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?


marked as duplicate by Nate Eldredge, Davide Giraudo, Asaf Karagila general-topology Jun 6 '14 at 22:39

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  • $\begingroup$ Well one direction is easy, every linear function defined on a finite dimensional Banach space is continuous. The other direction isn't immediately clear to me yet. $\endgroup$ – Alex Schiff Jun 6 '14 at 15:25
  • $\begingroup$ Why is this so? $\endgroup$ – C-Star-W-Star Jun 6 '14 at 15:27
  • $\begingroup$ Because the weak topology and the strong topology are always equivalent on a finite dimensional Banach space. $\endgroup$ – Alex Schiff Jun 6 '14 at 15:31
  • $\begingroup$ Ok ^^ but isn't that precisely the question? $\endgroup$ – C-Star-W-Star Jun 6 '14 at 15:35
  • $\begingroup$ See here for a proof that every linear functional on a finite dimensional banach space is continuous. $\endgroup$ – Alex Schiff Jun 6 '14 at 15:49

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.

  • $\begingroup$ 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 $\endgroup$ – C-Star-W-Star Jun 6 '14 at 16:02
  • $\begingroup$ Yes it is right. $\endgroup$ – Davide Giraudo Jun 6 '14 at 16:06
  • $\begingroup$ But it might happen that one of the functionals is in fact continuous? $\endgroup$ – C-Star-W-Star Jun 6 '14 at 16:19
  • 1
    $\begingroup$ Sorry, one more caveat: I think all we can say is $\Vert L_n\Vert\ge n$, unless I'm missing something. $\endgroup$ – David Mitra Jun 6 '14 at 19:18
  • $\begingroup$ 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). $\endgroup$ – Davide Giraudo Jun 6 '14 at 19:20

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