we can always extend functions on subspaces

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Lemma

If V is a normed vector space and M⊂V is a subspace and u:M→C is linear such that |u(t)|≤C||t|| for all t∈M, and

Proof (of Lemma)

First, note that if t′∈M′(=M+Cx) then there exists unique t∈M and a∈C such that t′=t+ax.

Thus, upon choosing λ∈C, the map

u′(t+ax)=u(t)+aλt∈M,a∈C

is well-defined on M′ and u′:M′→C is linear.

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

∀t∈M,a∈C|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 a≠0. When a≠0, we have

|u′(t)+aλ|≤||t+ax||⟺||u(t−a)−λ||≤||t−a−x||∀t∈M⟺|u(t)−λ|≤||t−x||∀t∈M

Since t−a∈M. We first show there exists some α∈R such that

|w(t)−α|≤||t−x||∀t∈M

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(t1−t2)∀t1,t2∈M≤|w(t1−t2)|≤||t1−t2||≤||t1−x||+||t2−x||⟹w(t1)−||t1−x||≤w(t2)+||t2−x||∀t1,t2∈M⟹supt∈Mw(t)−||t−x||≤w(t2)+||t2−x||∀t2∈M⟹supt∈Mw(t)−||t−x||≤inft∈Mw(t)+||t−x||

So we can choose an α between these two quantities such that for all t∈M we have

w(t)−||t−x||≤α≤w(t)+||t−x||⟹−||t−x||≤α−w(t)≤||t−x||⟹|w(t)−α|≤||t−x||

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

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