If f(x) = 0 for every bounded linear map f, is x = 0?

Click For Summary

Discussion Overview

The discussion revolves around the question of whether a vector \( x \) in a complex vector space \( X \) must be zero if it is known that \( f(x) = 0 \) for every bounded linear functional \( f: X \to \mathbb{C} \). The scope includes theoretical aspects of functional analysis, particularly the implications of the Hahn-Banach theorem and the properties of normed spaces.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants suggest that if \( f(x) = 0 \) for every bounded linear functional \( f \), then \( x \) must be zero, referencing the Hahn-Banach theorem.
  • Others propose that the identity map being a bounded linear map supports the argument but question if this is the intended meaning of the original question.
  • A participant mentions a more complex solution applicable only to Hilbert spaces, involving the Riesz representation theorem and properties of the inner product.
  • There is a discussion about the necessity of the Hahn-Banach theorem, with some noting that it requires the axiom of choice and is not trivial in arbitrary normed spaces.
  • Another participant points out that if an inner product exists, it can be used to define the functional directly, potentially avoiding the need for Hahn-Banach.
  • Concerns are raised about the non-triviality of the dual space in arbitrary normed spaces and the implications of the Hahn-Banach theorem in this context.

Areas of Agreement / Disagreement

Participants express varying levels of agreement regarding the necessity of the Hahn-Banach theorem and the implications of the identity map. There is no consensus on the necessity of invoking Hahn-Banach in all cases, particularly in relation to inner product spaces versus general normed spaces.

Contextual Notes

Participants note that the discussion hinges on the definitions of bounded linear functionals and the properties of normed spaces, with some indicating that the argument may depend on whether the space is complete or has an inner product.

AxiomOfChoice
Messages
531
Reaction score
1
Suppose you're looking at a complex vector space X, and you know that, for some x in X, you have f(x) = 0 for every linear map on X. Can you conclude that x = 0? If so, how?

This seems easy, but I can't think of it for some reason.

(EDIT: Assume it holds for every CONTINUOUS (i.e., bounded) linear map on X. This may, or may not, make a difference.)
 
Last edited:
Physics news on Phys.org
The identity map is linear and bounded. :smile:
 
Well, you can take f the identity map on X. Then f(x)=0 means that x=0. But I'm quite sure that this is not what you meant...
 
Well the identity function is a bounded linear map, so yes.

If you mean bounded linear functional, this is still true by Hahn-Banach, because we can send x to ||x||, and send each vector of the form ax to a||x||, where a is in C. By Hahn-Banach, this extends to a bounded linear functional.
 
ebola1717 said:
Well the identity function is a bounded linear map, so yes.

If you mean bounded linear functional, this is still true by Hahn-Banach, because we can send x to ||x||, and send each vector of the form ax to a||x||, where a is in C. By Hahn-Banach, this extends to a bounded linear functional.

Yes, perhaps I've phrased it badly. I'm talking about X being a normed space, and I'm talking about f(x) = 0 for every bounded linear functional f: X \to \mathbb C.

Perhaps I screwed up by writing "map" instead of "functional"...sorry, guys :frown:
 
ebola1717 said:
Well the identity function is a bounded linear map, so yes.

If you mean bounded linear functional, this is still true by Hahn-Banach, because we can send x to ||x||, and send each vector of the form ax to a||x||, where a is in C. By Hahn-Banach, this extends to a bounded linear functional.

So, just following up...if one has f(x) = 0 for every bounded linear functional f: X \to \mathbb C, then we have x = 0, but one has to invoke Hahn-Banach to show that? Ok...no wonder I couldn't think of it :redface:
 
I didn't think of it either, but it's a really nice solution. Definitely the simplest one. I have a more complicated one that only works for Hilbert spaces, if you're interested. It's based on the fact that there's a unique vector x_f such that the map y\mapsto\langle x_f,y\rangle is equal to f (the Riesz representation theorem). I'm also using that (\ker f)^\perp=\mathbb Cx_f.

Edit: I thought of a way to avoid the Hahn-Banach theorem, but it requires an inner product. Ebola's solution goes like this: Suppose that x\neq 0. Then ax\mapsto a\|x\| is a bounded linear functional on the 1-dimensional subspace \mathbb Cx. Let's call this functional f_0. By the Hahn-Banach theorem, there's a linear functional f:X\rightarrow\mathbb C such that f|_{\mathbb Cx}=f_0 and \|f\|=\|f_0\|. Since f(x)=f_0(x)=\|x\|\neq 0, we have a contradiction.

If there's an inner product, we don't have to invoke Hahn-Banach. We just define f explicitly: f(y)=\langle x,y\rangle.
 
Last edited:
It shouldn't come as a surprise that Hahn-Banach is needed. For an arbitrary normed vector space, it isn't even clear a priori that the dual spaces is non-trivial. If it were, then your condition would be satisfied trivially! You use Hahn-Banach to show that the dual space is non-trivial.
 
this seems non trivial as hahn banach uses the axiom of choice as i recall.
 
  • #10
Yes, it's definitely not trivial if we're dealing with an arbitrary normed space or Banach space. However, if there's an inner product on the space, we can use the inner product instead of the Hahn-Banach theorem.

If we're dealing with an arbitrary normed space X, I guess we would have apply the argument based on the Hahn-Banach theorem to the completion of X.

The proof I've seen of the Hahn-Banach theorem relies on Zorn's lemma, which is equivalent to the AoC.
 
  • #11
@Fredrik: Hahn-Banach is already a result about normed vector spaces, not necessarily complete. So no need to consider the completion.

Hahn-Banach is strictly weaker than AoC: in ZF, AoC implies H-B, but not the other way around.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K