Let $E$ be a Banach vector space and for every $n\in \mathbb{N}$ let $\varphi_n$ be an element of $E^{\ast}$. Prove that the following assertions are equivalent:

a) For every sequence $(x_n)$ which tends to $0$, $\sum_{n=1}^{\infty}\varphi_n(x_n)$ is convergent.

b) $\sum_{n=1}^{\infty}\varphi_n$ is absolutely convergent.

For $(b)\Rightarrow (a)$ I think the following argument is valid: If $x_n\to 0$, in particular $(x_n)$ is a bounded sequence. Let $M$ be an uper bound. Therefore $\sum_{n}\left |\varphi_n(x_n)\right |\leq \sum_{n}\left \|\varphi_n\right \|\left \|x_n\right \|\leq M\sum_{n}\left \|\varphi_n\right \|$, which is finite by the assumption.

Now, what about $(a)\Rightarrow (b)$? One thing we can say is that the convergence of $\sum_{n=1}^{\infty}\varphi_n(x_n)$ is an absolute convergence: if $x_n\to 0$ then taking $y_n=\text{sgn}(\varphi(x_n))x_n$ (where $\text{sgn}(x)=1$ if $x\geq 0$ and $\text{sgn}(x)=-1$ if $x<0$), we have that $y_n\to 0$ and $\sum_{n=1}^{\infty}\varphi_n(y_n)=\sum_{n=1}^{\infty}|\varphi_n(x_n)|$. (If the field is $\mathbb{C}$ we can take a rotation).

I do not know how to continue. How would you proceed?

  • $\begingroup$ For $(b)\Rightarrow (a),$ I think you need to be more precise. What do you mean by $(x_n)$ is bounded? In your proof, it seems that you are using the fact that the scalar sequence $(\|x_n\|)$ is bounded. $\endgroup$
    – Idonknow
    May 7, 2018 at 14:47

1 Answer 1


Suppose $\sum_{n=1}^{\infty} ||\phi_n||=\infty$. Then there exists a sequence $\{a_n\}$ of positive numbers decreasing to 0 such that $\sum_{n=1}^{\infty} a_n||\phi_n||=\infty$. [ There is an exerecise in Rudin's Principles which tells you how to choose $a_n$'s]. There exist unit vectors $y_n$ such that $\phi_n (y_n) >\frac 1 2 ||\phi_n||$. Let $x_n=a_ny_n$. Then $x_n \to 0$ but $\sum \phi_n(x_n) >\sum \frac {a_n} 2 ||\phi_n||=\infty$ contradicting a).

  • $\begingroup$ To save others looking for the reference, exercise 3.11ii in Principles gives that $a_n = \sum_{i=1}^n \|\varphi_n\|$ works, and shows that the resulting series is not Cauchy, hence not convergent. $\endgroup$
    – B. Mehta
    May 8, 2018 at 21:42
  • $\begingroup$ @B. Mehta Thanks! $\endgroup$ May 10, 2018 at 4:56

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