MHB Show that 1/f_n -> 1/f uniformly

[evinda]

Hey!

I am looking at the following exercise:
Let $f_n: I \to \mathbb{R}$ a sequence of functions,that do not get zero anywhere.We suppose that $f_n \to f$ uniformly and that there is a $M>0$ such that $f(x) \geq M, \forall x \in I$.Show that $\frac{1}{x_n} \to \frac{1}{f}$ uniformly.

That's what I have tried:
Let $\epsilon>0.$
Since $f_n \to f$ uniformly $\exists n_0$ such that $\forall n \geq n_0: sup_{x \in I} \{f_n-f\} \leq \epsilon $.

For these $n \text{ and } \forall x \in I$, we have:
$$|\frac{1}{f_n(x)}-\frac{1}{f(x)}|=|\frac{f(x)-f_n(x)}{f_n(x) \cdot f(x)}| \leq \frac{sup_{x \in I} \{f_n-f\}}{|f_n(x)| |f(x)|}$$

$$M \leq |f(x)|=|f(x)-f_n(x)+f_n(x)| \leq |f_n(x)-f(x)|+|f_n(x)| \leq \epsilon + |f_n(x)| \Rightarrow |f_n(x)| \geq M-\epsilon \Rightarrow \frac{1}{|f_n(x)|} \leq \frac{1}{M-\epsilon}$$

So, $$sup_{x \in I} \{|\frac{1}{f_n(x)}-\frac{1}{f(x)}|\} \leq \frac{\epsilon}{M \cdot (M- \epsilon)} \to 0 $$

So, $$\frac{1}{f_n} \to \frac{1}{f} \text{ uniformly }$$
Could you tell me if it is right? Do I have to take maybe a specific $\epsilon$? (Thinking)

 

[Evgeny.Makarov]

The idea is correct, but there are a couple of remarks.

$$|f_n(x)| \geq M-\epsilon \Rightarrow \frac{1}{|f_n(x)|} \leq \frac{1}{M-\epsilon}$$

This holds only if $M-\epsilon>0$.

$$sup_{x \in I} \{|\frac{1}{f_n(x)}-\frac{1}{f(x)}|\} \leq \frac{\epsilon}{M \cdot (M- \epsilon)} \to 0 $$

$\dfrac{\epsilon}{M \cdot (M- \epsilon)}$ tends to 0? But it does not even depend on $n$...

 

[evinda]

The idea is correct, but there are a couple of remarks.

This holds only if $M-\epsilon>0$.

$\dfrac{\epsilon}{M \cdot (M- \epsilon)}$ tends to 0? But it does not even depend on $n$...

What could I do to improve my idea?

 

[Evgeny.Makarov]

What could I do to improve my idea?

In the beginning of the proof, you are given an $\epsilon>0$. You need to find an $N$ such that \[
\|1/f_n-1/f\|=\sup_{x\in I}|1/f_n(x)-1/f(x)|<\epsilon\qquad(*)
\]
for all $n>N$. Now, one of the first steps of the proof is instantiating the definition of uniform continuity of $f_n$ with some potentially different number $\epsilon'$:
\[
f_n\to f\text{ uniformly}\implies
\forall\epsilon'\,\exists n_0\,\forall n>n_0\;\|f_n-f\|\le\epsilon'\qquad(**)
\]
You know that whatever $\epsilon'$ you use, in the end you get an upper bound of the form
\[
\|1/f_n-1/f\|\le\frac{\epsilon'}{M(M-\epsilon')}.
\]
In order to obtain (*), we must have
\[
\frac{\epsilon'}{M(M-\epsilon')}\le\epsilon
\]
It is possible to solve this inequality for $\epsilon'$, but we can simplify it and at the same time ensure that $M-\epsilon'>0$, which will solve the second problem later. Unlike $\epsilon$, which was given, we can select $\epsilon'$. Let us do it so that $\epsilon'\le M/2$, i.e., $M-\epsilon'\ge M/2>0$. Then $1/(M-\epsilon')\le 2/M$, so
\[
\frac{\epsilon'}{M(M-\epsilon')}\le\frac{2\epsilon'}{M^2}.
\]
To make the right-hand side $\le\epsilon$, we make the second restriction: $\epsilon'\le\dfrac{\epsilon M^2}{2}$. So, if
\[
\epsilon'=\min\left(\frac{M}{2},\frac{\epsilon M^2}{2}\right)
\]
and we find $n_0$ by instantiating (**) with $\epsilon'$, then according to your calculations in post #1, we have
\[
\|1/f_n-1/f\|\le\frac{\epsilon'}{M(M-\epsilon')}\le\frac{2\epsilon'}{M^2}\le\epsilon.
\]
Choosing $\epsilon'\le M/2$ also allows concluding
\[
|f_n(x)| \geq M-\epsilon' \implies \frac{1}{|f_n(x)|} \leq \frac{1}{M-\epsilon'}
\]
because $M-\epsilon'>0$.

 

[evinda]

In the beginning of the proof, you are given an $\epsilon>0$. You need to find an $N$ such that \[
\|1/f_n-1/f\|=\sup_{x\in I}|1/f_n(x)-1/f(x)|<\epsilon\qquad(*)
\]
for all $n>N$. Now, one of the first steps of the proof is instantiating the definition of uniform continuity of $f_n$ with some potentially different number $\epsilon'$:
\[
f_n\to f\text{ uniformly}\implies
\forall\epsilon'\,\exists n_0\,\forall n>n_0\;\|f_n-f\|\le\epsilon'\qquad(**)
\]
You know that whatever $\epsilon'$ you use, in the end you get an upper bound of the form
\[
\|1/f_n-1/f\|\le\frac{\epsilon'}{M(M-\epsilon')}.
\]
In order to obtain (*), we must have
\[
\frac{\epsilon'}{M(M-\epsilon')}\le\epsilon
\]
It is possible to solve this inequality for $\epsilon'$, but we can simplify it and at the same time ensure that $M-\epsilon'>0$, which will solve the second problem later. Unlike $\epsilon$, which was given, we can select $\epsilon'$. Let us do it so that $\epsilon'\le M/2$, i.e., $M-\epsilon'\ge M/2>0$. Then $1/(M-\epsilon')\le 2/M$, so
\[
\frac{\epsilon'}{M(M-\epsilon')}\le\frac{2\epsilon'}{M^2}.
\]
To make the right-hand side $\le\epsilon$, we make the second restriction: $\epsilon'\le\dfrac{\epsilon M^2}{2}$. So, if
\[
\epsilon'=\min\left(\frac{M}{2},\frac{\epsilon M^2}{2}\right)
\]
and we find $n_0$ by instantiating (**) with $\epsilon'$, then according to your calculations in post #1, we have
\[
\|1/f_n-1/f\|\le\frac{\epsilon'}{M(M-\epsilon')}\le\frac{2\epsilon'}{M^2}\le\epsilon.
\]
Choosing $\epsilon'\le M/2$ also allows concluding
\[
|f_n(x)| \geq M-\epsilon' \implies \frac{1}{|f_n(x)|} \leq \frac{1}{M-\epsilon'}
\]
because $M-\epsilon'>0$.

I understand..Thank you very much! :)

 

[evinda]

In the beginning of the proof, you are given an $\epsilon>0$. You need to find an $N$ such that \[
\|1/f_n-1/f\|=\sup_{x\in I}|1/f_n(x)-1/f(x)|<\epsilon\qquad(*)
\]
for all $n>N$. Now, one of the first steps of the proof is instantiating the definition of uniform continuity of $f_n$ with some potentially different number $\epsilon'$:
\[
f_n\to f\text{ uniformly}\implies
\forall\epsilon'\,\exists n_0\,\forall n>n_0\;\|f_n-f\|\le\epsilon'\qquad(**)
\]
You know that whatever $\epsilon'$ you use, in the end you get an upper bound of the form
\[
\|1/f_n-1/f\|\le\frac{\epsilon'}{M(M-\epsilon')}.
\]
In order to obtain (*), we must have
\[
\frac{\epsilon'}{M(M-\epsilon')}\le\epsilon
\]
It is possible to solve this inequality for $\epsilon'$, but we can simplify it and at the same time ensure that $M-\epsilon'>0$, which will solve the second problem later. Unlike $\epsilon$, which was given, we can select $\epsilon'$. Let us do it so that $\epsilon'\le M/2$, i.e., $M-\epsilon'\ge M/2>0$. Then $1/(M-\epsilon')\le 2/M$, so
\[
\frac{\epsilon'}{M(M-\epsilon')}\le\frac{2\epsilon'}{M^2}.
\]
To make the right-hand side $\le\epsilon$, we make the second restriction: $\epsilon'\le\dfrac{\epsilon M^2}{2}$. So, if
\[
\epsilon'=\min\left(\frac{M}{2},\frac{\epsilon M^2}{2}\right)
\]
and we find $n_0$ by instantiating (**) with $\epsilon'$, then according to your calculations in post #1, we have
\[
\|1/f_n-1/f\|\le\frac{\epsilon'}{M(M-\epsilon')}\le\frac{2\epsilon'}{M^2}\le\epsilon.
\]
Choosing $\epsilon'\le M/2$ also allows concluding
\[
|f_n(x)| \geq M-\epsilon' \implies \frac{1}{|f_n(x)|} \leq \frac{1}{M-\epsilon'}
\]
because $M-\epsilon'>0$.

Hello! (Smile)

I am looking again at the exercise..

At the end, when we have :

$$\frac{\epsilon'}{(M- \epsilon')M} \leq \frac{\epsilon' 2}{M^2}$$

couldn't we take $\epsilon'=\frac{M^2 \epsilon}{2}$ ? (Thinking)

 

[Evgeny.Makarov]

How is it different from what I did?

 

[evinda]

How is it different from what I did?

I just wanted to know if we could take,instead of the inequality,the equality.. (Blush)

 

[Evgeny.Makarov]

I did use equality:
\[
\epsilon'=\min\left(\frac{M}{2},\frac{\epsilon M^2}{2}\right)
\]
If the second number is smaller, then $\epsilon'=\frac{\epsilon M^2}{2}$. But the reasoning I suggested also uses the fact that $\epsilon'\le\frac{M}{2}$, which is the reason for using $\min$.

In fact, what do you mean by taking inequality? The two inequalities
\[
\epsilon'\le\frac{M}{2},\quad
\epsilon'\le\frac{\epsilon M^2}{2}\qquad(+)
\]
are constraints on $\epsilon'$ that are used in the proof. The logic of the proof says that given $\epsilon$, we choose an $\epsilon'$ and then an $n_0$ based on $\epsilon'$ in a way that guarantees $\|1/f_n-1/f\|\le\epsilon$ for all $n>n_0$. We have to choose a specific $\epsilon'$; we can't choose all $\epsilon'$ satisfying (+). So, an obvious choice is to select the smaller of the two values. We could also choose the half of the smaller and so on.

 

[evinda]

I did use equality:
\[
\epsilon'=\min\left(\frac{M}{2},\frac{\epsilon M^2}{2}\right)
\]
If the second number is smaller, then $\epsilon'=\frac{\epsilon M^2}{2}$. But the reasoning I suggested also uses the fact that $\epsilon'\le\frac{M}{2}$, which is the reason for using $\min$.

In fact, what do you mean by taking inequality? The two inequalities
\[
\epsilon'\le\frac{M}{2},\quad
\epsilon'\le\frac{\epsilon M^2}{2}\qquad(+)
\]
are constraints on $\epsilon'$ that are used in the proof. The logic of the proof says that given $\epsilon$, we choose an $\epsilon'$ and then an $n_0$ based on $\epsilon'$ in a way that guarantees $\|1/f_n-1/f\|\le\epsilon$ for all $n>n_0$. We have to choose a specific $\epsilon'$; we can't choose all $\epsilon'$ satisfying (+). So, an obvious choice is to select the smaller of the two values. We could also choose the half of the smaller and so on.

A ok.. I understood it..thank you very much! :)