Edit: Definition of uniform metric: Let $\Lambda $ be a nonempty set and $\bar{d}$ be the standard bounded metric on $\mathbb{R}$. Define a metric $\bar{\rho}$ on ${\mathbb{R}}^{\Lambda}$
by $\bar{\rho}(x,y)= lub${$\bar{d}(x_{\alpha}, y_{\alpha}): \alpha \in \Lambda$}.$\bar{\rho}$ is a metric called uniform metric on $\mathbb{R}^{\Lambda}$.
Consider the problem asked in my end term examination.
Let $X$ be a set , $(Y,d)$ be a metric space, and suppose $F$ is a subset of $Y^X$ such that if $f,g \in F$ then $\{d(f(x),g(x)): x \in X\}$ is bounded. Let $\rho$ be be the sup metric on $F$, and let $\bar{\rho}$ be the uniform metric on $Y^{X}$. Prove that if $f,g \in F$, then $\bar{\rho}(f,g)= \min\{\rho(f,g),1\}$.
On the LHS of what we have to prove, we have a term which is always less than $1$, and on RHS is also less than $1$ due to minimum with $1$. But on LHS is the least upper bound of $$\frac{d(f,g)}{1+d(f,g)}$$ and on RHS is the min of $\{d(f,g) ,1\}$.
I don't have any idea on how should I prove them equal.
Do you mind helping me?
Best Answer
I am assuming that the uniform metric on $Y^X$ is defined as $$\bar\rho(f,g) := \sup_{x \in X} \min(d(f(x),g(x)),1), \quad f,g \in Y^X. $$ Then, what we need to show amounts to showing $$ \sup_{x \in X} \min(d(f(x),g(x)),1) = \min(\sup_{x\in X} d(f(x),g(x)),1).$$ Choose $f,g \in F$ and let's distinguish two cases.