In my course of functional analysis, we always work with a Hausdorff locally convex space when we're creating the weak* and weak topology. My professor once stated that there is a reason for this assumption and gave the example of $L^p([0,1])$ for $0<p<1$ (let me write $L^p$ for brevity). He showed that $(L^p)'=0$, but I didn't fully get the point of why he did this.
What is the relation between this fact and the weak topology on $L^p$? It seems that I'm missing something rather easy.
The problem is that $L^p$, for $p \in (0,1)$ is not locally convex.
Your professor showed that its dual is zero, hence it wouldn't make sense to talk about weak topology as it is defined using elements in the dual space.
An easy way to convince yourself such spaces are not locally convex is to define the $L^p$ norm in $\mathbb R^2$ and draw the unit ball, for $p<1$, $p=1$ and $p>1$. You will see that for $p<1$ the unit ball is not a convex set.