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 $.
  • $\begingroup$ Nobody? Please help me out, because I don't understand even the question :( $\endgroup$ Oct 23, 2012 at 0:38
  • $\begingroup$ In statement of Darboux Theorem it should be $f´(c)=\alpha$ $\endgroup$ Apr 11, 2020 at 5:01

3 Answers 3


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.

  • $\begingroup$ you are great, thank you so much :) $\endgroup$ Oct 23, 2012 at 11:09
  • $\begingroup$ 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? $\endgroup$ Oct 23, 2012 at 12:25
  • $\begingroup$ @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. $\endgroup$
    – davidlowryduda
    Oct 23, 2012 at 14:36
  • $\begingroup$ @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 $(x_0,\ 1]$. $\endgroup$
    – EuYu
    Oct 23, 2012 at 15:47
  • $\begingroup$ @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. $\endgroup$
    – davidlowryduda
    Oct 23, 2012 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 $[0,1]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.