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

[evinda]

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

 

[stainburg]

What's the difference? Is it that you forgot a '$$\forall x \in A$$' at the end?(Giggle)

 

[evinda]

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)

 

[stainburg]

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

 

[evinda]

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

 

[stainburg]

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.

 

[evinda]

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

Ok,thanks! :)