Given $ a=(a_n)_{n\in \mathbb{N}}\in l^1(\mathbb{N}) $, define for each $ c=(c_n)_{n\in \mathbb{N}}\in c^0(\mathbb{N}) $ $$ \Lambda_a(c)=\sum_{n=1}^{\infty}a_nc_n. $$ I want to show that $ \Lambda:a\mapsto \Lambda_a $ defines a linear map from $ l^1(\mathbb{N}) $ to $ c^0(\mathbb{N})^* $, where $ c^0(\mathbb{N})^* $ denotes the dual of $ c^0(\mathbb{N}) $ (see the definition in my answer).
Here is my answer (in my answer, $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$):
Since $ a\in l^1(\mathbb{N}) $, I don't need to show that $ \Lambda $ is a map FROM $ l^1(\mathbb{N}) $. Since $$ c^0(\mathbb{N})^*=L(c^0(\mathbb{N}),\mathbb{K})=\{ \lambda:c^0(\mathbb{N})\to \mathbb{K}:\lambda \text{ is linear and bounded} \}, $$ to show that $ \Lambda $ is a map TO $ c^0(\mathbb{N})^* $ is the same as showing that $ \Lambda_a $ is linear and bounded.
$ \Lambda_a $ is linear:
For $ k\in \mathbb{K} $ and $ c\in c^0(\mathbb{N}) $: \begin{align*} \Lambda_a(kc)&= \sum_{n=1}^{\infty}a_nkc_n\\ &=k\sum_{n=1}^{\infty}a_nc_n\\ &=k\Lambda_a(c) \end{align*}
For $ c,d\in c^0(\mathbb{N}) $: \begin{align*} \Lambda_a(c+d)&= \sum_{n=1}^{\infty}a_n(c_n+d_n)\\ &=\sum_{n=1}^{\infty}(a_nc_n+a_nd_n)\\ &=\sum_{n=1}^{\infty}a_nc_n+\sum_{n=1}^{\infty}a_nd_n\\ &=\Lambda_a(c)+\Lambda_a(d) \end{align*}
$ \Lambda_a $ is bounded:
Since $ a=(a_n)_{n\in \mathbb{N}}\in l^1(\mathbb{N}) $, we know that $$ \|(a_n)_{n}\|_1:=\left( \sum_{n=1}^{\infty}|a_n|^1 \right)^{1/1}= \sum_{n=1}^{\infty}|a_n|< +\infty $$ Let $$ C:=\sum_{n=1}^{\infty}|a_n|$$ Also, from one of the examples in class, we have that $$ \|c\|_0=\sup_{n\in \mathbb{N}}|c_n| $$ defines a norm in $ c^0(\mathbb{N}) $ who's metric is uniformly convergent, so $ \|c\|_0<\infty $. Hence, \begin{align*} \|\Lambda_a(c)\|&=\left\|\sum_{n=1}^{\infty}a_nc_n\right\|\\ &\le|\sum_{n=1}^{\infty}|a_n|\|c\|_0\\ &=C\|c\|_0 \end{align*} and so $ \Lambda_a $ is bounded. We have showed that $ \Lambda $ is a map from $ l^1(\mathbb{N}) $ to $ c^0(\mathbb{N})^* $. It remains to show that this map is linear. We'll use $ c\in c^0(\mathbb{N})$ as a test-function:
For $ k\in \mathbb{K} $ and $ a\in l^1(\mathbb{N}) $, \begin{align*} \Lambda(ka)(c)=&\Lambda_{ka}(c)\\ &=\sum_{n=1}^{\infty}ka_nc_n\\ &=k\sum_{n=1}^{\infty}a_nc_n\\ &=k\Lambda_a(c)\\ &=k\Lambda(a)(c)\\ \end{align*}
Is my answer correct? Is it unnecessary long?