# Homework Help: Question about a proof

1. Nov 7, 2009

### hth

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

Show that, if f(0)=0 and |f'(x)| ≤ M |f(x)| for 0 ≤ X ≤ L, show that on 0 ≤ X ≤ L that f(x) ≡ 0.

2. Relevant equations

3. The attempt at a solution

I'm having a hard getting this one started (as of right now, this seems a little over my head). At first, it seems like I need to find some polynomial where f(0)=0. Of course there are a lot of candidates for f(0) = 0.

Now, I know that |f(x)| = |∫0 to x of f'| ≤ M ∫0 to x of |f|, but can't seem to do anything else with it that progresses me further.

Any help is truly appreciated.

2. Nov 7, 2009

### Quantumpencil

You'd be done if you could show that f'(x) = C, for any constant C right? for then f(0) would imply that f(x)=0.

3. Nov 7, 2009

### jgens

Nope, he would need to show that $f'(x)=0$ in order to prove that $f(x)=0$. At the moment, I can't think of any way to do that however.

4. Nov 7, 2009

### Quantumpencil

right, blah, that's what I meant.

Mean value theorem would be my guess. I might try to work though it.

5. Nov 7, 2009

### JG89

Use the MVT multiple times. You should be able to construct a sequence that goes to 0, and "squeezes" f(x) to 0.

6. Nov 7, 2009

### Quantumpencil

Yeah. So here is Rudin's Hint for doing this problem:

If you fix x_o < L, then using mean value you know that

f'(c)(x - a)=f'(x)- f(a), for some c < x_o. and x in 0<x<L If S is the supremum of f(x), and S' is the supremum of f'(x), you've shown

f(x)<S'(x - a) < SM(x - a), x<x_o

From there, I think you can go here:

f(x)<SM(x_o - a). Now, this can be done for any choice of x, so why not take the supremum of f(x) on the interval (0, x_o). Then you obtain S<SM(x_o - a).

7. Nov 7, 2009

### aPhilosopher

I thought I had a proof but I tried to do it in my head and made a stupid mistake.

It is interesting to note that |f'(x)| ≤ M |f(x)| implies that |f'(x)| ≤ (M + N)|f(x)| with 0 ≤ N. If we let U = M + N for arbitrary N then we can work with a constant that's arbitrarily large instead of M. The trick would be to get it under something on the GT side.

Last edited: Nov 7, 2009
8. Nov 7, 2009

### aPhilosopher

updated my last message

9. Nov 8, 2009

### hth

Thx for the help guys. I'll work with your ideas, see what I can come up with and post it back here.

10. Nov 8, 2009

### hth

And here we go:

Proof.

Suppose M > 0. Then, take a point x_0 and fix x_0 < L.

Now, let S be the supremum of f and S' be the supremum of f' for a ≤ x ≤ x_0.

For any such x, f(x) - F(a) = = f'(c)(x-a) by Mean Value Theorem where x < c < a.

So, |f(x)| ≤ S'(x-a) ≤ S'(x_0 - a) ≤ SM (x_0 - a).

Hence, S = 0 if M(x_0 - a) < L. Therefore, f(x) = 0 on 0 ≤ X ≤ L.

End of proof.

Edit: Any comments on this will be appreciated.

Last edited: Nov 8, 2009
11. Nov 8, 2009

### hth

Note that M has to be > 0 or else it kind of unravels itself.

12. Nov 8, 2009

### Quantumpencil

So, |f(x)| ≤ S'(x-a) ≤ S'(x_0 - a) ≤ SM (x_0 - a).

Hence, S = 0 if M(x_0 - a) < L. Therefore, f(x) = 0 on 0 ≤ X ≤ L.

I don't think this is obvious enough to just state. It works if (M*x_o-a) is less than one, but this requires a specific choice of x_o, and perhaps not the one which equals b.

13. Nov 8, 2009

### hth

Alright, here's a different approach I took:

Note: ck's are coefficients in the interval [0, 1/M].

Take any point, x, in the interval [0, 1/2M].

By MVT, f(x) = f(x) - f(0) = f'(ck)(x-0) = f'(ck)x for ck < x.

So, |f(x)| ≤ M |f'(ck)| |x| by the condition on f stated in the question.

Using the condition on f(ck) again,

|f(ck)| ≤ M |f(ck1)| |ck| then,

|f(x)|≤ M2 |f(ck1)| |x| |ck|

Repeatedly using MVT, you obtain

|f(x)| ≤ Mk |f(ck)| |x| |c1| |c2| etc.. where the c's are contained within [0, 1/2M].

So, in general,

|f(x)| ≤ (1/2)k |f(ck)|

Since f is continuous on [0, 1/2M] f is bounded and as k approaches infinity f(x) goes to zero.

Therefore, f(x) = 0.

Note: You'd use the same argument for [1/2M, 1/M] which is too much for me to type out.

Last edited: Nov 9, 2009
14. Nov 8, 2009

### Quantumpencil

Yeah, I like that one.

15. Nov 8, 2009

### hth

Alright, thank you.

Last edited: Nov 9, 2009
16. Nov 9, 2009

### hth

Can I get someone else to double check that proof for me? There's a couple of things I'm not 100% confident about:

i.) That the intervals, [0,1/2M], [1/2M, 1/M] and [0, 1/M] are correct. Especially for 0 <= X <= L.

ii.) The part where I state |f(x)| ≤ (1/2)^k |f(c_k)|. Is that really OK?