Editing Hahn–Banach theorem (section)

==Notes==

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* <ref group=note name="GeometricIllustration">Geometric illustration:
The geometric idea of the above proof can be fully presented in the case of <math>X = \R^2, M = \{(x, 0) : x \in \R\}.</math>

First, define the simple-minded extension <math>f_0(x, y) = f(x),</math> It doesn't work, since maybe <math>f_0 \leq p</math>. But it is a step in the right direction. <math>p-f_0</math> is still convex, and <math>p-f_0 \geq f-f_0.</math> Further, <math>f-f_0</math> is identically zero on the x-axis. Thus we have reduced to the case of <math>f = 0, p \geq 0</math> on the x-axis.

If <math>p \geq 0</math> on <math>\R^2,</math> then we are done. Otherwise, pick some <math>v \in \R^2,</math> such that <math>p(v) < 0.</math>

The idea now is to perform a simultaneous bounding of <math>p</math> on <math>v + M</math> and <math>-v+M</math> such that <math>p \geq b</math> on <math>v+M</math> and <math>p \geq -b</math> on <math>-v+M,</math> then defining <math>\tilde f(w + rv) = rb</math> would give the desired extension.

Since <math>-v+M, v+M</math> are on opposite sides of <math>M,</math> and <math>p < 0</math> at some point on <math>v+M,</math> by convexity of <math>p,</math> we must have <math>p \geq 0</math> on all points on <math>-v+M.</math> Thus <math>\inf_{u\in -v + M} p(u)</math> is finite.

Geometrically, this works because <math>\{z : p(z) < 0\}</math> is a convex set that is disjoint from <math>M,</math> and thus must lie entirely on one side of <math>M.</math>

Define <math>b = -\inf_{u\in -v + M} p(u).</math> This satisfies <math>p\geq -b</math> on <math>-v+M.</math> It remains to check the other side.

For all <math>v+w \in v+M,</math> convexity implies that for all <math>-v+w' \in -v+M, p(v+w) + p(-v +w') \geq 2p((w+w')/2) = 0,</math> thus 
<math>p(v + w) \geq \sup_{u\in -v + M} -p(u) = b.</math>

Since during the proof, we only used convexity of <math>p</math>, we see that the lemma remains true for merely convex <math>p.</math>
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'''Proofs'''

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* <ref group=proof name="ProofNormOfExtensionIsLarger">The map <math>F</math> being an extension of <math>f</math> means that <math>\operatorname{domain} f \subseteq \operatorname{domain} F</math> and <math>F(m) = f(m)</math> for every <math>m \in \operatorname{domain} f.</math> Consequently,
<math display=block>\{|f(m)| : \|m\| \leq 1, m \in \operatorname{domain} f\} = \{|F(m)|: \|m\| \leq 1, m \in \operatorname{domain} f\} \subseteq \{|F(x)\,| : \|x\| \leq 1, x \in \operatorname{domain} F\}</math> and so the [[supremum]] of the set on the left hand side, which is <math>\|f\|,</math> does not exceed the supremum of the right hand side, which is <math>\|F\|.</math> In other words, <math>\|f\| \leq \|F\|.</math>
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