A space of bounded functions from a set $X$ to a metric space $Y$ is itself a metric space, with distance defined by $$d(f,g):= \underset{x \in X}{\sup} \hspace{1mm} d_Y(f(x),g(x))$$
QuestionS:
1) Is the distance function $d(f,g)$ is the same as $d_Y$? If it is not the same, then is $d_Y$ is any arbitrary distance function as long as it satisfies the axioms of the metric? Wonder if someone could explain this with an example.
2) Is uniform convergence an inherent property of the above distance function $d(f,g)$?
3) Lastly, in the inequality below, does the notation $d_Y(f_n(x),f_m(x))$ means that its convergence depends on $x$ while $d(f_n,f_m) $ does not depend on $x$?
$$d_Y(f_n(x),f_m(x)) \le d(f_n,f_m) \rightarrow 0$$
Thank you!
I think you are mainly just confused about what $d_Y$ means. Here $(Y,d_Y)$ is a metric space, so $d_Y$ denotes the distance function of $Y$. This distance function is fixed as soon as you say "$Y$ is a metric space".
So $d$ and $d_Y$ are totally different kinds of objects: the inputs to $d_Y$ are elements of $Y$, while the inputs to $d$ are functions $X\to Y$.
I don't really know what your second question is supposed to mean, but convergence with respect to this metric $d$ is the same thing as uniform convergence. That is, if $(f_n)$ is a sequence of functions $X\to Y$ and $f$ is some other function $X\to Y$, then $(f_n)$ converges to $f$ with respect to the metric $d$ iff $(f_n)$ converges uniformly to $f$.
For your last question, I again don't really know what you mean. The notation $$d_Y(f_n(x),f_m(x)) \le d(f_n,f_m) \rightarrow 0$$ is really a juxtaposition of two statements: the statement that $d_Y(f_n(x),f_m(x)) \le d(f_n,f_m)$ and the statement that $d(f_n,f_m) \rightarrow 0$. These two statements together (assuming the first is true for all $n$ and $m$) imply that $d_Y(f_n(x),f_m(x))\to 0$ as well. Here $f_n(x)$ and $f_m(x)$ are points of $Y$ (depending on a choice of a specific point $x\in X$), while $f_n$ and $f_m$ are entire functions $X\to Y$.