If $ f\left(0\right)=f'\left(0\right)=f'\left(1\right)=0 $ and $f\left(1\right)=1$, then $ \exists x\in[0,1]:|f''\left(x\right)|\geq4 $ Let $f$ be twice differentiable function, and assume $$ \begin{cases}
f\left(0\right)=0\\
f\left(1\right)=1\\
f'\left(0\right)=0\\
f'\left(1\right)=0
\end{cases} $$
I want to prove that there exists $x_0 \in [0,1]$ such that $ |f''\left(x_{0}\right)|\geq4 $.
Now, I know that this question already been asked before, but I am intrested in a solution that does not use integrals nor taylor expansions (I am familier with those solutions). I want to find a solution based solely on Lagrange mean value theorem.
Here's my work:
Using Lagrange's theorem in $(0,1/2)$ and then in $(1/2,1)$ yields existence of points $c_1 \in (0,1/2)$ and $c_2 \in (1/2,1)$ such that $ f'\left(c_{1}\right)=2f\left(\frac{1}{2}\right) $ and $ f'\left(c_{2}\right)=2\left(1-f\left(\frac{1}{2}\right)\right) $.
Now, using Lagrange's theorem on $f'$ in $(0,c_1)$ and $(c_2,1)$ yields existence of $\theta_1 \in (0,c_1)$ and $\theta_2 \in (c_2,1) $ such that $$ \begin{cases}
f''\left(\theta_{1}\right)=\frac{f'\left(c_{1}\right)}{c_{1}}=\frac{2f\left(\frac{1}{2}\right)}{c_{1}}\\
f''\left(\theta_{2}\right)=\frac{-f'\left(c_{2}\right)}{1-c_{2}}=\frac{2\left(1-f\left(\frac{1}{2}\right)\right)}{1-c_{2}}
\end{cases} $$
I thought that if I'll assume that $ f''\left(\theta_{1}\right)<4 $ and $ f''\left(\theta_{2}\right)>-4$ will lead me to a contradiction (and thus one of them must be false and that's the point we want), but I actually could not reach a contradiction.
Any help would be appreciated.
Thanks in advance.
 A: Suppose there doesn't exist $x_{0}$ such that such that $$|f^{''}(x_{0})|\geq 4$$, then for all $x\in[0,1]$ we have $$|f^{''}(x)|<4$$ but then $$-4<f^{''}(x)<4$$ so integrating give $$-4x<f'(x)<4x$$ and $$-2x^2<f(x)<2x^2$$, Define $$\boxed{g(x)=f(x)-(dx^{k}-rx^{j})}$$ where $k,j$ are natural numbers now $g(1)=g(0)=0$ such that $d-r=1$ Hence there exist $c\in(0,1)$ such that $$\boxed{f'(c)=dkc^{k-1}-jrc^{j-1}}$$ then $$\boxed{f'(c)=r(kc^{k-1}-jc^{j-1})+kc^{k-1}}$$ Now we need $$kc^{k-1}-jc^{j-1}>0$$ for some $j>k>N$ where $N$ is a natural number as the function $$g_{c}(x)=xc^{x-1}$$ for $c\in(0,1)$ is decreasing on the positive real axis after certain time moving towards the right. Thus by, Archimedean property there exist such $r$(Possibly large) with $r(kc^{k-1}-jc^{j-1})>4c$ thus $f'(c)>4c$ a contradiction!!!
A: My previous answer seems correct, but got downvoted. Probably it’s because the use of Lagrange remainder theorem was not allowed. Here is an attempt using only Lagrange mean value theorem. Please let me know if there is any error if you plan to downvote it.
Lemma. Let $g$ be a twice differentiable function on $[0,a].$ Assume that $$g(0)=g’(0)=0,g(a)\geq 0.$$ Then $\exists \xi\in (0,a)$ such that $g’’(\xi)\geq 0.$
Proof. By Lagrange mean value theorem, $\exists b\in (0,a)$ such that $$\frac{g(a)-g(0)}{a-0}=g’(b),$$ hence $g’(b)\geq 0.$ Applying Lagrange mean value theorem again, $\exists \xi\in (0,b)$ such that $$\frac{g’(b)-g’(0)}{b-0}=g’’(\xi),$$ hence $g’’(\xi)\geq 0,$ as required.
Now to prove the OP’s statement, define $g(x)$ on $[0,1/2]$ as follows. $$g(x)=\left\{\begin{array}{cc}f(x)-2x^2&{\rm if~}f(1/2)\geq 1/2\\ 1-f(1-x)-2x^2&{\rm if~}f(1/2)<1/2.\end{array}\right.$$ It’s easy to check that $$g(0)=g’(0)=0,g(1/2)\geq 0.$$ It follows from the Lemma that there exists $\xi\in (0,1/2)$ such that $g’’(\xi)\geq 0,$ hence if $f(1/2)\geq 1/2,$ one has $$f’’(\xi)-4\geq 0\Rightarrow f’’(\xi)\geq 4.$$ If $f(1/2)<1/2,$ one has $$-f’’(1-\xi)-4\geq 0\Rightarrow f’’(1-\xi)\leq -4$$
$$\Rightarrow |f’’(\eta)|\geq 4,$$ where $\eta=1-\xi\in (1/2,1).$ QED
