- #1
JG89
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Homework Statement
Suppose f is a differentiable function. Prove that if f(0) = 0 and |f'(x)| <= |f(x)| then f(x) = 0 for x in (0,1).
Homework Equations
The Attempt at a Solution
The actual question asks me to prove f(x) = 0 for all real x, but if I can prove f(x) = 0 for x in (0,1) then the rest follows easily.
So:
Assume there exists a c in (0,1) such that f(c) > 0. Then by the MVT there exists a c* in (0,c) such that [tex] \frac{f(c) - f(0)}{c-0} = \frac{f(c)}{c} = f'(c*) [/tex]. We know that [tex] f'(c*) \le f(c*) [/tex]. So [tex] \frac{f(c)}{c} \le f(c*) \Leftrightarrow f(c) \le cf(c*) < f(c*) [/tex] since c is in (0,1).
It follows that f(c) < f(c*) whenever c* < c. But this means that the function is decreasing. However, f(0) = 0 and from our assumption that f(c) > 0, this must mean that f must be increasing towards f(c), which is a contradiction. The proof for assuming f(c) < 0 is similar.
I think the proof is a bit shaky in the last paragraph. How is it?
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