0
$\begingroup$

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.

Thanks a lot in advance.

$\endgroup$
0
$\begingroup$

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: \begin{equation} \sup_{t\in[0,\infty]} |f_n(t) - f(t)| = \sup_{t\in[0,\infty)} |f_n(t) - f(t)| \end{equation}


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.

$\endgroup$
0
$\begingroup$

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 $^X\Bbb R.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.