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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 $\frac{f(q) - f(c)}{q-c} <1$

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

1 Answer 1

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As user10354138 mentioned,

$$ f(x) = \begin{cases} x\sin(\frac{\pi}{2}x^{-1}) \quad & x \neq 0 \\ 0 \quad & x = 0 \end{cases} $$

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 $f(1) = 1$.

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