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 $\frac{f(q) – f(c)}{q-c} <1$
Best Answer
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 $f(1) = 1$.