Let $\ell^1:=\{f:\mathbb{N}\to\mathbb{C}\colon ||f||_1:=\sum_{n=1}^\infty |f(n)|<\infty\}$ provided with $||\cdot||_1$. Show, that every $f\in\ell^1$ defines a continuous linear functional $S_f: c_0\to\mathbb{C}$ with $S_f(g)=\sum_{n=1}^\infty g(n)f(n)$ and it is $||S_f||_\text{op}=||f||_1$ for every $f\in \ell^1(\mathbb{N})=\ell^1$
It is easy to see, that $S_f$ is linear: $(\ell^1, ||\cdot||_1)$ is a normed $\mathbb{C}$-vectorspace. Let $g, h\in c_0$ and $\mu, \lambda\in\mathbb{C}$, then:
$S_f(\mu g+\lambda h)=\sum_{n=1}^\infty f(n)(\mu g+\lambda h)(n)=\sum_{n=1}^\infty f(n)(\mu g)(n)+f(n)(\lambda h)(n)\\=\mu\sum_{n=1}^\infty f(n)g(n)+\lambda\sum_{n=1}^\infty f(n)h(n)$ since the sum converges.
We get the desired $\mu S_f(g)+\lambda S_f(h)$.
Now I want to show, that $S_f$ is continuous. Is it enough to show that $S_f$ is continuous at 0?
Else I can show, that for every sequence $(g_n)_n$ in $c_0$ with $g_n\to 0$ it is $S_f(g_n)\to 0$ or that $S_f$ is bounded. Therefore $||S_f(g)||_1\leq c||g||$ for some $c\geq 0$ and every $g\in c_0$.
Unfortunatly nothing really seems to work for me... I wanted to show, that $S_f$ is bounded. I started like this:
$||S_f(g)||_1=||\sum_{n=1}^\infty f(n)g(n)||_1$ and now I already start to struggle. How do I proceed from here. Do I get a "double sum"?
$\sum_{n=1}^\infty\vert\sum_{n=1}^\infty f(n)g(n)\vert$ but this does not make sense, because it does not converge...
And how can I show, that $||S_f||_\text{op}=||f||_1$, where I experience the same issue as above.
Hints are appreciated. Thanks in advance.
The norm on $\mathbb C$ is just the modulus and I assume that you are using the supremum norm on $c_0$ so to show that $S_f$ is bounded, you need to bound $\lvert S_f(g) \rvert$ in terms of $\|g\|_\infty.$ This much is trivial. We see $$\lvert S_f(g) \rvert = \left \lvert \sum^\infty_{n = 1} f(n) g(n) \right \rvert \le \sum^\infty_{n=1} \lvert f(n)\rvert \lvert g(n) \rvert \le \| g\|_\infty \sum^\infty_{n=1} \lvert f(n) \rvert = \|f\|_1 \|g\|_\infty.$$ This shows that $S_f$ is bounded with $\|S_f\|_{\text{op}} \le \|f \|_1$. In order to prove that $\|S_f\|_{\text{op}} = \|f\|_1$, you need to shows that the above inequality is sharp. To do so, take $g_k$ defined by $g_k(n) = \text{sign}(f(n)),$ for $n \le k$ and $g_k(n) = 0$ for $n > k$. Then we see $$\lvert S_f(g_k)\rvert = \sum^k_{n=1} \lvert f(n) \rvert $$ and this becomes arbitrarily close to $\|f\|_1$ as $k\to \infty$ so since each $g_k$ has $\|g_k \|_\infty = 1$, this shows that $\|S_f\|_{\text{op}} = \|f\|_1$.