Does \(g \circ f_n \to g \circ f\) Uniformly on \(A\)?

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Discussion Overview

The discussion revolves around a mathematical exercise concerning the uniform convergence of compositions of functions. Participants explore the implications of uniform convergence of a sequence of functions \(f_n\) to a function \(f\) on a set \(A\), and how this affects the uniform convergence of the composed functions \(g \circ f_n\) to \(g \circ f\), given that \(g\) is continuous on the interval \([a,b]\).

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a solution involving the uniform convergence of \(f_n\) to \(f\) and the continuity of \(g\), leading to the conclusion that \(g \circ f_n\) converges uniformly to \(g \circ f\).
  • Another participant questions the necessity of explicitly stating certain conditions, such as the requirement for \(\forall x \in A\) in the context of uniform convergence.
  • Several participants discuss whether it is necessary to re-establish parameters like \(\epsilon\) and \(n_0\) when making claims about the convergence of \(g(f_n(x))\) to \(g(f(x))\).
  • There is a suggestion that the notation and conditions could be simplified, but others emphasize the importance of clarity in stating conditions for uniform convergence.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of certain formalities in the proof. While some advocate for a more concise approach, others argue for the importance of explicitly stating all conditions to avoid ambiguity. No consensus is reached on the best approach to present the argument.

Contextual Notes

Participants highlight the importance of maintaining rigor in mathematical arguments, particularly regarding the conditions under which uniform convergence is established. There is an emphasis on the need for clarity in notation and assumptions, especially when dealing with limits and continuity.

Who May Find This Useful

This discussion may be useful for students and practitioners in mathematics, particularly those studying real analysis and the properties of uniform convergence and continuity.

evinda
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Hi! :)
I am looking at this exercise: Let $f_n \to f$ uniformly on $A$. If $f_n: A \to [a,b], \forall n \in \mathbb{N}$,show that $f:A \to [a,b]$.If,$g:[a,b] \to \mathbb{R}$ is continuous at $[a,b]$,show that $g \circ f_n \to g \circ f$ uniformly on $A$.

That's the solution that the assistant of the professor gave us:

Firstly,we have $a \leq f_n(x) \leq b \Rightarrow \lim_{n \to +\infty} a \leq \lim_{n \to +\infty} f_n(x) \leq \lim_{n \to +\infty} b \Rightarrow a \leq f(x) \leq b$.

$g$ is continuous at $[a,b]$,so it is also uniformly continuous at this interval.So, $\forall \epsilon>0, \exists \delta=\delta(\epsilon)>0$ such that $\forall x,y \in [a,b]$ with $|x-y|< \delta \Rightarrow |g(x)-g(y)| < \epsilon$.
Let $\epsilon'>0$ .$\exists n_0 \in \mathbb{N}$ such that $\forall n \geq n_0$ we have $|f_n(x)-f(x)|< \epsilon' , \forall x \in A$.
We choose $\epsilon'=\delta$.Let now $\epsilon>0$.For $\delta>0, \exists n_0$ such that $\forall n \geq n_0, |f_n(x)-f(x)|< \delta, \forall x \in A \Rightarrow |g(f_n(x))-g(f(x))|< \epsilon $ .
Taking the supremum we have $\sup_{x}|g(f_n(x))-g(f(x))|< \epsilon \Rightarrow \sup |g(f_n(x))-g(f(x))| \to 0$.
But...is it necessary to write it like that:

We choose $\epsilon'=\delta$.Let now $\epsilon>0$.For $\delta>0, \exists n_0$ such that $\forall n \geq n_0, |f_n(x)-f(x)|< \delta, \forall x \in A \Rightarrow |g(f_n(x))-g(f(x))|< \epsilon $ .

Couldn't I just do it like that? :confused:
We pick $\epsilon'=\delta$.Since $f_n,f \in [a,b]$ we have that $|f_n-f|< \delta \Rightarrow |g(f_n(x))-g(f(x))|< \epsilon$.
 
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What's the difference? Is it that you forgot a '$$\forall x \in A$$' at the end?(Giggle)
 
Do I have to set again an $\epsilon$ and a $n_0$, such that the conditions are satisfied?Or can I just say that because of the fact that $f_n,f \in [a,b]$ and $|f_n-f| < \delta$,we have that $|g(f_n(x))-g(f(x))|< \epsilon$ ? (Wondering)
 
evinda said:
Do I have to set again an $\epsilon$ and a $n_0$, such that the conditions are satisfied?Or can I just say that because of the fact that $f_n,f \in [a,b]$ and $|f_n-f| < \delta$,we have that $|g(f_n(x))-g(f(x))|< \epsilon$ ? (Wondering)

No but if you like:cool:
 
So,can I say it like that?

Firstly,we have $a \leq f_n(x) \leq b \Rightarrow \lim_{n \to +\infty} a \leq \lim_{n \to +\infty} f_n(x) \leq \lim_{n \to +\infty} b \Rightarrow a \leq f(x) \leq b$.

$g$ is continuous at $[a,b]$,so it is also uniformly continuous at this interval.So, $\forall \epsilon>0, \exists \delta=\delta(\epsilon)>0$ such that $\forall x,y \in [a,b]$ with $|x-y|< \delta \Rightarrow |g(x)-g(y)| < \epsilon$.
Let $\epsilon'>0$ .$\exists n_0 \in \mathbb{N}$ such that $\forall n \geq n_0$ we have $|f_n(x)-f(x)|< \epsilon' , \forall x \in A$.

We pick $\epsilon'=\delta$.Since $f_n,f \in [a,b]$ we have that $|f_n-f|< \delta \Rightarrow |g(f_n(x))-g(f(x))|< \epsilon$.

Taking the supremum we have $\sup_{x}|g(f_n(x))-g(f(x))|< \epsilon \Rightarrow \sup |g(f_n(x))-g(f(x))| \to 0$.
 
evinda said:
So,can I say it like that?

Firstly,we have $a \leq f_n(x) \leq b \Rightarrow \lim_{n \to +\infty} a \leq \lim_{n \to +\infty} f_n(x) \leq \lim_{n \to +\infty} b \Rightarrow a \leq f(x) \leq b$.

$g$ is continuous at $[a,b]$,so it is also uniformly continuous at this interval.So, $\forall \epsilon>0, \exists \delta=\delta(\epsilon)>0$ such that $\forall x,y \in [a,b]$ with $|x-y|< \delta \Rightarrow |g(x)-g(y)| < \epsilon$.
Let $\epsilon'>0$ .$\exists n_0 \in \mathbb{N}$ such that $\forall n \geq n_0$ we have $|f_n(x)-f(x)|< \epsilon' , \forall x \in A$.

We pick $\epsilon'=\delta$.Since $f_n,f \in [a,b]$ we have that $|f_n-f|< \delta \Rightarrow |g(f_n(x))-g(f(x))|< \epsilon$.

Taking the supremum we have $\sup_{x}|g(f_n(x))-g(f(x))|< \epsilon \Rightarrow \sup |g(f_n(x))-g(f(x))| \to 0$.

You should say $$f(x) \in [a,b]$$, $$|f_n(x)-f(x)|< \delta$$ and so on. Others are OK.
 
stainburg said:
You should say $$f(x) \in [a,b]$$, $$|f_n(x)-f(x)|< \delta$$ and so on. Others are OK.

Ok,thanks! :)
 

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