Show that $f_n$ converges uniformly to $F$ on $D$ iff $\lim\limits_{n\rightarrow \infty} \sup\limits_{x\in D} \vert f_n(x)-F(x)\vert=0$

Question

I must show that $f_n$ converges uniformly to $F$ on $D$ iff $\lim\limits_{n\rightarrow \infty} \sup\limits_{x\in D} \vert f_n(x)-F(x)\vert=0$, and $f_n \rightarrow F$ pointwise on $D$. From what I can see, if $\sup\limits_{x\in D} \vert f_n(x)-F(x)\vert=0$ , then $f_n$ is already $\epsilon$-close to F - so a limit as $n \rightarrow \infty$ should also yield $0$?

Answer 1

(1) Suppose $f_n \to F$ uniformly on $D$. Here $\lim_n \sup_{x \in D} |f_n(x)- F(x)| = 0$ can be reformulated to say;

$$\epsilon>0,\exists N_0 \in \mathbb{N} \ s.t \ \sup_{x \in D} |f_n(x) - F(x)|< \epsilon, \forall x \in D, n \geq N_0$$

Notice that if for uniform convergence of $f_n \to F$ we have a $N_1$ such that $|f_n(x) - F(x)|< \epsilon$ for all $x \in D$ and $n \geq N_1$. Therefore, choosing $n \geq \max\{N_0,N_1\}$ then;

$$\sup_{x \in D} |f_n(x) - F(x)| < \epsilon$$

and we are done. Try and see if you can do the reverse direction on your own and comment if you have a problem.