# $E$ a Banach Space and $(\varphi_n)\subset E^{\ast}$ such that $x_n\to0\Rightarrow \sum \varphi_n(x_n)<\infty$

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?

• 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. Commented May 7, 2018 at 14:47

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).
• 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. Commented May 8, 2018 at 21:42