I have a function $f:[-1,1]\rightarrow\mathbb{R}$ which can derives twice and I have $f(0)=0$ and $f(-1)+f(1)>0$. I want to prove that exists $x_0\in(-1,1)$ for which is true that: $$f''(x_0)>\frac{1}{2}(f(-1)+f(1))$$
So, using the Mean Value Theorem (in $f$) I get that:
- exists $x_1\in(-1,0)$ for which is true that $f'(x_1)=-f(-1)$
- exists $x_2\in(0,1)$ for which is true that $f'(x_2)=f(1)$
Now, using again the Mean Value Theorem (in $f'$) I get that: exists $x\in(x_1,x_2)$ for which is true that: $$f''(x)=\frac{f(1)+f(-1)}{x_2-x_1}>0$$ But how can I prove the thing that I want?
Put $g(x)=\frac{f(x)+f(-x)}{2}$, with $x\in [0,1]$. We have $g(0)=0$, $g(1)=\frac{f(1)+f(-1)}{2}$, hence there exists $c\in (0,1)$ such that $g(1)=g(1)-g(0)=g^{\prime}(c)$. Now $2g^{\prime}(c)=f^{\prime}(c)-f^{\prime}(-c)=2cf^{\prime\prime}(d)$, with $-c<d<c$, hence $\frac{f(1)+f(-1)}{2}=cf^{\prime\prime}(d)$. Now as $\frac{f(1)+f(-1)}{2}>0$ and $c>0$, we have $f^{\prime\prime}(d)>0$, and as $c<1$, we get $\frac{f(1)+f(-1)}{2}<f^{\prime\prime}(d)$.