First, we define a metric. Let $$d^\infty(f,g) = \sup \{\rho(f(x),g(x))| x \in X\}$$ for some topological space $X$. Now let $Y$ be a metric space, and let $C_b(X,Y)$ denote the space of bounded continuous functions from $X$ to $Y$.
My question is to determine whether or not there are two spaces so that the topology generated by the metric is the compact-open topology on this space of bounded continuous functions.
So far, I have been able to show that if $X$ is compact, then the topology is the same. In light of this, it makes sense that we should only be looking at non-compact spaces. I have been trying to consider what happens in cases with spaces I am familiar with, like maps from $\mathbb{R} \rightarrow \mathbb{R}$, but this has turned up nothing.
As a next approach, I wanted to produce two metrics on this space, and show they do not produce equivalent topologies (while still looking at maps from $\mathbb{R} \rightarrow \mathbb{R}$). The idea was to write $\mathbb{R} = \bigcup_{n \in \mathbb{N}} [-n,n],$ and then use a metric defined by $$D_n (f,g) =\max\{ |f(x)-g(x)| : x \in [-n,n]\}$$ and then set $D(f,g)= \sum \frac{D_n (f,g)}{2^n (1+D_n (f,g))}$. This metric should be okay because all the maps in consideration are bounded, so it's easy to see this sum is well-defined.
The only reason I suspect something like this will work at all is because I have been told that this metric induces the compact-open topology, and so I thought it might be easier to compare two metrics and try to show that they do not produce equivalent topologies, by trying to violate the inequalities for equivalent metrics. For my purpose, this means it is enough to show that there do not exist numbers $k, K, k<K$ so that $kD(f,g) \leq d^\infty (f,g) \leq KD(f,g)$ for all $f,g$.
Intuitively, my (limited) experience with analysis makes me believe such functions can exist, but I need to find them.
Edit: I think I can, without loss of generality, just take one of the functions in each distance to be identically $0$, so that these distances simplify. In the first case, $d^\infty (f,0) = \sup \{\rho(f(x),0)) |x\in \mathbb{R} \}$, each $D_n = \max \{|f(x)|, x\in [-n,n]$ and so if we rename those values $\max_n f(x)$, then $D = \sum \frac{\max_n f}{2^n (1+ \max_n f)}$
Edit 2: After staring at these simplified metrics (formed by subtracting $g(x)$ from both functions and replacing $f(x)-g(x)$ with just $f(x)$ again, now I'm not sure eveything's right with this metric. Why don't constant maps show this? In one metric, constant maps with large image are actually close to 1, while in the other, they are just arbitrarily large. This makes me suspicious that something's wrong with the metric, but I can't see what.
Since $Y$ is a metric space, the compact-open topology (which is also defined for a mere topological space $Y$) is the same as the topology of compact convergence, in which a sequence of maps $(f_n)_n$ converges to $f$ if and only if for each compact subset $K$ of $X$, the restrictions $f_n|_K$ converge uniformly to $f|_K$, i.e. with respect to the sup-metric restricted to $K$.
For $X=Y=\Bbb R$, the maps $$ f_n(x) = \begin{cases} 0, &\text{ if } x\le n \\ x-n, &\text{ if } n\le x\le n+1\\ 1, &\text{ if }n+1\le x \end{cases} $$ for $n\in\Bbb N$, converge compactly to $f=0$ since on any bounded set they are eventually $0$. But they do not converge uniformly since $d^\infty(f_n,f)=1$ for any $n$.