Understanding the weak topology, and showing a function is weakly continuous My first trouble is understanding the weak topology. I understand this definition: Given a vector space $X$, every functional in its dual $f \in X^*$ has an associated seminorm $|f|$, and the weak topology is the coarsest topology such that all semi-norms $|f|$  are continuous. But my question is how can we compare this to other topologies on $X$, for example

Question 1: If $X$ is a normed vector space, can we say that the norm topology contains the weak topology? And if so, who is to say that the norm topology is such that the semi-norms  $|f|$ associated to the functionals $ f \in X^*$ are norm-continuous?

Also comes the question of showing a function is weakly continuous,  which is so much more unintuitive to me than showing norm-continuity.  For example,

Question 2: Given $X$ normed vector space and bounded linear operator $L: X \to X$, show that the map $f : x \mapsto \| Lx \|$ is weakly continuous

First I tried to prove it by showing that $f \in X^*$, so that $|f|$ is a semi-norm that's continuous in the weak topology by definition... But it's not possible!! Then I tried to show that the preimage of basis elements of the weak topology on $\mathbb{R}$ are weak-open in $X$, but I cant wrap my head around what weak-open sets would be. So now I figure maybe we exploit relationships between norm topology and weak topology, but not sure how 
 A: One goes about understanding a topology by looking at the basis, the "simplest" open sets of which all other sets are composed as unions.  For a metric space, the basis consists of the balls, of course.  And what's even nicer about topological vector spaces is that you only have to study what the basis sets look like at the origin, since topological vector spaces are shift-invariant.  
The norm-topology's basis at the origin consists of the balls $ B(0,r) $.  Easy enough.  What do basis sets look like in the weak topology?  Take a bounded linear functional $ f \in X^* $, and then look at $ \{ x \in X : |f(x)| < r \} $.  You can hopefully imagine, this is "the space that is sandwiched between two hyperplanes".  Now, this isn't quite the basis.  To get a basis (at the origin), you need to consider all possible intersections of these "sandwiches".  But the important thing is, you only get to "sandwich" in a finite number of directions $ f_1, \ldots, f_N $.  So, in the infinite dimensional case, there's always some direction that you fail to sandwich.  Hence, these basis sets are not bounded.  This is counterintuitive, because in $ \mathbb{R}^3 $ (for example), you obtain a bounded set once you "sandwich" along the x-axis, y-axis, and z-axis (or in 3 directions of your choice).  
Question 1  Let's first get clear on why the norm-topology contains the weak topology.  Let $ U $ be open in the weak topology.  Fix a point $ x_0 \in U $.  We can find a basis set $ V $ with $ x \in V \subseteq U $.  Recall, by "basis set", I mean a finite number of sandwiches, but we can be rigorous about this: for some numbers $ m_1 < M_1, \ldots m_N < M_N $ and $ f_1, \ldots f_N \in X^* $, the basis set $ V $ can be expressed 
$$
V = \{ x \in X : m_1 < f_1(x) < M_1, \ldots, m_N < f_N(x) < M_N \}
$$
Now, it's pretty straightforward to find a ball $ B(x_0, r_0) $ that fits inside of $ V $.  Do you see how?  It relies on the boundedness of hte functionals $ f_1, \ldots f_N $.  Thus we have
$$
x_0 \in B(x_0, r_0) \subseteq V \subseteq U
$$
which, you'll recall, shows that $ U $ is open in the norm topology (make sure you see why). 
Question 1, Part b You then asked why $ x \mapsto |f(x)| $ is norm-continuous.  But this is a composition of two continuous functions, the absolute value function and $ f \in X^* $ (which is continuous by definition).  
Question 2 You asked about showing that $ g : x \mapsto \|Lx\| $ is weakly continuous, where $ L:X \rightarrow X $ is a bounded linear map.  Surprisingly, it is not continuous with respect to the weak topology (in general).  Indeed, pick $ X $ to be your favorite infinite-dimensional space, let's say $ X = \mathcal{l}^1(\mathbb{N}) $ and $ L = Identity : X \rightarrow X $.   Then I claim $ g : x \mapsto \|Lx\| $ is not weakly continuous, since indeed $ g^{-1}((-a,a)) = B(0,a) $ is bounded, whereas weakly open sets in $ \mathcal{l}^1 $ are not bounded (!).  
Addendum Fundamentally, the thing that's surprising here is that the norm $ x \mapsto \|x\| $ is not weakly continuous.  To try to wrap our heads around this, consider the weak basis we discussed above, composed of finite intersections of "hyperplane sandwiches."  This basis of the weak topology tells us that, in order for a function $ f : X \rightarrow \mathbb{R} $ to be weakly continuous, it is only allowed to change in a finite number of directions.  The norm function, on the other hand, changes in all directions that you walk away from the origin.  
