Making sense of weak topology and set of continuous maps 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,

*

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


*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 $X^*$.

 A: 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".
A: 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.
