# 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.

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 $$-4 so integrating give $$-4x and $$-2x^2, Define $$\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$$ a contradiction!!!
Lemma. Let $$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