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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    $\begingroup$ You're right. A linear functional on an infinite-dimensional normed vector space needs not be continuous. $\endgroup$ – Julien Feb 11 '13 at 20:25
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    $\begingroup$ Maybe you should take a look at this question: math.stackexchange.com/questions/63569/… $\endgroup$ – Elmar Zander Feb 11 '13 at 20:30
  • $\begingroup$ 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? $\endgroup$ – user58514 Feb 11 '13 at 20:31
  • $\begingroup$ The proposition in question is on this website, perso.crans.org/lecomte/Math/WeakTopologies.pdf Proposition 7 (page 5) (statement 1) $\endgroup$ – user58514 Feb 11 '13 at 20:38
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    $\begingroup$ 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.) $\endgroup$ – Harry Altman Feb 11 '13 at 21:01

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