Let $f \in \cal{C}^2[-\epsilon, \epsilon]$. I want to show that $$|f'(0)|^2 \leq \frac{4}{\epsilon^2}\|f\|^2_{\infty} + 4\|f\|_{\infty}\|f''\|_{\infty}$$

Using Taylor's theorem, it is easy to deduce $$|f'(0)| \leq \frac{2}{\epsilon}\|f\|_{\infty} + \frac{\epsilon}{2}\|f''\|_{\infty}$$ whence it follows that $$|f'(0)|^2 \leq \frac{4}{\epsilon^2}\|f\|^2_{\infty} + 2\|f\|_{\infty}\|f''\|_{\infty} + \frac{\epsilon^2}{4}\|f''\|^2_{\infty}$$ Any hints on how to finish the proof?


Using Taylor's formula as you suggested, you get $$f(x)=f(0)+f'(0)x+\frac12 f''(c)x^2$$ and so $$|f'(0)|\le \frac1{|x|}|f(x)|+ \frac1{|x|}|f(0)|+\frac12 |f''(c)||x|\le \frac2{|x|}\Vert f\Vert_\infty+\frac12 \Vert f''\Vert_\infty |x|. $$ Since this is true for every $x\in [-\varepsilon,\varepsilon]\setminus\{0\}$ you have that $$|f'(0)|\le \inf_{x\in [-\varepsilon,\varepsilon]\setminus\{0\}}\left(\frac2{|x|}\Vert f\Vert_\infty+\frac12 \Vert f''\Vert_\infty {|x|}\right).$$ So now you minimize the function $g(x)=\frac{a}{x}+b x$ over $x\in (0,\varepsilon]$. You have $g'(x)=-\frac{a}{x^2}+b\ge 0$ for $x\ge \sqrt\frac{a}{b}$. If $\sqrt\frac{a}{b}\le \varepsilon$ then $g$ has a minimum at $\sqrt\frac{a}{b}$ and so $|f'(0)|\le 2\frac{a}{\sqrt\frac{a}{b}}+b \sqrt\frac{a}{b}=2\sqrt{ab} $. If $\sqrt\frac{a}{b}> \varepsilon$ then $|f'(0)|\le 2\frac{a}{\varepsilon}+b \varepsilon\le 2\frac{a}{\varepsilon}+b \sqrt\frac{a}{b}=2\frac{a}{\varepsilon}+\sqrt{ab}$.

  • $\begingroup$ thanks. I fixed it $\endgroup$ – Gio67 Jun 4 '17 at 18:38
  • 1
    $\begingroup$ Indeed: that shows it. Thank you! $\endgroup$ – Ignatius Jun 4 '17 at 18:40

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