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

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.)

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Fredrik
Staff Emeritus
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The identity map is linear and bounded.

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.

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

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

Fredrik
Staff Emeritus
Gold Member
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: $f(y)=\langle x,y\rangle$.

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Landau
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.

mathwonk
Homework Helper
this seems non trivial as hahn banach uses the axiom of choice as i recall.

Fredrik
Staff Emeritus
Gold Member
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.

Landau
@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.