Functional Analysis proof of uniform convergence

Question

I was told that it is suppose to be a proof involving epsilon I am just hoping to get a few pointers.

I guess I'm just confused on the notation of P_R(f) and trying to understand what it is. Should it be treated as a function?

Answer 1

First the fact that $f_n$ converges uniformly to $f$ means that $\forall\epsilon_1>0,\exists N_1\in\mathbb{N},\forall t\in[-R,R] : n\geq N_1\implies \lvert f_n(t) - f(t)\rvert\leq\epsilon_1$

Let us consider $p_{R}(f_n-f)=sup_{t\in[-R,R]}\lvert f_n(t) -f(t)\rvert$.

We want to show that : $\forall\epsilon_2>0,\exists N_2\in\mathbb{N} : n\geq N_2\implies \lvert sup_{t\in[-R,R]}\lvert f_n(t) -f(t)\rvert\rvert = sup_{t\in[-R,R]}\lvert f_n(t) -f(t)\rvert\leq\epsilon_2$

My idea is to fix $\epsilon_2$ and go to the definition of uniformly convergent sequence to put $\epsilon_1= \epsilon_2$ ! Indeed, this will give us a $N_1\in\mathbb{N}$ such that for all $t\in[-R,R] : $

$n\geq N_1\implies \lvert f_n(t)-f(t)\rvert\leq\epsilon_1$

However, since it is for all $t\in[-R,R]$ it also includes $sup_{t\in[-R,R]}\lvert f_n(t) -f(t)\rvert$. So we have $n\geq N_1\implies sup_{t\in[-R,R]}\lvert f_n(t) -f(t)\rvert\leq \epsilon_1=\epsilon_2 $

To show the converse. We have that $\forall\epsilon_2>0,\exists N_2\in\mathbb{N} : n\geq N_2\implies sup_{t\in[-R,R]}\lvert f_n(t) -f(t)\rvert\leq\epsilon_2$

and we want to show that $\forall\epsilon_1>0,\exists N_1\in\mathbb{N},\forall t\in[-R,R] : n\geq N_1\implies \lvert f_n(t) - f(t)\rvert\leq\epsilon_1$

My idea : fix $\epsilon_1$ and put $\epsilon_2=\epsilon_1$. As before, it will give us a $N_2\in\mathbb{N}$ such that $n\geq N_2\implies sup_{t\in[-R,R]}\lvert f_n(t) -f(t)\rvert\leq\epsilon_2$

however it is clear that $\forall t\in[-R, R], \lvert f_n(t) - f(t)\rvert\leq sup_{t\in[-R,R]}\lvert f_n(t) -f(t)\rvert$

So we have that $n\geq N_2 \implies\lvert f_n(t) - f(t)\rvert\leq sup_{t\in[-R,R]}\lvert f_n(t) -f(t)\rvert \leq\epsilon_2=\epsilon_1$

I hope this will help !

I put this as an "answer" but don't hesitate to vote against it if something is wrong or not clear or comment