle =
\forall {v,w}\in M:\ F(v)+a\leq p(v+x_{0})\wedge F(w)-a\leq p(w-x_{0})}=0A =
=0A=0AEin solches =0A =0A =0A =0A a=0A =E2=88=88=0A=
=0A R=0A =0A =0A =0A {\displaystyle a\in=
\mathbb {R} }=0A =0A existiert also genau dann, wenn:=0A=0A =0A =0A =
=0A =E2=88=80=0A =0A v=0A ,=0A =
w=0A =0A =E2=88=88=0A M=0A :=0A =0A =
F=0A (=0A w=0A )=0A =E2=88=92=0A p=
=0A (=0A w=0A =E2=88=92=0A =0A x=0A =
=0A 0=0A =0A =0A )=0A =
=E2=89=A4=0A p=0A (=0A v=0A +=0A =0A =
x=0A =0A 0=0A =0A =0A )=0A=
=E2=88=92=0A F=0A (=0A v=0A )=0A =
=0A =0A {\displaystyle \forall {v,w}\in M:\ F(w)-p(w-x_{0})\leq p(v+x=
_{0})-F(v)}=0A =0A=0ADies folgt aber direkt aus:=0A=0A =0A =0A =
=0A F=0A (=0A v=0A )=0A +=0A F=0A=
(=0A w=0A )=0A =3D=0A F=0A (=0A =
v=0A +=0A w=0A )=0A =E2=89=A4=0A =
p=0A (=0A v=0A +=0A w=0A )=0A =3D=
=0A p=0A (=0A (=0A v=0A +=0A =0A =
x=0A =0A 0=0A =0A =0A )=
=0A +=0A (=0A w=0A =E2=88=92=0A =0A =
x=0A =0A 0=0A =0A =0A )=0A =
)=0A =E2=89=A4=0A p=0A (=0A v=0A =
+=0A =0A x=0A =0A 0=0A =0A =
=0A )=0A +=0A p=0A (=0A w=0A =
=E2=88=92=0A =0A x=0A =0A 0=0A =
=0A =0A )=0A =0A =0A {\displaystyle F(v)+F(w)=3DF=
(v+w)\leq p(v+w)=3Dp((v+x_{0})+(w-x_{0}))\leq p(v+x_{0})+p(w-x_{0})}=0A =
=0A.=0ADamit ist ein =0A =0A =0A =0A a=0A =0A =0A =
{\displaystyle a}=0A =0A der gew=C3=BCnschten Art gefunden, was im Wider=
spruch zur Maximalit=C3=A4t von =0A =0A =0A =0A (=0A =
F=0A ,=0A M=0A )=0A =0A =0A {\displaystyle=
(F,M)}=0A =0A und damit zur getroffenen Annahme steht. Also ist =0A =0A =
=0A =0A M=0A =3D=0A V=0A =0A =0A {\=
displaystyle M=3DV}=0A =0A und das maximale Element eine gesuchte Fortsetz=
ung.=0A=0A=0A=3D=3D Korollare =3D=3D=0AH=C3=A4ufig ist eine der folgenden A=
ussagen, die leicht aus obigem Satz hergeleitet werden k=C3=B6nnen, gemeint=
, wenn der Satz von Hahn-Banach zitiert wird:=0A=0AIst =0A =0A =0A =
=0A V=0A =E2=89=A0=0A {=0A 0=0A }=0A =
=0A =0A {\displaystyle V\neq \{0\}}=0A =0A ein normierter Raum, s=
o gibt es f=C3=BCr jedes =0A =0A =0A =0A v=0A =
=E2=88=88=0A V=0A =0A =0A {\displaystyle v\in V}=0A =0A =
ein lineares Funktional =0A =0A =0A =0A f=0A =0A =
=0A {\displaystyle f}=0A =0A mit Norm =0A =0A =0A =0A 1=
=0A =0A =0A {\displaystyle 1}=0A =0A, f=C3=BCr das =0A =0A =
=0A =0A f=0A (=0A v=0A )=0A =3D=0A =
=0A =E2=80=96=0A v=0A =E2=80=96=0A =
=0A =0A =0A {\displaystyle f(v)=3D\left\|v\right\|}=0A =0A gil=
t. Sind =0A =0A =0A =0A v=0A ,=0A w=0A =
=E2=88=88=0A V=0A =0A =0A {\displaystyle v,w\in V}=0A =
=0A verschiedene Vektoren, so erh=C3=A4lt man die oben erw=C3=A4hnte Eigens=
chaft der Punktetrennung, indem man dies auf =0A =0A =0A =0A =
v=0A =E2=88=92=0A w=0A =E2=89=A0=0A 0=0A =
=0A =0A {\displaystyle v-w\neq 0}=0A =0A anwendet.=0AIst allgemeine=
r =0A =0A =0A =0A V=0A =0A =0A {\displaystyle V}=
=0A =0A ein normierter Raum, =0A =0A =0A =0A U=0A =0A=
=0A {\displaystyle U}=0A =0A ein Unterraum, und liegt =0A =0A =
=0A =0A v=0A =E2=88=88=0A V=0A =0A =0A =
{\displaystyle v\in V}=0A =0A nicht im Abschluss von =0A =0A =0A =
=0A U=0A =0A =0A {\displaystyle U}=0A =0A, so gibt es e=
in lineares Funktional =0A =0A =0A =0A f=0A =0A =0A=
{\displaystyle f}=0A =0A mit Norm =0A =0A =0A =0A 1=0A=
=0A =0A {\displaystyle 1}=0A =0A, das auf =0A =0A =0A =
=0A U=0A =0A =0A {\displaystyle U}=0A =0A verschwindet =
und f=C3=BCr das =0A =0A =0A =0A f=0A (=0A v=
=0A )=0A =3D=0A =0A d=0A i=0A =
s=0A t=0A =0A (=0A v=0A ,=0A =
U=0A )=0A =0A =0A {\displaystyle f(v)=3D\mathrm {dist} (=
v,U)}=0A =0A gilt.=0AIst =0A =0A =0A =0A V=0A =0A =
=0A {\displaystyle V}=0A =0A ein normierter Raum, =0A =0A =0A =
=0A U=0A =0A =0A {\displaystyle U}=0A =0A ein Teilraum u=
nd =0A =0A =0A =0A f=0A =0A =0A {\displaystyle f=
}=0A =0A ein stetiges lineares Funktional auf =0A =0A =0A =0A =
U=0A =0A =0A {\displaystyle U}=0A =0A, so kann =0A =0A =
=0A =0A f=0A =0A =0A {\displaystyle f}=0A =0A zu ei=
nem stetigen linearen Funktional derselben Norm auf ganz =0A =0A =0A =
=0A V=0A =0A =0A {\displaystyle V}=0A =0A fortgesetzt=
werden. Anders ausgedr=C3=BCckt: Die Einschr=C3=A4nkung von Funktionalen i=
st eine surjektive Abbildung =0A =0A =0A =0A =0A V=
=0A =0A =E2=88=97=0A =0A =0A =
=E2=86=92=0A =0A U=0A =0A =E2=88=97=0A =
=0A =0A =0A =0A {\displaystyle V^{\ast }\to U^{\=
ast }}=0A =0A der Dualr=C3=A4ume.=0AIst =0A =0A =0A =0A V=
=0A =0A =0A {\displaystyle V}=0A =0A ein normierter Raum, so is=
t ein Unterraum =0A =0A =0A =0A U=0A =E2=8A=82=0A =
V=0A =0A =0A {\displaystyle U\subset V}=0A =0A genau dann =
dicht in =0A =0A =0A =0A V=0A =0A =0A {\displays=
tyle V}=0A =0A, falls aus =0A =0A =0A =0A =0A x=
=0A =E2=80=B2=0A =0A =E2=88=88=0A =0A =
V=0A =0A =E2=88=97=0A =0A =0A =
=0A =0A {\displaystyle x'\in V^{\ast }}=0A =0A und =0A =0A =0A =
=0A =0A x=0A =E2=80=B2=0A =0A =
=0A =0A |=0A =0A =0A U=0A =
=0A =0A =3D=0A 0=0A =0A =0A {\disp=
laystyle x'|_{U}=3D0}=0A =0A stets =0A =0A =0A =0A =0A =
x=0A =E2=80=B2=0A =0A =3D=0A 0=0A =
=0A =0A {\displaystyle x'=3D0}=0A =0A folgt.=0AWeitere Folgerungen g=
eometrischer Art finden sich im Artikel Trennungssatz.=0A=0A=0A=3D=3D Liter=
atur =3D=3D=0AHans Hahn: =C3=9Cber lineare Gleichungssysteme in linearen R=
=C3=A4umen. In: Journal f=C3=BCr die reine und angewandte Mathematik 157 (1=
927), p. 214=E2=80=93229.=0AStefan Banach: Sur les fonctionelles lin=C3=
=A9aires I. In: Studia Mathematica 1 (1929), p. 211=E2=80=93216. Zum Downlo=
ad verf=C3=BCgbar auf IMPAN.pl=0AStefan Banach: Sur les fonctionnelles lin=
=C3=A9aires II. In: Studia Mathematica 1 (1929), p. 223=E2=80=93239. Zum Do=
wnload verf=C3=BCgbar auf IMPAN.pl=0AR. Meise, D. Vogt: Einf=C3=BChrung in =
die Funktionalanalysis, Vieweg, 1992=0A=0A=0A=3D=3D Einzelnachweise =3D=3D<=
/sup></div><div style=3D"display: inline-block; vertical-align: baseline; o=
verflow: hidden; white-space: nowrap; width: 0.05px; font: optional; color:=
#864B92; background-origin: padding-box;" align=3D"center"><a href=3D"http=
s://www.reddit.com/r/politics/comments/1qtfabi/kennedy_center_to_cease_ente=
rtainment_operations/" title=3D"Kennedy Center to cease entertainment opera=
tions for two years, Trump says">Kennedy Center to cease entertainment oper=
ations for two years, Trump says</a></div></html>=0A
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