Suppose that $f''$ exists on [0,1] and that $f(0)=0=f(1)$. Suppose also that $|f''(x)|\leq K$ for $x\in(0,1)$. Prove that $|f'(1/2)|\leq K/4$ and that $|f'(x)|\leq k/2$ for $x\in(0,1)$.
I'm trying to use a Taylor polynomial and the mean value theorem, but I cannot get the bounds to work out.
Use some "unconventional" reading of the linear Taylor formula, \begin{align} 0=f(0)&=f(x)+f'(x)(0-x)+\frac12f''(x_0)(0-x)^2\\ 0=f(1)&=f(x)+f'(x)(1-x)+\frac12f''(x_1)(1-x)^2 \end{align} and combine to eliminate $f(x)$ $$ 0=-f(0)+f(1)=f'(x)-\frac12f''(x_0)x^2+\frac12f''(x_1)(1-x)^2 $$ giving the estimate $$ |f'(x)|\le\frac K2·(x^2+(1-x)^2) $$ for $x\in[0,1]$. Thus $$ |f(\tfrac12)|\le \frac K4 $$