# Show something using the Mean Value Theorem

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 $$f'(c) = \alpha$$.
• Nobody? Please help me out, because I don't understand even the question :( Oct 23 '12 at 0:38
• In statement of Darboux Theorem it should be $f´(c)=\alpha$ Apr 11 '20 at 5:01

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.

• you are great, thank you so much :) Oct 23 '12 at 11:09
• can you explain why it is not possible that $f'(c)> \sup f'(y)$ ? And you forgot to write all the absolute value symbols right? Oct 23 '12 at 12:25
• @Hempo: Ah, I did forget a few absolute values. There's more or less one around everything. And $f'(c)$ can't be greater than $\sup f'(y)$ because it's the sup, and so is at least as big as any other. Oct 23 '12 at 14:36
• @mixedmath A small question. If we pick $x_0$ such that $x_0M < 1$ then the supremum is taken over $y\in[0,\ x_0]$. Certainly the function has to be zero over that interval, but I'm a bit confused as to whether this argument shows the function is $0$ over $(x_0,\ 1]$.
– EuYu
Oct 23 '12 at 15:47
• @Euyu: I've only written down the proof for $[0,x_0]$, but it's easy to extend this. In particular, the key idea is that we've shown $f(x_0) = 0$ now. So if we apply the MVT again, we can see that $|f(x + x_0) - f(x_0)| \leq x_0 M \sup |f(y)|$ too, and repeating by translation. Oct 23 '12 at 18:38

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.

Investigate the inequality in two different parts: right-hand side and left-hand side. This is for the solution of the first part. Let us first write what we know. $$x\leq x_0\in [0,1], y\in[0,x_0], f(0)\space=\space0$$ We can write mean value theorem$$\Biggl|\frac{f(x)\space-\space f(0)}{x-0}\Biggl|\space\leq\space \operatorname{sup}|f'(y)|\space \Rightarrow \space \frac{|f(x)|}{x}\space \leq \space \operatorname{sup}|f'(y)|$$ Since we know $$x\space\geq\space 0$$ we can multiply the inequality by $$x$$ on both sides $$|f(x)|\space\leq\space x \operatorname{sup}|f'(y)|\space \Rightarrow\space |f(x)|\space\leq\space x_0 \operatorname{sup}|f'(y)|$$ As seen we successfully covered the left-hand side. At this point we can start to cover right-hand side. Indeed right-hand side is relatively easy to cover. The answer is hidden in the question. Let us again write what we know. $$\operatorname{sup}|f'(x)|\space\leq\space M\operatorname{sup}|f(x)|$$ for some $$M\gt 0$$. If we change the variables and multiply by $$x_0$$ on both sides, we get $$x_0\operatorname{sup}|f'(y)|\space\leq\space Mx_0\operatorname{sup}|f(y)|$$ Now we can combine two part to obtain the whole inequality $$|f(x)|\space\leq\space x_0 \operatorname{sup}|f'(y)|\space\leq\space Mx_0\operatorname{sup}|f(y)|$$ For the second part we can start this way$$x_0\frac{|f(x)|}{x}\space\leq\space x_0\operatorname{sup}|f'(y)|\space\leq\space Mx_0 \operatorname{sup}|f(y)|\space\Rightarrow\space \frac{x_0}{x}\space|f(x)|\space\leq\space Mx_0 \operatorname{sup}|f(y)|$$ $$\frac{x_0}{y}\space|f(y)|\space\leq\space Mx_0\space\operatorname{sup}|f(y)|\space\Rightarrow\space \frac{1}{y}\space|f(y)|\space\leq\space M\operatorname{sup}|f(y)|\space\Rightarrow\space\frac{1}{y}\space\operatorname{sup}|f(y)|\space\leq\space M\operatorname{sup}|f(y)|$$ We reached the core of the question $$\frac{1}{y}\space\leq\space M$$, we can deduce that $$Mx_0\space\geq\space 1$$ Let us take the case $$M\space\lt\space 1$$ $$0\space\leq\space Mx_0\space\leq\space M\space\lt\space 1\space\Rightarrow\space Mx_0\space\lt\space 1$$ Since this is a contradiction, $$f(x)$$ should be zero-function on $$[0,1]$$.