Making sense of weak topology and set of continuous maps

Question

It is common knowledge, for Banach spaces $X$ and $Y$, the space of bounded linear operators $\mathcal{B}(X,Y)$ is exactly the space of continuous linear maps. The weak topology on $X$ is then defined as the coarsest topology such that every norm-continuous linear functional in $X^*:=\mathcal{B}(X, \mathbb{F})$ remains continuous, where $\mathbb{F}=\mathbb{R};\mathbb{C}$ is the underlying (complete) field.

But, in general, if $X$ and $Y$ are general topological spaces, we cannot expect modifying the topology on $X$ won't result in changes of the set of continuous functions to $Y$. A smaller topology would generally result in less continuous functions.

So my questions are,

1. under what circumstances in general, can we have the phenomenon where changing the topology of $X$ doesn't change the set of continuous functions?

2. In the context of infinitely dimensional Banach spaces, do there exist weak continuous nolinear functionals? That is, under the weak topology, do we have $C(X, \mathbb{F})=X^*$, when Banach space $X$ has infinite dimensions?

Update:

After some thoughts, it seems I omitted something. The set of norm-continuous functions $C(X, Y)$ is much larger than the set of norm-continuous linear functions $\mathcal{B}(X,Y)$. When we modify the topology on $X$, in general, the size of $C(X, Y)$ cannot remain the same. But weak topology is about preserving the continuity of the linear functionals in the continuous dual $X^*$.

Answer 1

I think that your first question is very broad and can have many interesting answers depending on the context. The point is that, since you are considering $C(X,Y)$, the answer depends a lot on what is the topology of the other space $Y$. If you allow the topology of $Y$ to vary too, you can essentially make the set $C(X,Y)$ to be either the set of constant functions or the set of all functions. But if the topology of $Y$ is fixed (and somewhat nontrivial) you could have interesting behaviours, see the answer by Jochen.

Concerning the second question, as (again) Jochen suggests in a few comments, you can construct many examples, e.g. by composing any linear functional $f\in X^*$ with any continuous function $\phi:\mathbb R\to\mathbb R$.

In the above example, though, you are basically reducing yourself to the finite-dimensional case, where weak and strong topologies coincide. In fact, you can view any non-trivial functional $\phi\in X^*$ as a linear projection onto a one-dimensional space. I think that constructing a "genuinely infinite-dimensional" example is a bit less trivial (if not impossible, I don't know).

Even though this is not completely related, I want to mention the case of convex functions: if you have a continuous convex functional $\phi:X\to\mathbb R$, this is authomatically lower-semicontinuous in the weak topology of $X$ (this is related to the fact that a strongly closed, convex subset of a Banach space $X$ is also weakly closed), but you cannot upgrade this statement to continuous functions. In fact (again, again), Jochen suggests the example of the norm $||\cdot\||$, which is convex and continuous, thus weakly lower-semicontinuous, but not weakly continuous (there exist sequences of unitary vectors that converge weakly to zero). If you want to look at some references, have a look at the book of H. Brezis "Functional Analysis, Sobolev Spaces and Partial Differential Equations".

Answer 2

Pavel Urysohn constructed a Hausdorff topological space $(X,\tau)$ such that every continuous $f:X\to\mathbb R$ is constant, so that $\tau$ and the trivial topology $\{\emptyset,X\}$ have the same continuous real-valued functions. Such strange things do not happen in completely regular spaces where a set $A$ is closed if and only if, for every $x\notin A$, there is a continuous real-valued function vanishing on $A$ with $f(x)=1$. Thus, two completely regular topologies on a set coincide if they have the same continuous real-valued functions.