we can always extend functions on subspaces

[[concept]]

[!themes] Topics

Evaluation Error: SyntaxError: Unexpected token '>'

at DataviewInlineApi.eval (plugin:dataview:19027:21)
at evalInContext (plugin:dataview:19028:7)
at asyncEvalInContext (plugin:dataview:19038:32)
at DataviewJSRenderer.render (plugin:dataview:19064:19)
at DataviewJSRenderer.onload (plugin:dataview:18606:14)
at DataviewJSRenderer.load (app://obsidian.md/app.js:1:1214378)
at DataviewApi.executeJs (plugin:dataview:19607:18)
at DataviewCompiler.eval (plugin:digitalgarden:10763:23)
at Generator.next (<anonymous>)
at fulfilled (plugin:digitalgarden:77:24)

Lemma

If V is a normed vector space and MV is a subspace and u:MC is linear such that |u(t)|C||t|| for all tM, and

Proof (of Lemma)

First, note that if tM(=M+Cx) then there exists unique tM and aC such that t=t+ax.

Thus, upon choosing λC, the map

u(t+ax)=u(t)+aλtM,aC

is well-defined on M and u:MC is linear.

WLOG, suppose C=1. We want to choose λC such that

tM,aC|u(t)+aλ|||t+aλ||

Once λ is fixed, u is our continuous extension. When a=0, this inequality holds regardless of λ (because it holds on M). Thus we only need to choose λ for a0. When a0, we have

|u(t)+aλ|||t+ax||||u(ta)λ||||tax||tM|u(t)λ|||tx||tM

Since taM. We first show there exists some αR such that

|w(t)α|||tx||tM

where w(t)=u(t)u(t)2=Re(u(t)). Now, note that

|w(t)|=|Re(u(t))||u(t)|||t||w(t1)w(t2)=w(t1t2)t1,t2M|w(t1t2)|||t1t2||||t1x||+||t2x||w(t1)||t1x||w(t2)+||t2x||t1,t2MsuptMw(t)||tx||w(t2)+||t2x||t2MsuptMw(t)||tx||inftMw(t)+||tx||

So we can choose an α between these two quantities such that for all tM we have

w(t)||tx||αw(t)+||tx||||tx||αw(t)||tx|||w(t)α|||tx||

Using the same process for the imaginary part of u(t) and repeating the argument with ix, we can find a λ such that the desired bound holds.

Note

To see this, we can verify that since the bound holds on both the real and imaginary components of x, it holds for all complex multiples of x

This defines our function

u(t+ax)=u(t)+aλ

on all of M+Cx. Thus we are done.

We use this lemma in our proof for the Hahn-Banach theorem

Strategy for using this lemma to prove Hahn-Banach

  1. Place a partial order on all continuous extensions of u
  2. Apply Zorn's lemma to this set, giving us a maximal element U
  3. Use this lemma to show that this maximal extension is defined on all of our desired space V

References

References

See Also

Mentions

Mentions

File Last Modified
Functional Analysis Lecture 5 2025-07-01
Hahn-Banach theorem 2025-07-01

Created 2025-07-01 Last Modified 2025-07-01