(Treves Exercise 5.4) Let $E$ be a TVS and $E^*$ its algebraic dual. Provide $E^*$ with the topology of pointwise convergence in $E$. A basis of neighborhoods of zero in this topology is provided by the sets $$W(S,\epsilon) = \left\{x^* \in E^* : \sup\limits_{x\in S} |x^*(x)| \leq \epsilon\right\} $$ as $S$ ranges over the family of finite subsets of $E$ and $\epsilon$ over the set of numbers $>$ 0. Prove that $E^*$ is complete.
This is what I have so far:
Let $\mathscr{F}$ be a Cauchy filter on $E^*$. Define $\Phi_x:E^* \rightarrow \mathbb{C}$ by $$\Phi_x(x^*) = x^*(x). $$ Then $\Phi_x$ is continuous and linear, which implies $\Phi_x$ is uniformly continuous. Hence $\Phi_x\mathscr{F}$ is a Cauchy filter on $\mathbb{C}$. But $\mathbb{C}$ is complete, so there exists a $\lambda_x \in \mathbb{C}$ such that $\Phi_x\mathscr{F} \rightarrow \lambda_x$. This defines a function $\varphi^*:E\rightarrow \mathbb{C}$ by $\varphi^*(x) = \lambda_x$.
How can I show that $\varphi^*$ is linear so that $\varphi^* \in E^*$? Also, how would I show $\mathscr{F} \rightarrow \varphi^*$?
You are absolutely on the right track. For $x,y\in E$ and $x^*\in E^*$ you have $\Phi_{x+y}(x^*) = x^*(x+y)=\Phi_x(x^*)+\Phi_y(x^*)=(\Phi_x+\Phi_y)(x^*)$. This yields the additivity of $\varphi^*$ and the same argument gives homogeneity. Hence $\varphi^*\in E^*$. Finally $\mathscr F$ converges to $\varphi^*$ pointwise.
If you know that cartesian products and closed subspaces of complete TVS are again complete you can argue without Cauch-Filters by showing that $E^*$ is a closed subspace of $\mathbb C^E$.