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 :( $\endgroup$ Commented Oct 23, 2012 at 0:38
  • $\begingroup$ In statement of Darboux Theorem it should be $f´(c)=\alpha$ $\endgroup$
    – user561334
    Commented 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 :) $\endgroup$ Commented 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? $\endgroup$ Commented 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
    Commented 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
    Commented 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
    Commented 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 $[0,1]$.


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