Show something using the Mean Value Theorem

Question

I've got an exercise about differentiability, mean value theorem and suprema.

To be honest I don't understand the structure of this question. Maybe you guys are so kind to help me out :)

Let $f: [0,1] \rightarrow \mathbb{R}$ be differentiable with $f(0) = 0$, and satisfying

$$|f'(x)|\le M|f(x)|, x\in[0,1] $$ for some $M>0$

a.) Use the Mean Value Theorem to show that for all $x \le x_0 \in[0,1], y\in[0,x_0]:$ $$ |f(x)|\le x_0 \text{ sup} |f'(y)|\le M x_0 \text{ sup} |f(y)|$$ b.) Use the previous part to show that f is the zero-function on [0,1].

(Hint: What happens if we choose $x_0$ such that M$x_0<1$?)

Known definitions, theorems:

• Mean Value Theorem (There is a c for which f'(c) equals f(b)-f(a)/(b-a) if [a,b] is the domain, and f is continuous on [a,b], differentiable on (a,b)
• Differentiable function means that that for all points c $\in$ A, the limit of f(x)-f(c)/(x-c) exists.
• Interior Extremum Theorem (Intermediate Value theorem). States that if f attains a maximum or minimum value on a open interval, then at some point c $\in$(a,b), f'(c)=0
• Darboux Theorem: If f is differentiable on an interval [a,b] and if $\alpha$ satisfies $f'(a) < \alpha < f'(b)$, then there exists a point c$\in$(a,b) where $f(c)=\alpha$

Answer 1

Let f:[0,1]→R be differentiable with f(0)=0, and satisfying

$$|f'(x)|\le M|f(x)|, x\in[0,1] $$ for some $M>0$

a.) Use the Mean Value Theorem to show that for all $x \le x_0 \in[0,1], y\in[0,x_0]:$ $$ |f(x)|\le x_0 \sup |f'(y)|\le M x_0 \sup |f(y)|$$ b.) Use the previous part to show that f is the zero-function on [0,1].

Let's begin with part a. So fix an $x_0$ in $(0,1)$. Suppose that there is an $x$ in $[0,x_0]$ such that $|f(x)| > x_0 \sup|f'(y)|$. Then in particular, $\dfrac{|f(x) - f(0)|}{x-0} = \dfrac{|f(x)|}{x} > \dfrac{x_0}{x}\sup f'(y) \geq \sup f'(y)$ as $x_0 \geq x$.

But then by the mean value theorem, there must exist a $c$ such that $f'(c) = \dfrac{f(x) - f(0)}{x - 0}$. This is a contradiction, as then $f'(c) > \sup f'(y)$.

The second inequality is much easier, relying just on using $|f(x)|\le x_0 \sup |f'(y)|\le M x_0 \sup |f(y)|$ and interpreting sup.

And then you can follow the hint for part b, and the answer falls out as $|f(x)| < \sup |f(y)|$ is nonsense.

Answer 2

There is another approach which starts with a number $x_0$ such that $0<Mx_0<1$. Note that by the given inequality we have $f(0)=0$. Clearly if $x\in(x, x_0]$ then we have by mean value theorem $$|f(x)| =|f(x) - f(0)|=x|f'(c_1)|\leq Mx|f(c_1)|$$ and applying the same argument repeatedly (with $c_i$ in place of $x$) and noting that at each step $0<c_n<x$ we get $$|f(x) |\leq (Mx) ^{n} |f(c_n) |\leq (Mx) ^{n} \sup|f(y) |$$ Since $0<Mx<1$ taking limit as $n\to\infty$ we get $f(x) =0$ for all $x\in[0,x_0]$. To go beyond $x_0$ apply the same argument on function $g$ given by $g(x) =f(x-x_0)$ and this will prove that $f(x) =0$ for all $x\in[0,2x_0]$. Continue in this manner till we reach to the interval $[0,kx_0]$ where $kx_0\geq 1$ and the proof is complete.