Exercise 21.7

Proof. $\text{(⇒)}$ First suppose that $(f_n)$ converges uniformly to $f$ in the functional sense. Consider any $𝜖>0$. Then, by the definition of uniform convergence there is an $N\in {\mathbb{Z}}_{\text{+}}$ where

for all $n>N$ and $x\in X$, noting that $d$ of course denotes the usual metric on $ℝ$. Now consider any $x_0\in X$ and an $n\ge N+1$ so that $n>N$. Then of course we have

from which it follows that

since $x_0$ was arbitrary. This shows that $f_n\in B_{\overline{\rho }}(f,𝜖)$. Since $n\ge N+1$ and $𝜖>0$ were both arbitrary, this shows that the sequence $(f_n)$ converges to $f$ as elements of the metric space ${ℝ}^X$.

$\text{(⇐)}$ Now suppose that $(f_n)$ converges to $f$ in the uniform metric space ${ℝ}^X$. Consider any $𝜖>0$ and let $\delta =\min <!--\; FUNCTION\; APPLICATION\; -->\left\{𝜖,1\right\}$. Then, since $B_{\overline{\rho }}(f,\delta )$ is a neighborhood of $f$ in the uniform metric space, there is an $N\in {\mathbb{Z}}_{\text{+}}$ where $f_n\in B_{\overline{\rho }}(f,\delta )$ for all $n\ge N$. So consider any $n>N$ and $x_0\in X$. Then clearly $f_n\in B_{\overline{\rho }}(f,\delta )$ so that $\overline{\rho }(f_n,f)<\delta$. Hence

so it has to be that $d(f_n(x_0),f(x_0))=\overline{d}(f_n(x_0),f(x_0))$ since

Therefore we have

Since $n>N$, $x_0\in X$, and $𝜖>0$ were arbitrary, this shows that $(f_n)$ converges to $f$ uniformly in the functional sense. □