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?