This is the second part where this questions was the first part.
As in part one, we have given $ a=(a_n)_{n\in \mathbb{N}}\in l^1(\mathbb{N}) $ and we define for each $ c=(c_n)_{n\in \mathbb{N}}\in c^0(\mathbb{N}) $ $$ \Lambda_a(c)=\sum_{n=1}^{\infty}a_nc_n. $$
In part one we showed that $ \Lambda:a\mapsto \Lambda_a $ defines a linear map from $ l^1(\mathbb{N}) $ to $ c^0(\mathbb{N})^* $, where $$c^0(\mathbb{N})^*=L\left(c^0(\mathbb{N}),\mathbb{K}\right)=\left\{ \lambda:c^0(\mathbb{N})\to \mathbb{K}:\lambda \text{ is linear and bounded} \right\}$$ and $ \mathbb{K} $ denotes $ \mathbb{R} $ or $ \mathbb{C} $
Now I want to show that $ \Lambda $ is bounded with $ \|\Lambda\|\le 1 $. Here is my answer:
I have already showed that $ \Lambda_a $ is bounded, so $$ \|\Lambda_a\|=\sup_{\|c\|_0=1}\|\Lambda_a(c)\| $$ is finite. Then, using $ c\in c^0(\mathbb{N})$ as a test-function, \begin{align*} \|\Lambda(a)(c)\|&=\|\Lambda_a(c)\|\\ &\le \sup_{\|c\|=1}\|\Lambda_a(c)\|\\ &=\sup_{\|c\|=1}\left\|\sum_{n=1}^{\infty}a_nc_n\right\|\\ &\le\sum_{n=1}^{\infty}|a_n|\cdot 1\\ &=\|a\|_1\\ &=C \end{align*} so $ \Lambda $ is bounded. Since $$ \|\Lambda a\|\le \|a\|_1 $$ we have that $ \|\Lambda\|\le 1 $.
My question is, as in part one, is my answer correct?