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Given a Banach space: $E$
and chosen a Hamel basis: $\mathcal{B}$

Any vector induces a (noncanonical) algebraic linear functional by: $$\delta:E\to E^*:\delta_b(b'):=\delta_{b,b'}\text{ defined linearly and extended linearly}$$ How to show that the induced linear functionals are continuous iff the Banach spaces is finite dimensional?

C-star-W-star
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1 Answers1

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Suppose that $\mathcal B$ contains a sequence $(b_k)_{k\geqslant 1}$ such that $\lVert b_k\rVert=1$ for each $k$. Define $$L_n(x)=\sum_{k=1}^nk\cdot \delta_{b_k}(x).$$ Then for each $x$, $\sup_{n\geqslant 1}|L_n(x)|$ is finite. Since $\lVert L_n\rVert\geqslant n$, the principle of uniform boundedness implies a contradiction.

Davide Giraudo
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  • Ok I need to clarify: Is it right when I say that finite dimensional all those functionals are continuous and infinite dimensional iff there is at least one being discontinuous? If so does it admittable to change the basis and then do the construction? Because I might (un)luckily start with a basis that goes through not giving any discontinuous functional? Besides, that's a really nice application of the uniform boundedness principle. =D – C-star-W-star Jun 06 '14 at 16:02
  • Yes it is right. – Davide Giraudo Jun 06 '14 at 16:06
  • But it might happen that one of the functionals is in fact continuous? – C-star-W-star Jun 06 '14 at 16:19
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    Sorry, one more caveat: I think all we can say is $\Vert L_n\Vert\ge n$, unless I'm missing something. – David Mitra Jun 06 '14 at 19:18
  • You are right. A priori nothing seems to prevent to have a norm greater than $n$ (it will depend on the norm of linear combinations of $b_k$'s). – Davide Giraudo Jun 06 '14 at 19:20