Prove that f(x) = 0 for all x in [a, b]

  • Thread starter Jamin2112
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  • #1
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Homework Statement



Suppose f is differentiable on [a, b], f(a) = 0, and there is a real number A such that |f '(x)
≤ A |f(x)| on [a, b]. Prove f(x) = 0 for all x in [a, b].

Homework Equations



Might need the Mean Value Theorem or a variation thereof.

The Attempt at a Solution



Suppose instead that there is an x0 in [a, b] such that |f '(x0)| > 0.
Then we have

0 < |f(x0)| = |f(x0) - f(a)| ≤ (x0 - a) |f '(x)| ≤ A(x0 - a)|f(x)|

for some x in (a, x0).

Am I close? Give me the vaguest of hints, as I am really not supposed to solicit help on this assignment.

 
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Answers and Replies

  • #2
Dick
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Homework Helper
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Homework Statement



Suppose f is differentiable on [a, b], f(a) = 0, and there is a real number A such that |f '(x)
≤ A |f(x)| on [a, b]. Prove f(x) = 0 for all x in [a, b].

Homework Equations



Might need the Mean Value Theorem or a variation thereof.

The Attempt at a Solution



Suppose instead that there is an x0 in [a, b] such that |f '(x0)| > 0.
Then we have

0 < |f(x0)| = |f(x0) - f(a)| ≤ (x0 - a) |f '(x)| ≤ A(x0 - a)|f(x)|

for some x in (a, x0).

Am I close? Give me the vaguest of hints, as I am really not supposed to solicit help on this assignment.


No, I don't think you are very close. I really can't figure out what that's supposed to mean. Thanks for confessing that you aren't supposed to solicit help. That's very honest. Keep it up. Here's a bit of general help. Try and prove it for a special case first. Take [a,b]=[0,1], f(0)=0 and f(1)=1 and A=1. If you can't prove it for a special case then you won't be able to prove it in general case. And the special case will make it easier to write down your thoughts. Think about subdividing the interval [0,1]. That's pretty vague and that's about as far as I want to go.
 
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  • #3
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as I am really not supposed to solicit help on this assignment.

This is considered cheating and that is not allowed here.
 

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