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This question already has an answer here:

Let $(V, \| \cdot \|)$ be a normed space and suppose that $(x_n)_{n\in\mathbb{N}} \subset V$ has the property $\sum_{n=1}^{\infty} |\phi(x_n)|<\infty$ for each $\phi \in V^*$. Prove that

$$ \sup_{\|\phi \| \leq 1} \sum_{n=1}^{\infty} |\phi (x_n) | < \infty. $$

I know that I'm supposed to show what work I have done. But I couldn't get started on this exercise. So it would be helpful if it could be pointed out where to start.

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marked as duplicate by Daniel Fischer functional-analysis Apr 15 '17 at 18:02

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The property $\sum_{n = 1}^\infty |\varphi(x_n)| < \infty$ tells us that for every $\varphi \in V^*$ the sequence $(\varphi(x_n))_{n \in \mathbb N} \in \ell^1(\mathbb N)$. Hence, we can define an operator $\Phi \colon V^* \to \ell^1$ by setting $$ \Phi(\varphi) := (\varphi(x_n))_{\mathbb{N}}. $$ For all $N \in \mathbb N$ we can furthermore define a sequence of Operators $$ \Phi_N(\varphi) = (\phi(x_1),\dots,\phi(x_N),0,\dots) \in \ell^1(\mathbb N). $$ Each of the $\Phi_N$ is continuous since $$\sum_{k = 1}^N |\varphi(x_k)| = \|\Phi_N(\varphi)\|_1 \leq \|\varphi\| \sum_{k = 1}^N |x_k|.$$ Furthermore, we have that $$ \lim_{N \to \infty} \Phi_N(\varphi) = \Phi(\varphi) $$ for all $\varphi \in V^*$. Now you can apply Banach-Steinhaus: $\Phi$ is bounded and thus $$ \sup_{\varphi \in V^*_1} \sum_{k = 1}^\infty |\varphi(x_k)| =\sup_{\varphi \in V^*_1} \| \Phi(\varphi)\|_1 = \|\Phi\| < \infty. $$

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Since $V$ is a normed vector space, $V^*=B(X \rightarrow \mathbb{R})$ is a Banach Space, hence one may apply the uniform boundedness principle to any family of continous linear operators in $B(V^* \rightarrow \mathbb{R})=(V^{*})^{*}$ (since $\mathbb{R}$ is a complete normed vector space). Consider the family $\mathcal{F}=\{T_n^{\epsilon}:n\in \mathbb{N},\epsilon\in \{-1,1\}^n$}, defined by $T_n^{\epsilon}(\phi) = \sum_{k=1}^n \phi(\epsilon_k x_k$). You may check by the triangle inequality that these linear functionals are continuous. Notice that $|T_n^{\epsilon}(\phi)|\leq \sum_{k=1}^\infty |\phi(x_k)| < \infty$. So for any $\textit{fixed}$ $\phi$, $|T_n^{\epsilon}(\phi)|$ is bounded by a constant $M_{\phi} = \sum_{k=1}^\infty |\phi(x_k)|$. This tells us that the family $\mathcal{F}$ is pointwise bounded, so by the uniform boundedness principle it is uniformly bounded by some constant $M$. Now fix $\phi$ with $||\phi||\leq 1$, and choose $\epsilon_i \in \{-1,1\}$ adequately so that $\phi(\epsilon_i x_i)=|\phi(x_i)|$. The fact that $|T_n^{\epsilon}(\phi)|\leq M ||\phi||$ then tells us that $\sum_{k=1}^n |\phi(x_k)| \leq M ||\phi|| \leq M$, and by letting $n \rightarrow \infty$ we get $\sum_{k=1}^\infty |\phi(x_k)| \leq M ||\phi|| \leq M$. Hence

$$ \sup_{||\phi||\leq 1} \sum_{k=1}^\infty |\phi(x_k)| \leq M $$

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    $\begingroup$ Interesting approach. It should also work for complex vector spaces. To this end, I think one has to chose $\epsilon \in \mathbb T^n$ and not in $\{-1,1\}^n$, i.e. you "chose $\epsilon_i \in \mathbb T$ adequately so that $\phi(\epsilon_i x_i) = |\phi(x_i)|$". $\endgroup$ – el_tenedor Apr 11 '17 at 20:47
  • $\begingroup$ That's right, the same argument follows through in the complex case if you replace $\{-1,1\}$ by $\mathbb{T}$. $\endgroup$ – SMJK Apr 11 '17 at 23:04

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