Can anybody explain this to me? (Analysis)

[Artusartos]

For x in [0, infinity), let $f_n(x)= \frac{x}{n}$...

Determine whether $f_n$ converges uniformly to f (the limit, which is equal to 0) on [0,1].

Answer:

Let $\epsilon > 0$ be given. Let $N= \frac{1}{\epsilon}$. Then for n>N, $| f_n(x) - 0 | = | \frac{x}{n} | \leq \frac{1/ \epsilon}{n} = \frac{1}{n \epsilon} < \epsilon$ as desired.

My questions:

1) Why is $| \frac{x}{n} | \leq \frac{1/ \epsilon}{n}$? How do we know that x is less than 1 over epsilon?

2) Why is $\frac{1}{n \epsilon} < \epsilon$?

3) Finally, how did they know that N was supposed to be 1/epsilon?

Thanks in advance

 

[hedipaldi]

This "proof" is false.This sequence does not converge uniformly in that interval,because the suprimum for x is not final so certainly does not converge to 0.

 

[Artusartos]

This "proof" is false.This sequence does not converge uniformly in that interval,because the suprimum for x is not final so certainly does not converge to 0.

But this answer is from here (page 49, exercise 24.2 b)

http://www.scribd.com/doc/70434268/Ross-Solutions

 

[HallsofIvy]

This "proof" is false.This sequence does not converge uniformly in that interval,because the suprimum for x is not final so certainly does not converge to 0.

I don't know what you mean by "not final". Since the sequence converges, it MUST converge uniformly on any closed and bounded (i.e. compact) interval.

 

[jbunniii]

But this answer is from here (page 49, exercise 24.2 b)

http://www.scribd.com/doc/70434268/Ross-Solutions

(I assume you mean page 47.) The proof for part (b) says "Claim: $f_n \rightarrow f$ uniformly on [0,1]." Part (c) shows by contradiction that it is not true on $[0,\infty)$.

I am not sure which part you are trying to solve:

For x in [0, infinity), let $f_n(x)= \frac{x}{n}$...

Determine whether $f_n$ converges uniformly to f (the limit, which is equal to 0) on [0,1].

 

[Artusartos]

(I assume you mean page 47.) The proof for part (b) says "Claim: $f_n \rightarrow f$ uniformly on [0,1]." Part (c) shows by contradiction that it is not true on $[0,\infty)$.

I am not sure which part you are trying to solve:

I was actually talking about part (b)...about why it converges uniformly on [0,1].

 

[HallsofIvy]

Oops! I misread the problem.

 

[jbunniii]

I read through the proof at your link, and it's a mess. My comments below:

For x in [0, infinity), let $f_n(x)= \frac{x}{n}$...

Determine whether $f_n$ converges uniformly to f (the limit, which is equal to 0) on [0,1].

Answer:

Let $\epsilon > 0$ be given. Let $N= \frac{1}{\epsilon}$. Then for n>N, $| f_n(x) - 0 | = | \frac{x}{n} | \leq \frac{1/ \epsilon}{n} = \frac{1}{n \epsilon} < \epsilon$ as desired.

My questions:

1) Why is $| \frac{x}{n} | \leq \frac{1/ \epsilon}{n}$? How do we know that x is less than 1 over epsilon?

In general, it's not true. All you know is that $0 \leq x \leq 1$. If $0 < \epsilon < 1$ (assumption not stated in the proof), then $x \leq 1 < 1/\epsilon$.

2) Why is $\frac{1}{n \epsilon} < \epsilon$?

This is certainly not true. Take $\epsilon = 1/10$, $N = 10$, and $n = 11$. Then
$$\frac{1}{n \epsilon} = \frac{1}{11 /10} = \frac{10}{11}$$
which is certainly not less than 1/10.

I suggest ignoring that "solution" altogether. The statement is true, but the proof is bogus.

 

[jbunniii]

I suggest starting with
$$|f(x) - 0| = \left|\frac{x}{n}\right| \leq \left|\frac{1}{n}\right| \ldots$$
where the inequality is true because $|x| \leq 1$.

 

[Artusartos]

I read through the proof at your link, and it's a mess. My comments below:


In general, it's not true. All you know is that $0 \leq x \leq 1$. If $0 < \epsilon < 1$ (assumption not stated in the proof), then $x \leq 1 < 1/\epsilon$.


This is certainly not true. Take $\epsilon = 1/10$, $N = 10$, and $n = 11$. Then
$$\frac{1}{n \epsilon} = \frac{1}{11 /10} = \frac{10}{11}$$
which is certainly not less than 1/10.

I suggest ignoring that "solution" altogether. The statement is true, but the proof is bogus.

So can I prove it like this?

$| f_n(x) - 0| = | \frac{x}{n} | \leq \frac{1}{n}$

So when is 1/n equal to epsilon? It is when n=1/epsilon...so n needs to be greater than 1/epsilon. Do you think my answer is correct?

 

[jbunniii]

So can I prove it like this?

$| f_n(x) - 0| = | \frac{x}{n} | \leq \frac{1}{n}$

So when is 1/n equal to epsilon? It is when n=1/epsilon...so n needs to be greater than 1/epsilon. Do you think my answer is correct?

Yes, if $n > 1/\epsilon$, then $1/n < \epsilon$, which is what you need. The key is that this choice of $n$ works for any $x \in [0,1]$.

 

[Artusartos]

Yes, if $n > 1/\epsilon$, then $1/n < \epsilon$, which is what you need. The key is that this choice of $n$ works for any $x \in [0,1]$.

Thanks a lot. I also have another question, if you don't mind...

From the same link that I gave...for exercise 24.6 (b), we are asked if f_n converges uniformly on [0,1]. And the solution (in the link) says that $| \frac{1-2xn}{n^2} | \leq \frac{1}{\sqrt{n}}$. I wasn't able to understand why that is true...

 

[jbunniii]

Thanks a lot. I also have another question, if you don't mind...

From the same link that I gave...for exercise 24.6 (b), we are asked if f_n converges uniformly on [0,1]. And the solution (in the link) says that $| \frac{1-2xn}{n^2} | \leq \frac{1}{\sqrt{n}}$. I wasn't able to understand why that is true...

That's because it is false. Consider $x = 1$, $n = 2$. Then
$$\left|\frac{1 - 2xn}{n^2}\right| = \left|\frac{1-4}{4}\right| = \frac{3}{4}$$
which is not less than $1/\sqrt{2} \approx 0.7071$.

Whoever wrote the solution manual you are using is just plain wrong.

You can obtain a bound that will work as follows:
$$|1 - 2xn| \leq |1| + |2xn| = 1 + 2n|x| \leq 1 + 2n \leq n + 2n = 3n$$
Try using this to solve the problem. You'll have to use a different N from what the solution chose, but it should work.

 

[Artusartos]

That's because it is false. Consider $x = 1$, $n = 2$. Then
$$\left|\frac{1 - 2xn}{n^2}\right| = \left|\frac{1-4}{4}\right| = \frac{3}{4}$$
which is not less than $1/\sqrt{2} \approx 0.7071$.

Whoever wrote the solution manual you are using is just plain wrong.

You can obtain a bound that will work as follows:
$$|1 - 2xn| \leq |1| + |2xn| = 1 + 2n|x| \leq 1 + 2n \leq n + 2n = 3n$$
Try using this to solve the problem. You'll have to use a different N from what the solution chose, but it should work.

Thanks a lot :)