Proving Maximal Derivative Bound for a Twice Continuously Differentiable Function on a Closed Interval

Question

Let $f \in C^2[a,b]$ be a function, where $f(a) = f(b) = 0$. Define $M = \max\limits_{x\in[a,b]}\{|f''(x)|\}$. I am trying to show that $$\max_{x\in[a,b]}|f'(x)| \leqslant \frac{M}{2}(b-a).$$

I have tried following process.

First we let $f(c)=\max\limits_{x\in[a,b]}\{f(x)\}$ (or $f(c)=\min\limits_{x\in[a,b]}\{f(x)\}$). By using the Fermat theorem we can show that $$ f'(c)=0. $$

By using Taylor's formula with Lagrange remainder, we can get that for $\forall i\in[a,b]$,

$$ f'(i)=f'(c)+f''(\xi)(i-c),\qquad \text{where }\xi\in(c,i)\text{ or }(i,c) $$

However, it is unclear how to proceed from here. Performing either summation or multiplication leads to the intermediate result

$$\max_{x\in[a,b]}|f'(x)| \leqslant M(b-a),$$ which is weaker than the desired conclusion.

I would appreciate any insights or proofs related to the inequality stated above. Any help would be valuable. Thank you!

Answer 1

After a few other attempts I managed to solve this problem.

Because $f(a) = f(b) = 0$, let's take $x \in (a, b)$ and expand around $x$ at $a$ and $b$:

$$ f(a) = f(x) + f'(x)(a-x) + \frac{f''(\xi)}{2}(a-x)^2,\\ f(b) = f(x) + f'(x)(b-x) + \frac{f''(\mu)}{2}(b-x)^2, $$

where $\xi\in (a,x),\ \mu\in(b,x)$. Therefore, $$ 0 = f(a) - f(b) = f(x) + f'(x)(a-x) + \frac{f''(\xi)}{2}(a-x)^2 - \left(f(x) + f'(x)(b-x) + \frac{f''(\mu)}{2}(b-x)^2\right). $$

Hence, using the triangle inequality, $$ |f'(x)| \leq \frac{\frac{M}{2}((x-a)^2+(x-b)^2)}{b-a}. $$ If we let $(x-a)^2+(x-b)^2=g(x)$, then it can be easily calculated that when $g'(x) = 0,\ x=\frac{a+b}{2}$. Simple calculations show that $g(\frac{a+b}{2})<g(a)=g(b)$, so the maximum is attained at either $a$ or $b$, therefore $$ |f'(x)| \leq \frac{\frac{M}{2}((b-a)^2+(b-b)^2)}{b-a} = \frac{\frac{M}{2}g(b)}{b-a} = \frac{M}{2}(b-a). $$

Answer 2

Are you sure this a correct assertion?

Here is a counterexample. Let $\alpha\in (0,1)$ be fixed and define $$ g(x) = \begin{cases} e^x & 0\leq x\leq \alpha, \\ e^{-x} & -\alpha\leq x\leq 0. \end{cases} $$

Although $g(x)$ is not $C^2$ at origin, we can use the smoothing technique to get a $C^2$ function $f(x)$ based on $g(x)$. Then obviously $$ \max_{x\in[-\alpha,\alpha]} |f’(x)| = e^\alpha\qquad \text{and} \qquad \max_{x\in[-\alpha,\alpha]} |f’’(x)| = e^\alpha. $$ However $e^\alpha \leq \frac{e^\alpha}{2} (\alpha - (-\alpha))$ can not hold in this case provided $\alpha\in (0,1)$.