Can anybody explain this to me? (Analysis)

1. Oct 23, 2012

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].

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?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 23, 2012

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.

3. Oct 23, 2012

Artusartos

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

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

4. Oct 23, 2012

HallsofIvy

Staff Emeritus
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.

5. Oct 23, 2012

jbunniii

(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:

6. Oct 23, 2012

Artusartos

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

7. Oct 23, 2012

HallsofIvy

Staff Emeritus

8. Oct 23, 2012

jbunniii

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.

9. Oct 23, 2012

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 $|x| \leq 1$.

10. Oct 23, 2012

Artusartos

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?

11. Oct 23, 2012

jbunniii

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 $x \in [0,1]$.

12. Oct 23, 2012

Artusartos

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...

13. Oct 23, 2012

jbunniii

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.

14. Oct 24, 2012

Artusartos

Thanks a lot :)