Let f be 2 times differentiable function in open interval (0,1) with following conditions:
1. lim(x-->0)f(x) = lim(x-->1)f(x) = 0
2.Exist M>0 so that |f ''(x)|<=M for all x in (0,1)
Prove that |f '(x)|<=M/2
Thank you all!
It looks like you have that $\displaystyle f(0)=f(1)=0$ thus by Rolle's theorem there exists some $\displaystyle c\in (0,1)$ such that $\displaystyle f'(c)=0$. Form the first degree taylor polynomial around that point $\displaystyle f'(x)=f'(c)+\frac{f''(\xi)}{2!}(x-c)^2=\frac{f''(\xi)}{2}(x-c)^2$. Now, combining $\displaystyle \left|f''(\xi)\right|\leqslant M$ and $\displaystyle (x-c)^2\leqslant 1$ we see that $\displaystyle \left|f'(x)\right|=\left|\frac{f''(\xi)}{2}(x-c)^2\right|\leqslant\frac{M}{2}$
Use the Taylor series with remainder $\displaystyle f(x+t) = f(x) + tf'(x) + \tfrac12t^2f''(\xi)$, where $\displaystyle \xi$ lies between x and x+t. This implies that $\displaystyle |tf'(x) +f(x) - f(x+t)| = |\tfrac12t^2f''(\xi)|\leqslant\tfrac12Mt^2$. There are now two separate cases to consider.
If f(x) and f'(x) have the same sign, take $\displaystyle t = 1-x$. Then $\displaystyle |tf'(x) +f(x)| = |(1-x)f'(x) +f(x)| \geqslant |(1-x)f'(x)|$. Also, $\displaystyle f(x+t) = f(1) = 0$. So the inequality in the previous paragraph tells us that $\displaystyle |(1-x)f'(x)| \leqslant \tfrac12M(1-x)^2$. Hence $\displaystyle |f'(x)|\leqslant \tfrac12M(1-x)\leqslant \tfrac12M$.
If f(x) and f'(x) have the opposite signs, take $\displaystyle t = -x$. Then $\displaystyle |tf'(x) +f(x)| = |-xf'(x) +f(x)| \geqslant |xf'(x)|$. Also, $\displaystyle f(x+t) = f(0) = 0$. So the inequality tells us that $\displaystyle |xf'(x)| \leqslant \tfrac12Mx^2$. Hence $\displaystyle |f'(x)|\leqslant \tfrac12Mx\leqslant \tfrac12M$.
The second term in that Taylor polynomial is $\displaystyle f''(\xi)(x-c)$, not $\displaystyle \frac{f''(\xi)}{2!}(x-c)^2$.