EDIT: Have just realised that I misunderstood your question. The first paragraph is stil valid and I am leaving the last three paragraphs to show where I misunderstood you. I thought that under initial topology you meant the finest l.c. topology on $E$.

An infinite dimensional metrisable l.c.s. is never complete for the weak topology. So any complete, infinite dimensional space wii do what you want (assuming I have now understood your question correctly). Sorry for the confusion.

THE SOLUTION TO MY UNDERSTANDING OF YOUR QUESTION.
Each vector space has a finest l.c. topology. It can be defined as that l.c. structure which is defined by ALL seminorms on $E$ or as the l.c. inductive limit
of the the finite dimensional subspaces. An example of an l.c. space for which this coincides with its weak topology is $\phi$, the space of finite sequences, with the natural inductive limit structure. Its topological and algebraic dual is $\omega$, the space of ALL sequences. For this see G. Köthe: Topological linear spaces.

If a l.c. space has a non continuous linear functional, then the above two topologies cannot coincide for the simple reason that they have distinct duals.

A host of such spaces can be found as follows: a separable l.c.s. whose linear dimension is uncountable automatically has a non continuous linear functional.

The last fact uses AC in the form of the existence of a Hamel basis. Work of Solovay, Schwartz and Garnir shows that without AC the situation is different.

everylinear map $X \to \mathbb{K}$ will be continuous? Can you provide a proof please, since I do not see this? $\endgroup$ – AlexE Feb 5 '14 at 8:40