1
$\begingroup$

I am trying to solve a problem which says:

Let $S \subset \mathbb R^{\mathbb N}$ and let $\{f_n\}_{n \in \mathbb N}$ a sequence of functions $f_n:S \to \mathbb R$ that converges uniformly to a function $f:S \to \mathbb R$. Prove that if $f_n$ is bounded for every $n$, then $f$ is bounded and $f_n$ is uniformly bounded.

I have an extremely basic problem with this exercise: I don't understand how these functions look like. I mean, I am used to work with sequences of functions $f_n:\mathbb R \to \mathbb R$, could someone give me an example of a function $f$ with domain $S$ and codomain $\mathbb R$?. I've interpreted that an element $x$ in the domain is a sequence, so $f:S \to \mathbb R$ would be a function that associates a sequence of real numbers to a real number, am I correct?

$\endgroup$

2 Answers 2

1
$\begingroup$

You are correct that these functions associate sequences of real numbers with real numbers.

It's hard to think about these "sequences of functions defined on sequences" at first, so here's the simplest example I could come up with: Let $\{f_n\}_{n \in \mathbb{N}}$ be our sequence where $f_n = \frac{1}{n}$. Functions in this sequence take any $x \in S$ to the constant $\frac{1}{n} \in \mathbb{R}$.

$\endgroup$
0
$\begingroup$

You have a sequence of real-valued functions $f_{n}$, all of which are defined on the same set in $\mathbb{R}^{n}$. For example, maybe all of the functions are defined on the spherical shell $S$ in $\mathbb{R}^{3}$ where $x^{2}+y^{2}+z^{2}=1$. Or Maybe they're defined everywhere on $\mathbb{R}^{3}$. All of the functions have a common domain, and all take their values in $\mathbb{R}$. $S$ is just a set of points, and each $f_{n}$ associates a real number with each one of those points. Don't try to interpret a point $s \in S$ as a finite sequence, even though it is. If you were working in $\mathbb{R}^{3}$, you would just think of each point as a spot in our real world, that just happens to be described by three numbers relative to some fixed "origin". The higher the dimension, the more numbers needed to pin down the location of the point.

Hope that helps.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .