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Let $X$ be a separable infinite dimensional Banach space and let $\{y_n\}\subset X$ be a dense sequence in the unit ball of $X$. Show that given any $x\in X$, there exists $(a_n)_{n=1}^\infty\subset \mathbb{C}$ such that $\sum_{n=1}^\infty |a_n|<\infty$ and $\sum_{n=1}^\infty a_ny_n=x$

So if $x=y_n$ for some $n$ then we are done with $(a_n)=(0,0,...,0,1,0,...)$ where $n$th coordinate being $1$. If not, we can first scale it to an element in the unit ball. Then there is a subsequence $\{y_{n_k}\}$ of $\{y_n\}$ such that the limit is $x$. But how do we make a series now?

EDIT: What if we take span$\{y_1\}$, span $\{y_1,y_2\}$,... Then can we say that $x$ should be in at least one of these spans since being dense and infinite dimensional?

Extremal
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Too long for comment, but I haven't thought this all the way through, and something doesn't feel quite right. So, more like an unhelpful hint.

Maybe you could construct as follows. Find $y_{N_0}$ such that $|x-y_{N_0}|<\frac{1}{2}$. Let $x_1 = x-y_{N_0}$. Now find $y_{N_1}$ with $N_1>N_0$ such that $|2x_1-y_{N_1}|<\frac{1}{2}$. This is possible because $2x_1$ is still in the unit ball. Now let $x_2 = x_1 -\frac{1}{2}y_{N_1}$. Now find $y_{N_2}$ with $N_2 > N_1$ such that $|4x_2-y_{N_2}|<\frac{1}{2}$. etc. Then your sequence $a_n$ is $0$ except it is equal to $\frac{1}{2^m}$ when $n=N_m$.

Callus - Reinstate Monica
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  • The choice of $N_m$ could be the same for different values of $m$, so perhaps you should let $a_n$ be the sum of $\frac{1}{2^m}$ over all $m$ where $n = N_m$. In that case we would still have $\sum{|a_n|}\le\sum\limits_{m}{\frac{1}{2^m}}$, so I think your approach would work. – Joey Zou Oct 19 '16 at 23:34
  • But now $\sum_{m=1}^\infty a_{N_m}y_{N_m}=x-x_1+x_1-x_2+x_2-...$. The RHS doesn't converge. – Extremal Oct 19 '16 at 23:57
  • @EpsilonDelta it does. Check the partial sum. $x_m\to 0$. – user251257 Oct 20 '16 at 00:04