What does point-wise convergence in the sup norm mean? I am starting out with functional analysis and learning about point-wise and uniform convergence. In particular, I need help with my proof that the space $$ \mathcal{B} = \left\{ f:X\rightarrow \mathbb{R} \, \middle| \, f \, \text{bounded} \right\}$$
of bounded real functions on an arbitrary space $X$ is complete in the sup norm $$ \lVert f \rVert_{\infty} := \sup_{x\in X} \lvert f\left(x\right) \rvert. $$
My book (Intro to Topology and Modern Analysis by Simmons) poses this proof as a problem before any discussion on convergence of function sequences, so I will not use terms like "point-wise" or "uniform" convergence in the proof itself.
My Proof
Let $\left(f_n\right)_{n\in \mathbb{N}} \subset \mathcal{B}$ be a Cauchy sequence.
Letting $\epsilon >0$, there is an $N\in \mathbb{N}$ such that $$ m,n\geq N \implies \lVert f_m - f_n\rVert_{\infty} < \epsilon. $$ Thus, by definition of the sup norm we have for every $x \in X$, $$ m,n \geq N \implies \lvert f_m\left(x\right) -f_n\left(x\right) \rvert \leq \lVert f_m-f_n\rVert_{\infty} < \epsilon $$ This shows that, for fixed $x$, the real sequence $\left(f_m\left(x\right)\right)_{m\in \mathbb{N}}$ is Cauchy and converges since $\mathbb{R}$ is complete. Hence, we define $f:X\rightarrow \mathbb{R}$ by $$ f\left(x\right) = \lim_{m\rightarrow \infty} f_m\left(x\right). $$
So far we have merely constructed a function $f$ to which our sequence $\left(f_m\right)_{m\in \mathbb{N}}$ may converge (see discussion after proof); it remains to show that it does converge as a bona fide sequence in the sup norm.
To this end, let $\epsilon >0$ and take $N$ in the Cauchy hypothesis so that for any $x\in X$and $m,n> N$ we have $\lvert f_m\left(x\right) - f_n\left(x\right) \rvert < \frac{\epsilon}{2}$. Then for this same $n>N$ we have $$ \lvert f\left(x\right) - f_n\left(x\right)\rvert = \lvert \lim_{m\rightarrow \infty} f_m\left(x\right) - f_n\left(x\right) \rvert = \lim_{m\rightarrow \infty} \lvert f_m\left(x\right) - f_n\left(x\right) \rvert < \frac{\epsilon}{2}  $$
(pulling the limit out of the absolute value required a small side lemma). Thus, we have $$  n>N \implies \lVert f-f_n\rVert_{\infty} = \sup_{x\in X} \lvert f\left(x\right) -f_n\left(x\right)\rvert \leq \frac{\epsilon}{2} < \epsilon $$ so that $\left(f_n\right)_{n\in \mathbb{N}}$ converges to our $f$ in the sup norm.
Proving the boundedness of the limit function $f$ is a whole other can of worms which is irrelevant to my confusion, so I will avoid typing it out here. Once the limit function is bounded, we know that the Cauchy sequence converges in $\mathcal{B}$ and so $\mathcal{B}$ is complete.
My Confusion
I was happy with this proof until I learned more about point-wise versus uniform convergence. My proof shows that the sequence $\left(f_n\right)_{n\in\mathbb{N}}$ converges in the sup norm to the $f$ we constructed. In doing so, I see in retrospect that I also proved that this is uniform convergence. This uniform convergence seemed to come naturally out of the proof since we used the sup norm: we found an $N\in \mathbb{N}$ for which the $\epsilon$ property holds for all $x\in X$. However, I didn't set out to prove uniform convergence, only convergence in the norm we had defined on the function space, so it looks like this convergence in this norm might imply uniform convergence.
My question is: where does the notion of point-wise convergence enter in? The construction of my limit function $f$ is done in the usual "point-wise" manner, and so I assume that by construction I can say that "$\left(f_n\right)_{n \in \mathbb{N}}$ converges point-wise to $f$," but I am bothered by when I specialize to a point $x\in X$, my "point-wise convergence" ends up using only the standard norm on $\mathbb{R}$ and not the sup norm.
Hence, I can imagine we might have function sequences which converge "point-wise" but not uniformly; how are we justified in saying that they "converge," then, since they won't converge as sequences in $\mathcal{B}$, where the norm is the sup norm?
 A: It seems to me that you are overthinking this problem. You proved that, for every $\varepsilon>0$, there is some $N\in\Bbb >N$ such that if $n\geqslant N$ and if $x\in X$, then$$\sup_{x\in X}\bigl|f(x)-f_n(x)\bigr|<\varepsilon;$$in other words, you proved that, for every $\varepsilon>0$, there is some $N\in\Bbb >N$ such that if $n\geqslant N$ and if $x\in X$, then $\|f-f_n\|_\infty<\varepsilon$, which is what you were supposed to proved.
Concerning your remark that “using only the standard norm on R and not the sup norm”, well… what's the sup norm on $\Bbb R$. On $\Bbb R^n$, you can define$$\|(x_1,\ldots,x_n\|_{\text{sup}}=\sup\{|x_1|,|x_2|,\ldots,|x_n|\}=\max\{|x_1|,|x_2|,\ldots,|x_n|\},$$but, when $n=1$, this is just the usual distance in $\Bbb R$. Besides, on $\Bbb R^n$ all norms are equivalent.
A: After focusing a bit more on the basic definitions, I think I've made sense of this on my own. I'll leave my notes here in this answer to help anyone else who hits upon this.
I will generalize the definitions of convergence that I was using to more general spaces. Let $\left(X, d_X\right)$ and $\left(Y,d_Y\right)$ be metric spaces and $f$ and $\left\{f_n\right\}_{n\in\mathbb{N}}$ be functions from $X$ to $Y$.
We say that "$\left(f_n\right)_{n\in\mathbb{N}}$ converges pointwise to $f$" if $\forall \epsilon>0$ and $x\in X$ we can find some $N \in \mathbb{N}$ so that $$ n>N \implies d_{Y}\left(f_n\left(x\right), f\left(x\right) \right) < \epsilon. $$ Notice that we don't use the metric on $X$ anywhere in this definition, since we are concerned with a single point in the domain; the metric on $Y$ allows us to quantify the distance between the function values (in my original problem, this was simply the usual metric on $\mathbb{R}$ since my functions were real-valued).
We say that "$\left(f_n\right)_{n\in\mathbb{N}}$ converges uniformly to $f$" if $\forall \epsilon >0$ we can find some $N\in \mathbb{N}$ so that for all $x\in X$ we have $$ n>N \implies d_Y\left(f_n\left(x\right), f\left(x\right) \right) < \epsilon. $$ Again we do not use the metric on $X$.
These two notions of convergence of function sequences are certainly not equivalent, though uniform convergence implies pointwise. My earlier proof proceeded by finding a pointwise limit of the sequence and showing that --- by the grace of its Cauchy-ness --- this convergence was also uniform. My confusion started when trying to consider these function sequences as sequences in a function space equipped with the uniform metric. My proof below shows that uniform convergence of a function sequence is equivalent to convergence of the sequence in a function space equipped with the uniform norm. Thus, I am happily convinced that these notions of convergence coincide and it makes sense to consider uniform convergence as true convergence of a sequence of elements (functions) in a (function) space.
Proof
First let $\left(f_n\right)_{n\in \mathbb{N}}$ converge uniformly to $f$, all functions from metric space $\left(X, d_X\right)$ to $\left(Y, d_Y\right)$ belonging to some function space, call it $\mathcal{C}$. Define the uniform metric on $\mathcal{C}$ to be the usual $$ d_{\mathcal{C}}\left(f,g\right) = \lVert f - g\rVert_{\infty} := \sup_{x\in X} d_Y\left(f\left(x\right), g\left(x\right)\right). $$
Let $\epsilon >0 $ arbitrary. Then by uniform convergence there exists some $N \in \mathbb{N}$ so that $$ \forall x\in X, n > N \implies d_Y\left(f_n\left(x\right), f\left(x\right) \right) < \frac{\epsilon}{2}. $$ Now let $$B = \left\{ d_Y\left(f_{n'}\left(x'\right), f\left(x'\right)\right) \, \middle| \, x' \in X, n' > N\right\}. $$ We have just shown that this set $B\subset \mathbb{R}$ is bounded, and as a subset of $\mathbb{R}$ it has a supremum $\beta = \sup B.$ But then $\beta \leq \frac{\epsilon}{2} < \epsilon$. Now, notice that since for a fixed n > N, $$ \left\{ d_Y\left(f_n\left(x\right), f\left(x\right)\right)\, \middle| \, x\in X  \right\} \subseteq B, $$ we thus have $$ n>N \implies \lVert f_n - f\rVert_{\infty} = \sup_{x\in X} d_Y\left(f_n\left(x\right), f\left(x\right)\right) \leq \beta < \epsilon. $$ That is, $\left(f_n\right)_{n\in \mathbb{N}}$ converges to $f$ in the uniform norm on $\mathcal{C}$.
Conversely, let $\left(f_n\right)_{n\in \mathbb{N}}$ converge to $f$ in the uniform norm on $\mathcal{C}$. Then let $\epsilon >0 $ and take $N \in \mathbb{N}$ so that $$ n>N \implies \lVert f_n - f\rVert_{\infty} = \sup_{x\in X} d_Y\left( f_n\left(x\right), f\left(x\right)\right) < \epsilon.$$ But then for arbitrary $x\in X$ we have $$ n>N \implies d_Y\left(f_n\left(x\right), f\left(x\right)\right) \leq \lVert f_n - f\rVert_{\infty} < \epsilon,$$ so that $\left(f_n\right)_{n\in\mathbb{N}}$ converges uniformly to $f$. $\blacksquare$
