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X is Banach space, $f_{i} \in X^{*},\forall 1\leq i<\infty$ and $ \sum_{i=1}^{\infty}| f_{i}(x)|<\infty,\forall x\in X$

how to prove the following lemma: $\forall F \in X^{**},\quad \sum_{i=1}^{\infty}| F(f_{i})|<\infty.$

Any comments and advice will be appreciated.

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1 Answer 1

up vote 5 down vote accepted

Apply the Banach-Steinhaus Theorem.

Let $\mathcal{F}$ be the collection of elements in $X^*$ of the form $\sum e_if_i$, where only finitely many of the $e_i$ are non-zero, and for each non-zero $e_i$ we have $|e_i| = 1$.

Note that for $f\in \mathcal{F}$ $$ |f(x)| = |\sum e_i f_i(x)| \leq \sum |f_i(x)| < \infty $$ and hence by the uniform boundedness principle $$ \sup_{f\in \mathcal{F}} \|f\|_{X^*} < \infty $$ that is, $\mathcal{F}$ is a bounded subset of $X^*$.

Now, for each $g\in X^{**}$ and $n\in \mathbb{N}$, there exists an element $f\in \mathcal{F}$ such that $$ \sum_{i = 1}^n |g(f_i)| = g(f) $$ Indeed, choose $e_i$ such that $e_i g(f_i)$ is positive and real for $i\in 1,\ldots,n$ and $0$ otherwise. Hence $$ \sum_{i = 1}^n |g(f_i)| = g(f) < \|g\|_{X^{**}} \|f\|_{X^*} \leq \|g\|_{X^{**}} \sup_{f\in\mathcal{F}} \|f\|_{X^*}$$ Note now that the right hand side depends only on the norm of $g$ and a constant that is independent of $g$ or $n$.

Now taking the limit on the left as $n\to\infty$ you have a sequence of increasing real numbers that is uniformly bounded, and hence converges to a finite limit.

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