# Problem in functional analysis.

I heard of this problem that caught my attention and I am curious now thus I would appreciate if I could have a hint or a solution.

Let $(x_n)$ a sequence in a normed space $X$ such that $\sum_{n=1}^{\infty} {|f(x_n)|}<+\infty$ for every $f\in X^*$. Prove that there exists $M>0$ such that $$\sum_{n=1}^{\infty} {|f(x_n)|} \leqslant M \|f\|$$ for every $f\in X^*$.

I have strong suspicions that the following fact is somehow used.

If $X,Y$ are Banach and $F:X\rightarrow Y$ a linear, bounded and onto operator then there exists $M>0$ such that for every $y\in Y$ there exists a $x\in X$ with $F(x)=y$ and $\|x\| \leqslant M \|y\|$.

So, my thought would be to define an operator $T:K \rightarrow X^*$ from the subspace of $l^1$ , $K=\{ (f(x_n))_n , f\in X^* \}$ with $T((f(x_n))_n)=f$ and then the result would follow from the above fact if I prove it's bounded. However, I don't think that this operator is well defined so the reasoning doesn't work. And I am not sure even if $K$ is a closed subset of $l^1$ to have the Banach space requirement so this reasoning fails. Am I even thinking in the right direction?

Thanks!

• Hint: Banach-Steinhaus. Partial sums. – Daniel Fischer Jul 18 '14 at 12:51
• You should consider the operator from $X^*$ to $\ell_1$ that maps $f\in X^*$ to $\bigl(f(x_n)\bigr)\in\ell_1$. You need to show this is bounded. Use Daniel's hint. – David Mitra Jul 18 '14 at 12:59
• Figured out now, thanks both! – Esoog Jul 18 '14 at 13:12
• You could also use Zabreiko's Lemma, which is really cool. I learned it in this post (first answer) math.stackexchange.com/questions/137673/… – PhoemueX Jul 18 '14 at 15:53
• Interesting, quite nice indeed, I've never heard of that one before. – Esoog Jul 18 '14 at 18:24