Proving Corollary to Hahn-Banach Theorem: The Uniqueness of the Zero Point

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Homework Statement


To clarify- this isn't a homework problem; its something that's stated as a corollary in my notes (as in the proof is supposed to be obvious) and I haven't yet managed to prove it- I'm probably just missing something! Would appreciate a hint or a link to where I might find the proof.

Let X be a normed vector space, t \in X such that for all g \in X', g(t) = 0. Then t = 0.

Homework Equations


Hahn-Banach Theorem (stated on my course as:)
Let M be a subspace of a normed vector space X. Let f \in M' . Then there exists g \in X' such that the norm of f (wrt M') is equal to the norm of g (wrt X'), and g is equal to f on M.


The Attempt at a Solution


Just a bit confused- this is equivalent to, the only point all linear functionals can vanish at is zero. Presumably I want to define some linear subspace to then use but the only one that jumps out at me is span(t) which then doesn't seem to give me anything. Any hints?
 
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You're going in the right direction. Assume that ##t\neq 0##. Then set ##M= span(t)##. Now define a suitable nonzero functional on ##M## and extend it by Hahn-Banach.
 
Possibly in the right direction, but unfortunately not at speed :P

I want to find a linear functional f defined on M s.t. f vanishes at t (so that its Hahn-Banach extension will satisfy the given property).. but if f vanishes at t then it vanishes on the whole of M, which is not particularly useful. I'm probably being incredibly slow but would appreciate a further hint, sorry! Also I have a lot of exams to revise for and don't really want to spend loads of time trying to prove this one little bit... Thank you in advance for any help!
 
You want to assume that ##t\neq 0## and you want to find a linear functional that does not vanish at ##t##. This would be a contradiction with the property that ##t## has (namely, that all the linear functionals vanish there.

So, can you find a linear functional ##f:M\rightarrow \mathbb{R}## with ##M=span(t)## such that ##f(t)\neq 0##?
 
Hmmn I think I was getting confused with the logic of what I was trying to do then. I can take f(at) = a |t| (where |t| is its norm). Then if every functional vanishes at t then the extension of f, g satisfies g(t) = 0 = |t| and by positive definiteness t is zero.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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