When does the uniform metric coincide with the sup metric on $C(X,Y)$ Suppose I have two topological spaces $X,Y$, and define the uniform metric defined on $C(X,Y)$ (continuous maps from $X$ to $Y$) to $p^{-}(f,g) = sup$ ${d^{-}(f(x),g(y))}$ where $d^{-} = min (1, d(x,y))$, and the sup metric to be the normal one.
My question is when are these two metrics equivalent topologically. If $X$ is compact, then it seems like  $p^{-}$ will coincide (not sure entirely why). Any insights appreciated.
 A: Your question is not precise.
(1) You should explicitly mention that $Y$ is a metric space (with metric $d$).
(2) If $C(X,Y)$ contains unbounded functions, then there does not exist a "normal" sup-metric. To assure that all functions are bounded, you need additional requirements, for example that $X$ is compact or that $d$ is a bounded metric. Introducing the metric $d^-$ is a standard "trick" to endow $Y$ with a bounded metric which is equivalent to $d$.
So let us assume that all functions in $C(X,Y)$ are bounded with respect to $d$. Only in that case $p(f,g) = \sup_{x \in X} d(f(x),g(x))$ is well-defined. We shall show that $ p^-,p$ induce the same topology. To do this, it suffices to show $p^-(f,g) = p(f,g)$ if $p^-(f,g) < 1$ or $p(f,g) < 1$. This means that $p^-$ and $p$ have the same open $\epsilon$-neighborhoods for $0 < \epsilon < 1$.
Both $p^-(f,g) < 1 $, $p(f,g) < 1$ say that $d(f(x),g(x)) < 1$ for all $x$. Therefore $d^-(f(x),g(x)) = d(f(x),g(x))$ which implies $p^-(f,g) = p(f,g)$.
Edited: It seems to be a good idea to clarify what a "bounded" function is. This is clear if we consider $Y = \mathbb{R}$ with the usual metric $d(x,y) = \lvert x - y \rvert$. In the general case we say that $f$ is bounded if $\delta(f) = \sup_{x,y \in X} d(f(x),f(y)) < \infty$ (that is, if the image $f(X)$ is a bounded subset of $Y$).
a) $p(f,g)$ is well-defined for all bounded functions $f,g$:
Choose any $z \in X$. Then 
$$d(f(x),g(x)) \le d(f(x),f(z)) + d(f(z),g(z)) +  d(g(z),g(x)) \\ \le \delta(f) + d(f(z),g(z)) + \delta(g) .$$
b) If $f$ is undounded and $g$ is bounded, then $\sup_{x \in X} d(f(x),g(x)) = \infty$:
There exist $x_n,y_n \in X$ such that the sequence $(d(f(x_n),f(y_n)))$ is unbounded. Since
$$d(f(x_n),f(y_n)) \le d(f(x_n),g(x_n)) + d(g(x_n),g(y_n)) + d(g(y_n),f(y_n)) \\
\le d(f(x_n),g(x_n)) + \delta(g) + d(g(y_n),f(y_n)) ,$$ at least one of the sequences $(d(f(x_n),g(x_n)))$, $(d(g(y_n),f(y_n)))$ must be unbounded. This implies $\sup_{x \in X} d(f(x),g(x)) = \infty$.
Let us conclude with the remark that bounded metrics $d_1,d_2$ on $Y$ may induce different toplogies on $C(X,Y)$ even if they are equivalent. As an example take $Y = \mathbb{R}$ and$d_1(x,y) = \min(1,\lvert x -y \rvert)$ and $d_2(x,y) = \big\lvert \frac{x}{1+\lvert x \rvert} - \frac{y}{1+\lvert y \rvert} \big\rvert$. Details are left as an exercise!
