Proving f(x) ≡ 0 with given conditions |f'(x)| ≤ M|f(x)| for 0 ≤ x ≤ L

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In summary: Or is that just a lower bound? I'm not sure.Once those are checked, I'll post the full proof.In summary, using mean value theorem and the condition on f, Rudin's hint, and the fact that f is continuous on [0, 1/2M], the homework statement can be solved for any x in the interval.
  • #1
hth
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Homework Statement



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.

Homework Equations



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.
 
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  • #2
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
Quantumpencil said:
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.

Nope, he would need to show that [itex]f'(x)=0[/itex] in order to prove that [itex]f(x)=0[/itex]. At the moment, I can't think of any way to do that however.
 
  • #4
right, blah, that's what I meant.

Mean value theorem would be my guess. I might try to work though it.
 
  • #5
Use the MVT multiple times. You should be able to construct a sequence that goes to 0, and "squeezes" f(x) to 0.
 
  • #6
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
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.
 
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  • #8
updated my last message
 
  • #9
Thx for the help guys. I'll work with your ideas, see what I can come up with and post it back here.
 
  • #10
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.
 
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  • #11
Note that M has to be > 0 or else it kind of unravels itself.
 
  • #12
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
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.
 
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  • #14
Yeah, I like that one.
 
  • #15
Alright, thank you.
 
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  • #16
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?
 

1. What is a proof in science?

A proof in science is a logical and systematic way of demonstrating the validity of a statement, theory, or hypothesis. It involves gathering evidence and using reasoning to support a claim.

2. How do scientists use proofs?

Scientists use proofs to support their claims and theories. They use evidence from experiments, observations, and data analysis to provide logical reasoning and demonstrate the validity of their ideas.

3. Can a proof be wrong?

Yes, a proof can be wrong if there are flaws in the evidence or reasoning used to support it. Scientists constantly review and revise their proofs as new information and evidence becomes available.

4. What is the difference between a proof and a theory?

A proof is a demonstration of the validity of a statement or idea, while a theory is a well-supported and widely accepted explanation for a natural phenomenon. A theory may be supported by multiple proofs, but a proof is not the same as a theory.

5. How do scientists ensure the validity of their proofs?

Scientists ensure the validity of their proofs by following the scientific method, which involves making observations, forming hypotheses, conducting experiments, and analyzing data. They also undergo peer review, where other scientists review and critique their proofs before they are published.

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