Two real analysis problems: proving constancy and a uniform convergence problem

[Mr.Miyagi]

The problem statement

Let f:[a,b]→\mathbb{R} be differentiable and assume that f(a)=0 and \left|f'(x)\right|\leq A\left|f(x)\right|, x\in [a,b].
Show that f(x)=0,x\in [a,b].

The attempt at a solution

It was hinted at that the solution was partly as follows. Let a \leq x_0 \leq b. For all x\in [a,x_0] \left|f(x)\right|\leq (x_0-a)\underset{[a,x_0]}{\sup}\left|f'(x)\right|\leq A(x_0-a)\underset{[a,x_0]}{\sup}\left|f(x)\right|. Then we should ask what happens for A(x_0-a)<1.

The first inequality follows from the mean value theorem and the second from the given equation. When A(x_0-a)<1, then \left|f(x)\right|< \underset{[a,x_0]}{\sup}\left|f(x)\right|. I do not see how this would imply the proposed relation.

EDIT: I have already solved the next problem.

The problem statement

Is the series \sum^{\infty}_{n=1} \frac{x^n(1-x)}{n} uniformly convergent on [0,1]?

The attempt at a solution

I suspect it is. I want to use the Weierstrass M-test, but obviously the sequence M_n=1/n doesn't work as a bound since \sum^{\infty}_{n=1} \frac{1}{n} does not converge. I can't figure how to produce a tighter bound than M_n=1/n.

Homework Statement

 

[Mr.Miyagi]

I just wanted to say that I have solved the second problem. By finding the maximum of x^n(1-x) by differentiating it and setting it to zero, we can get the inequality x^n(1-x) < 1/n. I feel a little silly for not considering that before.

I am still breaking my brain over the first problem, though. If anyone has any suggestion on how to solve it, I'd love to hear from you.

 

[Bacle2]

What's A, Sensei? Is it a positive constant?

 

[Bacle2]

Maybe this will help: Since f(0)=f(1)=0, check to see what happens with values like
1/4, 1/2 and 3/4; if these values do not go towards 0, the series cannot converge uniformly. Similarly, try to see where the maximum is for the genersl product x(1-x)ⁿ to get an idea of how convergence happens.

 

[Mr.Miyagi]

A is indeed a positive constant. Sorry for the ambiguity.

Also, I have solved the second problem, so I don't need help with that anymore.

 

[JG89]

First prove it for the special case where [a,b] = [0,1]. Then let g: [0,1] \rightarrow \mathbb{R} be defined by g(x) = f(a + (b-a)x). Using your special case proof, conclude that g = 0, from which your theorem follows.

 

[Mr.Miyagi]

Alright, that is a nice strategy. But I still have to prove the special case, which I still can't seem to do.

I have made the following attempt at proving the general case in one go.

Choose c such that a<c<b and A/(c-a)>1. Let us assume for contradiction that there exists an x_0\in [a,c] satisfying f(x_0)\neq 0. By the mean value theorem there then exists an x_1\in (a,x_0) satisfying |f'(x_1)|=|f(x_0)|/(x_0-a). But |f(x_1)|\geq A|f'(x_1)|=A|f(x_0)|/(x_0-a)> |f(x_0)|>0 We can repeat this procedure to construct a sequence (x_n). In general there exists an x_n\in (a,x_{n-1}) satisfying |f(x_n)|>0.

Now if I can prove that the sequence (x_n) converges to a, then I am done. This is because then \lim x_n=a but \lim f(x_n)\neq f(a), which implies that f is not continuous, which contradicts the hypothesis of f being differentiable. This would allow me to say that for all x\in [a,c] , f(x)=0. But then I can apply the same argument to the interval [c, a+2(c-a)] (the interval of length (c-a) which lies next to [a,c]) and so on.

The problem is, I can't prove that \lim x_n=a.