If you know that $\|f''\|_\infty
I am assuming that $f$ is differentiable on $[0,1]$ (twice differentiable on $[0,1]$ in the sequel). You can show that if $\|f'\|_\inftyf(0)+L$ for some $x\in(0,1]$ and show that this contradicts the Mean Value Theorem).
We can probably adapt this to your problem where $f'$ plays the role of $f$ and $f''$ the role of $f'$. Using it straight away we have
$$\|f'\|_\infty\leq f'(0)+K.$$
Now we don't know $f'(0)$ but I suppose but we could just use any $f'(x)$ for $x\in(0,1)$ if we got one. We can do this using the Mean Value Theorem if we know that $f(a)=f_u(x)$ and $f(b)=f_l(x)$ (or we could probably do with estimates --- let's run with estimates).
Suppose that we have $m\leq f(x)\leq M$ (just take $m=f_l(x)$ and $M=\max\{|f_u(x)|,|f_l(x)|\}$.)
Now, can we say anything about
$$\frac{M-m}{x_M-x_m},$$
where $x_M$, $x_m$ are points that are 'close' to the maxima: $f(x_m)\sim f_m(x)$ and $f(x_M)\sim f_M(x)$. Well we have a problem if the max and min are close together...
I'm stuck at this point but I hope this hint helps --- any more information on the problem (how far apart are the maxima/ minima found, is there a local max/ min inside $[0,1]$ so that we could take $f'(a)=0$ there and work from this? Two pieces of information about $f$ or two about $f'$ will do.) and we might be able to take these problems to fruition.