# Use mean value theorem and intermediate value theorem to bound the derivative

Suppose I am given the property that $$f(1)=1$$ and $$f(0)=0$$, is there any way put some bounds on the value of it's derivative?

My attempt:

By the mean value theorem with some $$c,z \in [0,1]$$

$$\frac{f(1) - f(c)}{1-c} = f'(z)$$

It is clear that $$\frac{f(1) - f(c)}{1-c}<1$$, hence $$f'(z)<1$$ but since there are infinitely many choices of C, would that suggest that the derivative will always be less than one in the interval of $$[0,1]$$?

Alternate arguement: By intermediate value theorem, $$f$$ must take all values between $$[0,1]$$ in that interval hence consider some point $$q$$ then again we can say $$\frac{f(q) - f(c)}{q-c} <1$$

• No. You can make sure $f$ wiggles a lot while still bounded. Commented May 7, 2021 at 11:54
• Example of that? @user10354138
– Babu
Commented May 7, 2021 at 11:59
• $f(x)=\begin{cases}x^m\sin(\frac\pi2 x^{-n}) & x\neq 0\\0 & x=0\end{cases}$ for suitable $m,n$ Commented May 7, 2021 at 12:02
• @user10354138 if you post that coutner example as an answer, I'll gladly accept it
– Babu
Commented May 7, 2021 at 19:48

$$f(x) = \begin{cases} x\sin(\frac{\pi}{2}x^{-1}) \quad & x \neq 0 \\ 0 \quad & x = 0 \end{cases}$$
will do. Then $$f(0) = 0$$, $$f(1) = 1$$, $$f$$ is continuous on $$[0, 1]$$, and $$f$$ is differentiable on $$(0, 1)$$. I mention those last two conditions because that's what the mean value theorem requires. The MVT does give us the existence of a point where $$f' = 1$$, but there's also a lot of points where $$f'$$ is very large. For $$x \in (0, 1)$$,
$$f'(x) = \sin(\frac{\pi}{2}x^{-1}) - \frac{\pi}{2}x^{-1} \sin(\frac{\pi}{2}x^{-1}).$$
With that formula in hand, there are many points near $$0$$ where $$f'$$ is as large as you want.
Alternatively, we can look at the sequence of functions $$f_{n}(x) = x^{n}$$, for $$n \in \mathbb{N}$$. Their derivatives are $$f_{n}'(x) = nx^{n - 1}$$. In particular, $$f_{n}'(1) = n$$. Individually, each derivative is bounded. However, as a sequence they show that there is no uniform bound one can place on the derivatives of all smooth functions $$f$$ with $$f(0) = 0$$ and $$f(1) = 1$$.