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 $\lim x_n=a$.

[itex][/itex]
[tex][/tex]