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I have the following problem: $X$ real Banach space $(\epsilon_n)_{n\geq1}$ a positive sequence converging to zero $(f_n)_{n\geq1}$ a sequence in $X^*$ the dual space of $X$ with the following property:

$\exists r>0$ st $\forall x\in B_r(0) \exists C(x)\in \mathbb{R}: f_n(x)\leq \epsilon_n||f_n|| + C(x), \forall n$

We claim that $(f_n)_{n\geq1}$ is bounded.

Here what I tried:

From the lecture I know that since X is a normed space it follow that $X^*$ is a Banachspace. So I was thinking now that all I have to show is that $(f_n)_{n\geq1}$ is a Cauchy sequence, so that if the sequence converges then it is also bounded. Alternatively I was also thinking to show that the sequence is continuous and than deduce the boundness. both ways seem me a little bit hard.

Am I thinking in the right way?

From the property, since it has to hold for all n, then $C(x)>0$.

sky90
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  • This is a generalization of the uniform boundedness principle, and I think the same idea should work: suppose the sequence $f_n$ is not bounded; pick $x_n$ (after passing to subsequence) so that $|f_n(x_n)|$ is very large but $|x_n|$ is very small. Let $x=\sum x_n$ and obtain a contradiction by considering $f_n(x)$. – user127096 Mar 30 '14 at 15:42
  • No need for completeness, a Cauchy sequence is bounded in any metric space. But it is much more difficult to show that a sequence is Cauchy than that it is bounded: moreover, you have no guarantees that it is true. Try another way. (PS: In the inequality, are you sure that you did not miss an absolute value?) – Giuseppe Negro Mar 30 '14 at 15:43
  • no abs value missing but the $\epsilon_n$ are assumed to be positive – sky90 Mar 30 '14 at 15:46
  • @cheapeffectivedietpills could you explain me please a little bit more what are you doing with $x_n$ and the subsequences passage? – sky90 Mar 30 '14 at 16:04

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