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Assuming we have a normed vector space V (assume infinite dimensional, as trivial if finite dimensional), then why does the norm topology make all linear functionals on V continuos?

I can't see how this is true. As a linear functional on a normed vector space is continuos iff bounded. And there is definitely a linear functional on an infinite dimensional vector space that I can make unbounded!

My brain is fried! What am I missing here?

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    You're right. A linear functional on an infinite-dimensional normed vector space needs not be continuous. – Julien Feb 11 '13 at 20:25
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    Maybe you should take a look at this question: http://math.stackexchange.com/questions/63569/weakest-topology-with-respect-to-which-all-linear-functionals-are-continuous – Elmar Zander Feb 11 '13 at 20:30
  • Okay, so what I am specifically asked to prove is that weakly open => norm open.

    Since the weakly open topology makes all linear functional continuos (by definition), this would have to imply that all linear functions w.r.t the norm topology are continuos?

     –  Feb 11 '13 at 20:31
  • The proposition in question is on this website, http://perso.crans.org/lecomte/Math/WeakTopologies.pdf

    Proposition 7 (page 5) (statement 1)

     –  Feb 11 '13 at 20:38
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    You've misunderstood the definition. The weak topology isn't the coarsest topology making all the linear functionals continuous; it's the coarsest one making the bounded linear functionals continuous. (I.e., making continuous the ones that were already continuous, but with as few open sets as possible now.) – Harry Altman Feb 11 '13 at 21:01
  • @HarryAltman This is not the definition given in the notes above? –  Feb 11 '13 at 21:17
  • But to add, I have concluded that the statement they are making is considering the continuos dual, rather than the algebraic one. In which case your definition coincides. But the weak topology can be applied to much more situations than the continuos dual! It can be considered when looking at any family of functions. The statement in question is only true when the family of functions is a subset of the continuos functions. –  Feb 11 '13 at 21:27
  • Yes -- I was talking about this particular case. Or rather, when someone says "the weak topology" when talking about a normed vector space. If someone is talking about a normed vector space $X$ and they talk about $X^*$, they mean the continuous dual. The algebraic dual is just not very relevant most of the time. – Harry Altman Feb 11 '13 at 22:14
  • Okay, thanks for your input Harry. Much appreciated! –  Feb 11 '13 at 22:22

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