# Homework Help: Uniform convergence

1. Nov 10, 2008

### dirk_mec1

1. The problem statement, all variables and given/known data
http://img384.imageshack.us/img384/1643/60357312ro9.png [Broken]

2. Relevant equations

http://img440.imageshack.us/img440/1935/11582858vp9.png [Broken]

3. The attempt at a solution
I didn't know if I had to use the definition of uniform convergence or its Cauchy variant but I choose the Cauchy variant. Moreover not all relevant equations need to be used I've put equations there which might be useful.

It not difficult to verify that the pointwise limit heads to zero. Now there holds with n > m > N:

$$|f_n - f_m | = |g(t)| |(1-t)^n - (1-t)^m| \leq ||g||_{\infty} |(1-t^n|) \leq M \cdot (1-t)^n \leq M \cdot (1-t)^N \leq M \cdot N \leq (M+1) \cdot N$$

with $$M = | |g||_{\infty}$$

Choose $$N = \frac{ \epsilon }{ M+1}$$

So it is uniform convergent.

Is this correct?

Last edited by a moderator: May 3, 2017
2. Nov 10, 2008

### Office_Shredder

Staff Emeritus
What is N? It just appeared as a power, and then a logarithm swept through while no one was looking and pulled it down to multiply with M... If N can be whatever the hell it wants to be, then for small N (1-t)N is certainly larger than N. It looks like you wanted N to be s.t. n,m>N implies |fn - fm| but in that case you most certainly don't get to pick what N is

3. Nov 10, 2008

### Dick

There's quite a lot wrong with that 'proof'. But the most important fact is that the theorem is untrue if g(0) is not equal to zero. And you never even used that.

4. Nov 11, 2008

### dirk_mec1

N is an natural number.

Nooo, because $$(1-t)\leq 1$$ for all t in [0,1]. Then certainly $$(1-t)^N \leq 1 \leq N$$ since N is a natural number.

Why is that?

Ok I'll try to explain my steps below.

Well g(0)=0 in this exercise so why even consider that?

I presume these steps are clear (take n<m)

$$|f_n - f_m | = |g(t)| |(1-t)^n - (1-t)^m| \leq ||g||_{\infty} |(1-t)^n|$$

I call M the sup of G: $$M=||g||_{\infty}$$

$$||g||_{\infty} |(1-t)^n| = M \cdot (1-t)^n \leq M \cdot (1-t)^N$$

n is natural number larger than N then this must be true because I only look at the (1-t)^n part this is smaller than (1-t)^N. ( take for example: (1/2)^6 < (1/2)^3 )

I explained this step above

$$\leq M \cdot N$$

I have to take into consideration that g=0 => M=0 so I add one up then this is still true, right?

$$\leq (M+1) \cdot N$$

5. Nov 11, 2008

### Dick

!? That's an amazing attitude to take towards a theorem premise! :)

6. Nov 11, 2008

### Office_Shredder

Staff Emeritus
So N is just a random natural number. But then your last step is

M is fixed. I can give you epsilon small enough that N<1.

Furthermore, M=/=0. For example, if g(x) = x, M=1. Just because g(0)=0 doesn't mean g is identically 0

7. Nov 11, 2008

### Dick

You might also notice that at t=0, (1-t)^n=1. That's going to make it difficult for a bound like M*|(1-t)^n| to make anything 'small'.

8. Nov 11, 2008

### dirk_mec1

I still don't understand what your point is.

I just want to use the definition a Cauchy sequence with a n,m>N ( but N should then be real right?)

Where do you get the idea that I presume g=0? I take into account that it might be zero.

Yes you're right... but can you please tell me where my reasoning in my proof is wrong?

9. Nov 11, 2008

### Office_Shredder

Staff Emeritus
We just pointed out like 5.

Just intuitively, this is for large N, and hence won't be satisfied for small N. If you pick N to satisfy the Cauchy condition, you don't get to later redefine N to be smaller than epsilon, because N will no longer satisfy the Cauchy condition.

You have to use the fact that g(0) = 0, so g(t)*(1-t)n = 0 when t is zero, which gets rid of the problem of 1-t = 1 at t=0. This was Dick's point.... g(0)=0 wasn't just thrown in there for laughs, whoever wrote the question was smart enough to realize it was a necessary condition for the conclusion to be true, and hence if your proof doesn't utilize the fact that g(0)=0, you should be thinking in the back of your mind that your proof isn't a very good one

10. Nov 11, 2008

### dirk_mec1

But I said that the pointwise limit is tending to zero. Which utilizes the fact that g(0) =0

because if $$n \rightarrow \infty$$ then $$g(t)(1-t)^n \rightarrow 0$$ for t in (0,1] but for t=0 it is also zero hence the pointwise limit is zero, do you get what I mean?

As for the cauchy sequence I just determined the N for the difference becomes smaller than epsilon I'm not redefing anything. But I presume you know this better than me so at what point is the proof wrong?

11. Nov 11, 2008

### Dick

Yes, you've got that the pointwise limit is zero. And that uses g(0)=0. Now you just have to show that it goes uniformly to zero. I.e. there is an N such that |f_n(x)|<epsilon for all n>N. I'll give a hint. Use that g(x) is continuous at zero to make sure |f_n(x)|<epsilon for x in [0,a] for some small value of a>0. Now find an N such that |f_n(x)|<epsilon on [a,1] for n>N. (Use that |g(x)| is bounded, like you did).

Last edited: Nov 11, 2008
12. Nov 11, 2008

### Office_Shredder

Staff Emeritus
So mention that in the proof! :p

With the Cauchy sequence thing, (you didn't state this explicitly but this is how I've been interpreting it), given epsilon, there exists N such that $$||f_n-f_m||<\epsilon$$ if n,m>N. Now N is fixed. You can't swap up what N is later on in the proof. For small epsilon, N is probably something in the hundreds or thousands, or even higher. Yet at the end of your post you put

Which doesn't make sense! You can't change what N is in the middle

13. Nov 11, 2008

### dirk_mec1

OK, thanks for the hint but I don't recognize the definition of uniform convergence, in my lecture notes my instructor wrote this:

http://img82.imageshack.us/img82/2342/52594065wx1.png [Broken]

So I presume you mean with |...| the infinity-norm, right?

Last edited by a moderator: May 3, 2017
14. Nov 11, 2008

### Dick

The pointwise limit is f(x)=0. That looks exactly like what I wrote. Showing the infinity norm of f_n(x) goes to zero is exactly the same thing as showing for all epsilon there exists an N such that for all n>N, |f_n(x)|<epsilon. Think about it.