Let $f$ be twice differentiable function on interval $I$. Let $M_0=\sup_{x\in I}|f(x)|, M_1=\sup_{x\in I}|f'(x)|, M_2=\sup_{x\in I}|f''(x)|$. Show that
a) if $I=[-a, a]$, then $$|f'(x)|\leqslant \frac{M_0}{a}+\frac{x^2+a^2}{2a}M_2$$ b) 1. if length of $I$ is not less than $2\sqrt{M_0/M_2}$, then $M_1\leqslant 2\sqrt{M_0M_2}$
- if $I=\mathbb{R}$, then $M_1\leqslant \sqrt{2M_0M_2}$
c) Constants $2$ and $\sqrt{2}$ from b) cannot be improved
d) if $f$ is $p$ times differentiable on $\mathbb{R}$ and $M_0$ and $M_p=\sup_{x\in\mathbb{R}}|f^{(p)}(x)|$ are finite, then for $1\leqslant k\leqslant p$ $M_k=\sup_{x\in\mathbb{R}}|f^{(k)}(x)|$ are finite and
$$M_k\leqslant2^{k(p-k)/2}M_0^{1-k/p}M_p^{k/p}$$
Solutions for b) and c) were given here and here.
d) was solved here.
Now I am trying to solve a):
Using Taylor's expansion we have
$$f(a)=f(x)+f'(x)(a-x)+\frac{1}{2}f''(\xi_1)(a-x)^2\\ f(-a)=f(x)+f'(x)(-a-x)+\frac{1}{2}f''(\xi_2)(-a-x)^2,$$ for some $\xi_1, \xi_2$. From this
$$f'(x)=\frac{f(a)-f(-a)}{2a}+\frac{1}{4a}\Bigl((x^2+a^2)(f''(\xi_1)-f''(\xi_2))-2ax(f''(\xi_1)+f''(\xi_2))\Bigr)$$
Using triangle inequality gives $|f'(x)|\leqslant \frac{M_0}{a}+\frac{x^2+a^2}{2a}M_2+|x|\cdot M_2$
How to eliminate that $|x|\cdot M_2$?
Taylor expansion about $x=x_0$ is given by $$f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2}(x-x_0)^2.$$ So, $$f(a)=f(x_0)+f'(x_0)(a-x_0)+\frac{f''(x_0)}{2}(a-x_0)^2,~~~~~~~~~~~~~~~(1)$$ and $$f(-a)=f(x_0)+f'(x_0)(-a-x_0)+\frac{f''(x_0)}{2}(-a-x_0)^2~~~~~~~~~~~~~~~(2)$$ Subtracting $(2)$ from $(1)$, we get $$f(a)-f(-a)=2a~f'(x_0)+\frac{1}{2}(-4ax_0)f''(x_0)$$ $$\implies f'(x_0)=\frac{f(a)-f(-a)}{2a}+\frac{1}{2a}(2ax_0)f''(x_0).$$ Using triangle inequality, we have $$|f'(x_0)|\leq\frac{M_0}{a}+\frac{1}{2a}(2ax_0)M_2$$ $$\implies |f'(x_0)|\leq\frac{M_0}{a}+\frac{1}{2a}(x_0^2+a^2)M_2~~~(\because (x_0-a)^2\geq 0).$$ Since $x_0$ was arbitrary, so $$|f'(x)|\leq\frac{M_0}{a}+\frac{1}{2a}(x^2+a^2)M_2.$$