I am unsure what this question is asking - bounded linear functionals!

Question

There has been a question very similar to this in a different post, but the person who asked the question put a disclaimer that he only wanted hints. I would prefer a more full answer to what the question is actually asking for.

Let $(X,\left\|\cdot\right\|)=(l^1,\left\|\cdot\right\|_1)$ and let $f \in X^*$. Prove that there exists a bounded sequence $(v_1,v_2,v_3,...,v_n...)$ of real numbers such that $$f(x)=f(a_1,a_2,a_3,...,a_n,...)= \sum^{\infty}_{n=1}a_n v_n$$ for all $x= (a_1,a_2,a_3,...,a_n,...)\in l^1$.

So my thoughts are as $x\in l^1$, we must have $\sum^{\infty}_{n=1}|a_n| < \infty$. As the sequence $(v_1,v_2,v_3,...,v_n...)$ is bounded, we must be able to say that $$\sum^{\infty}_{n=1}a_nv_n \leq \sum^{\infty}_{n=1}|a_n|v_n \leq \|v_n\|_{\infty} \sum^{\infty}_{n=1}|a_n| \leq \infty.$$ However, I am not too sure if this answers the existence question!

Link to previous question: Existence of $\{a_n\}$ s.t $f(\vec{v})=\sum_{n \in \mathbb{N}} a_n v_n$ for continuous linear functions.

Answer 1

For $n\in \mathbb N$ let $e_n=(\delta_{n,i})_{i\in \mathbb N}$ where $\delta_{n,i}$ is the Kronecker delta: That is, $\delta_{n,n}=1,$ and $\delta_{n,i}=0$ if $n\ne i.$

Each $e_n\in l_1$ with $\|e_n\|_1=1.$ Let $v_n=f(e_n)$ for each $n\in \mathbb N.$

For $x=(x_i)_{i\in \mathbb N}\in l_1$ and for $n\in \mathbb N$ let $x|_n=(x'_i)_{i\in \mathbb N}$ where $x'_i=x_i$ for $i<n$ and $x'_i=0$ for $i\geq n.$ We have $x_n\in l_1$ and $$\lim_{n\to \infty}\|x-x|_n\|_1=0.$$

Now $f\in l_1^*$ so $f$ is a bounded linear functional so $f$ is continuous. From the continuity of $f$ we have $$\lim_{n\to \infty}\|x-x|_n\|=0 \implies \lim_{n\to \infty}|f(x)-f(x|_n)|=0.$$ Observe that $f(x|_n)=\sum_{i<n}v_ix_i.$ $$\text {Therefore }\quad f(x)=\lim_{n\to \infty}f(x|_n)=\lim_{n\to \infty}\sum_{i<n}v_ix_i=\sum_{n=1}^{\infty}v_ix_i.$$

We have $\|f\|<\infty.$ Therefore for every $i$ we have $|v_i|=|f(e_i)|\leq \|f\|\cdot \|e_i\|_1=\|f\|.$

Remark: We actually have $\|f\|=\sup_{i\in \mathbb N} |v_i|.$

Answer 2

We know that the dual of $\ell^1$ is $\ell^\infty$ [See Kreyszig for details]. Now for any $x\in \ell^1$ we have that $$f(x)=\sum_{n=1}^\infty a_nf(e_n),$$ where $e=\{(\delta_i^k)\}$ is the standard Schauder basis for $\ell^1$ (Why is this representation valid?). Hence choosing $\{f(e_n)\}_{n\in \mathbb N}$ to be your sequence will do the trick, because $f\in\ell^\infty$, so $|f(e_n)|\leq\|e_n\|_1\|f\|_\infty<\infty$ for every $x\in X$, because $\|e_n\|_1=1$ for each natural number $n$. We have now constructed the desired sequence, so it obviously exists.

For your clarification, your attempt indeed does not prove existence of the desired sequence. What you have done is shown some properties the desired sequence must have, assuming it exists, which is not equivalent to proving existence.