# Uniform convergence with supremum norm

Let $$X$$ an arbitrary set and $$(f_n)$$ a sequence of functions, $$f_n:X\to \mathbb{R}\,\,n\in\mathbb{N}$$. We say that the sequence $$(f_n)$$ converges uniformly to a function $$f:X\to \mathbb{R}$$ if given any $$\epsilon>0$$ there is a positive integer $$N=N(\epsilon)$$ such that $$|f_n(x)−f(x)|<\epsilon \qquad \text{for every}\, n>N \,\text{and for every}\, x\in X.$$

Some authors also introduce the so called "supremum-norm": given any function $$g:X\to \mathbb{R}$$, we define $$\left\lVert g\right\rVert_{\infty}:=\sup_{x \in X}|g(x)|.$$ It is trivial to show that $$0\leq\left\lVert g\right\rVert_{\infty}\leq +\infty$$ for every $$g$$ and $$\left\lVert g\right\rVert_{\infty}<+\infty$$ if and only if $$g$$ is bounded.

They also show that given $$(f_n)$$ a sequence of functions, $$f_n:X\to \mathbb{R}\,\,n\in\mathbb{N}$$, then $$(f_n)$$ converges uniformly to a function $$f:X\to \mathbb{R}$$ if, and only if, the (numerical) sequence $$\{\left\lVert f_n-f\right\rVert_{\infty}\}_n$$ converges to $$0$$.

My doubt is the following: without specifying anything about boundedness of $$f_n,f,$$ then $$\{\left\lVert f_n-f\right\rVert_{\infty}\}_n$$ is a sequence in the extended half line $$[0,+\infty]$$.

How is defined convergence to $$0$$ (or to any other $$l\in [0,+\infty]$$) in this case? I think this is important in order to understand the equivalence between the two definitions of uniform convergence.

I must say that this way of trying to see uniform convergence is a bad way.

The convergence $$f_n \to f$$ in this norm is the same: it converges if and only if the numerical sequence $$\sup_{t\in[0,\infty]} |f_n(t) - f(t)| \to 0$$. However, the definition of your functions gets more convoluted, since now you need to specify the values $$f_n(\infty)$$ and $$f(\infty)$$, and check convergence of $$|f_n(\infty) - f(\infty)|$$.

If the functions are actually continuous, you can set $$f_n(\infty) := \lim_{t\to\infty}f_n(t)$$ to still have continuous functions. Note that in this way you've actually bounded your $$f_n$$, and it doesn't make any difference whether you're working in the usual real line or in the extended one, since: $$$$\sup_{t\in[0,\infty]} |f_n(t) - f(t)| = \sup_{t\in[0,\infty)} |f_n(t) - f(t)|$$$$

Topological note: Your example could actually be replaced by uniform convergence in $$[0,1]$$, since the extended real line $$[0,\infty]$$ is actually homeomorphic to $$[0,1]$$. For example, take the function:

\begin{align} \phi: (0,1] &\to [0,\infty)\\ t &\mapsto\phi(t) = \frac{1}{t}-1 \end{align}

and extend it to $$t=0$$ by continuity as $$\phi(0) = \lim_{t \to 0}\phi(t)=\infty$$. Similarly, we can take:

\begin{align} \psi: [0,\infty)&\to (0,1]\\ s&\mapsto\frac{1}{s+1} \end{align}

and again by continuity extend it to $$t = \infty$$ as $$\psi(\infty) = \lim_{t\to\infty}\phi(t)=0$$. This way, $$\phi$$ and $$\psi$$ are continuous functions that are inverses of each other, hence a homeomorphism.

We can define $$\|f\|=\min (1,\sup_{x\in X}|f(x)|),$$ so that $$d(f,g)=\|f-g\|$$ is a metric on $$^X\Bbb R.$$

Then $$(f_n)_n$$ converges uniformly to $$f$$ iff $$\|f_n-f\|\to 0$$ iff $$f_n\to f$$ in the metric space $$^X\Bbb R.$$