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

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In summary, the conversation discusses the proof of $f_n \to f$ uniformly on $A$. It is shown that if $f_n$ maps from $A$ to $[a,b]$ for all $n \in \mathbb{N}$, then $f$ also maps from $A$ to $[a,b]$. Additionally, it is proved that if $g:[a,b] \to \mathbb{R}$ is continuous, then $g \circ f_n \to g \circ f$ uniformly on $A$. The conversation also includes a discussion about the notation and details of the proof.
  • #1
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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  • #2
What's the difference? Is it that you forgot a '\(\displaystyle \forall x \in A\)' at the end?(Giggle)
 
  • #3
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)
 
  • #4
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:
 
  • #5
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 $\sup_{x}|g(f_n(x))-g(f(x))|< \epsilon \Rightarrow \sup |g(f_n(x))-g(f(x))| \to 0$.
 
  • #6
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 \(\displaystyle f(x) \in [a,b]\), \(\displaystyle |f_n(x)-f(x)|< \delta\) and so on. Others are OK.
 
  • #7
stainburg said:
You should say \(\displaystyle f(x) \in [a,b]\), \(\displaystyle |f_n(x)-f(x)|< \delta\) and so on. Others are OK.

Ok,thanks! :)
 

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

What does it mean for a sequence of functions to converge uniformly?

Uniform convergence means that the limit of the sequence of functions is the same for all points in the domain, regardless of how close those points are to each other. In other words, the rate at which the functions approach their limit is consistent across the entire domain.

How is uniform convergence different from pointwise convergence?

Pointwise convergence means that the limit of the sequence of functions is the same for each individual point in the domain. This is different from uniform convergence, where the limit is consistent across the entire domain.

Why is uniform convergence important in mathematical analysis?

Uniform convergence is important because it allows us to interchange limits and integrals, which is a crucial tool in mathematical analysis. It also helps us prove continuity and differentiability of functions, as well as establish the convergence of series.

How is the uniform convergence of gof_n->gof related to the uniform convergence of f_n->f and g_n->g?

If both f_n and g_n converge uniformly to f and g respectively, then the composition gof_n will also converge uniformly to gof. This is because uniform convergence is preserved under composition, meaning that if two sequences of functions converge uniformly, their composition will also converge uniformly.

Can uniform convergence be proven using the ε-N definition of limit?

Yes, uniform convergence can be proven using the ε-N definition of limit. However, it requires some additional steps compared to pointwise convergence. In order to prove uniform convergence, we must show that for any given ε, there exists an N such that for all n>N, the distance between the nth term of the sequence and the limit is less than ε for all points in the domain.

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