Relating derivative to original function

[linearfish]

Homework Statement

Let [a, b] be an interval in R, and f:[a, b] -> R be differentiable on [a, b]. Assume f(a) = 0, and that there is a number M such that
\left|f'(x)\right| \leq M \left|f(x)\right|
for all x in [a, b]. Prove that f is identically 0 on [a, b].

Homework Equations

Mean Value Theorem?

The Attempt at a Solution

I've tried this problem multiple times and I keep hitting a wall. I've tried to approach it through contradiction, letting c be some point in [a, b] such that f(c) > 0, but it always seems that we could find an M sufficiently large such that this is true.

 

[linearfish]

I found some hints in Rudin and I think I've almost got it but I can't justify one step:

Fix x₀ in [a,b], let M₀ = sup|f(x)| and M₁ = sup|f'(x)| for a < x < x₀. (Should be less than or equal to).

Then for any such x,
\left|f(x)\right| \leq M_1 (x_0 - a) \leq A (x_0 - a) M_0

The middle step I can justify with the MVT and the next one with the given relation between f and its derivative. Now how do I show that M₀ must be 0. I believe it has something to do with how we choose x₀ (we can choose it to be very small). Thanks.

 

[morphism]

\left|f(x)\right| \leq M_1 (x_0 - a) \leq A (x_0 - a) M_0

Should that A be an M?

Anyway, suppose for a contradiction that M₀>0. We may also assume that M>0 (why?). By definition, we have

M_0 \leq M (x_0 - a) M_0.

What does this tell us?

 

[linearfish]

Yes, that should be an M, thank you.

I can see why M > 0. If M < 0 we would have

0 \leq \left| f&#039;(x) \right| \leq M \left| f(x) \right| \leq 0

and we are done.

Now if M₀ > 0, then we must have

M (x_0 - a) \geq 1

for all x in (a, x0), but this fails if we choose x to be a + 1/(2M).

M (a + \frac{1}{2M} - a) = M(\frac{1}{2M}) = \frac{1}{2} \geq 1.

Is this the idea? Thanks for the help.

 

[morphism]

Yup, that's the idea. (That last \geq should really be a &lt; though!)

 

[linearfish]

Right, I meant that as a contradiction. Thanks again for the help.