Why does the weak topology make the functions in $E'$ continuous? Let $E$ be a Hausdorff locally convex topological vectorspace. Consider the source $p_f\colon E\to\mathbb{R}$, where $p_f(x)=\lvert f(x)\rvert$ and $f\in E'$, the continuous dual of $E$. The maps $p_f$ are semi-norms, so they generate an initial topology on $E$, which we denote by $\sigma(E, E')$. This is the coarsest topology that makes all $p_f$ continuous. Check, I get this.
Now, I am supposed to show that $\sigma(E, E')$ is the coarsest topology that makes all linear maps in the dual $E'$ continuous. But aren't the maps in $E'$ continuous by definition? Or is there a typo in my notes and do they mean the algebraic dual $E^*$? In my notations
$$E^*:=\{f\colon E\to\mathbb{R} \mid f \text{ linear}\},\\
E':=\{f\colon E\to\mathbb{R} \mid f \text{ linear and continuous}\}.$$
Probably this is really easy, but I'm completely confused for the moment.
 A: I finally get it thanks to a bit of sleep and my teacher's explanation. First of all, the notation was indeed correct. For a Hausdorff locally convex space $(E, \tau)$, we define $$E':=\{f\colon (E, \tau)\to \mathbb{R} \mid f\text{ is linear and continuous}\}.$$
The subtility is that these maps can become discontinuous for another topology on $E$.
Now, note that $f$ is continuous if and only if $p_f$ is continuous. It is sufficient to show that this equivalence is true for $0$. In order to do this, simply write down the inverse images of an open neighbourhood of $0$:
$$p_f^{-1}([0, a[)=\{x\in E \mid \lvert f(x)\rvert=p_f(x)<a\}=\{x\in E \mid -a<f(x)<a\}=f^{-1}(]-a, a[).$$
It follows that the initial topologies for the sources $(f\colon E\to \mathbb{R})_{f\in E'}$ and $(p_f\colon E\to \mathbb{R})_{f\in E'}$ are exactly the same, so $\sigma (E, E')$ is also initial for $(f\colon E\to \mathbb{R})_{f\in E'}$, which means by definition that it is the coarsest topology that makes all the $f\in E'$ continuous.
