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Suppose $f(x) \in C^2([a, b])$ with $f(a) = f(b) = 0$. Show that $$\max_{x \in [a, b]} |f'(x)| \leq \frac{(b-a)^2}{2} \max_{x \in [a, b]}|f''(x)|.$$

My work: Suppose that $f'(x_0) = \max_{x \in [a, b]} |f'(x)|$. One can expand $f(x)$ at $x_0$ as follows: $$f(a) - f(x_0) = f'(x_0) (a - x_0) + \frac{1}{2}f''(\xi_1) (a-x_0)^2,~\xi_1\in[a, x_0]$$ $$f(b) - f(x_0) = f'(x_0) (b - x_0) + \frac{1}{2}f''(\xi_2) (b-x_0)^2,~\xi_2\in[x_0, b]$$ The difference of the two equations leads to $$f'(x_0) (b-a) = \frac{1}{2}\Big[f''(\xi_1)(a-x_0)^2 - f''(\xi_2)(b-x_0)^2\Big].$$ I have no idea how to reach the term $\max |f''(x)|$.

Arctic Char
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zengc
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1 Answers1

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The statement is false.

Let $a>0$ and define $f : [-a,a] \rightarrow \mathbb R : x \mapsto (a^2 - x^2) = (a-x)(a+x).$

Then clearly $f \in C^2[-a,a]$ with $f(a) = f(-a) = 0$.

Since $f'(x) = -2x$ and $f''(x) = -2$ we have

$$ \max_{x \in [-a,a]} \vert f'(x)\vert = \max_{x \in [-a,a]} \vert -2x \vert = 2a$$

and $$ \max \vert f''(x)\vert = \max \vert -2 \vert = 2. $$

But then we cannot have $$ 2a = \max \vert f' \vert \leq \frac{(a - (-a))^2}{ 2} \max \vert f''\vert = \frac{(2a)^2}{2} 2 = (2a)^2$$

for $0<a < \frac{1}{2}.$

Digitallis
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