# Homework Help: Relating derivative to original function

1. Aug 7, 2008

### linearfish

1. The problem statement, all variables and given/known data
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].

2. Relevant equations
Mean Value Theorem?

3. 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.

2. Aug 7, 2008

### 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.

3. Aug 7, 2008

### morphism

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?

4. Aug 8, 2008

### linearfish

Yes, that should be an M, thank you.

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

$$0 \leq \left| f'(x) \right| \leq M \left| f(x) \right| \leq 0$$

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.

5. Aug 8, 2008

### morphism

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

6. Aug 8, 2008

### linearfish

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