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].
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.