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Suppose that $f(x)$ is a $C^2$ function on $\mathbb{R}$ such that $\left| f(x) \right| \leq A$ and $\left| f''(x) \right| \leq C $ for $x \in \mathbb{R}$. Prove that $\left| f'(x)\right| \leq \sqrt{2AC}$.

math_rookie
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2 Answers2

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Let $h > 0$ and $x\in \Bbb R$. Then $$f(x + h) = f(x) + \int_0^{h} f'(x + t)\, dt$$ and by integration by parts, $$\int_0^h f'(x + t)\,dt = f'(x)h + \int_0^h (h - t)f''(x + t)\, dt.$$ By the second mean value theorem for integrals, $$\int_0^h (h - t)f''(x + t)\, dt = f''(c)\int_0^h (h - t)\, dt = \frac{f''(c)}{2}h^2$$ for some $c\in (x, x+h)$. Thus $$f(x + h) = f(x) + hf'(x) + \frac{f''(c)}{2}h^2$$ which implies $$f'(x) = \frac{f(x+h) - f(x)}{h} - \frac{f''(c)}{2}h.$$ Similarly,

$$f(x - h) = f(x) - hf'(x) + \frac{f''(d)}{2}h^2$$

for some $d\in (x - h,x)$. Hence

$$\frac{f(x + h) - f(x - h)}{2h} = f'(x) + \frac{f''(c) - f''(d)}{4}h.$$

By hypothesis and the triangle inequality,

$$\lvert f'(x)\rvert \le \frac{A}{h} + \frac{Ch}{2}.$$ Since this inequality holds for all $h > 0$, $$\lvert f'(x)\rvert \le \inf_{h > 0} \left\{\frac{A}{h} + \frac{Ch}{2}\right\} = 2\sqrt{\frac{AC}{2}} = \sqrt{2AC}.$$ Since $x$ was arbitrary, the result follows.

kobe
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  • I'm having trouble at the integration by parts section. How can we say that $\int_0^h f'(x+t)dt = f'(x)h + \int_0^h (h-t)f''(x+t)dt$? – math_rookie Jan 30 '15 at 00:11
  • Let $u = f'(x + t)$ and $v = t - h$ in the integration by parts formula $\int u, dv = uv - \int v, du$. – kobe Jan 30 '15 at 00:12
  • Oh, so you're integrating $\int_0^h (h−t)f″(x+t)dt$ by parts and rearranging. I get it now. – math_rookie Jan 30 '15 at 00:20
  • Actually I integrated $\int_0^h f'(x + t), dt$ by parts. It's of the form $\int_0^h u, dv$ with $u = f'(x + t)$ and $v = t - h$. But yes, you can perform integration by parts on $\int_0^h (h - t)f''(x + t), dt$ to get the same equation. – kobe Jan 30 '15 at 00:23
  • @kobe I was able to show that the bound $\sqrt{2AC}$ is indeed correct. ;-)) - Mark – Mark Viola Jan 27 '16 at 00:01
  • @Dr.MV oh now you tell me! : - ) Thanks. Nice answer, btw. I did forget to subtract a corresponding equation for $f(x - h)$ to make it work. I'll make an edit. – kobe Jan 27 '16 at 02:35
  • You're welcome. My pleasure. And well done. +1 – Mark Viola Jan 27 '16 at 02:54
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Since $f''(x)$ is assumed to be continuous we have

$$f'(x+x')-f'(x)=\int_{x}^{x+x'} f''(t)\,dt \tag 1$$

If we integrate $(1)$ with respect to $x'$, then we obtain

$$\begin{align} \int_{0}^{h} \left(f'(x+x')-f'(x)\right)\,dx'&=f(x+h)-f(x)-f'(x)h\\\\ &=\int_{0}^{h}\int_{x}^{x+x'}f''(t)\,dt\,dx'\\\\ &=\int_{x}^{x+h} f''(t)(x+h-t)\,dt \tag 2 \end{align}$$

Rearranging $(2)$ reveals that

$$f'(x)=\frac{f(x+h)-f(x)}{h}-\frac1h\int_{x}^{x+h} f''(t)(x+h-t)\,dt \tag 3$$

We can replace $h$ with $-h$ in the preceding analysis and obtain

$$f'(x)=\frac{f(x-h)-f(x)}{-h}+\frac1h\int_{x}^{x-h} f''(t)(x-h-t)\,dt \tag 4$$

whereupon adding $(3)$ and $(4)$ yields

$$2f'(x)=\frac{f(x+h)-f(x-h)}{h}+\frac1h\left(\int_{x}^{x-h} f''(t)(x-h-t)\,dt-\int_{x}^{x+h} f''(t)(x+h-t)\,dt\right) \tag 5$$

Now, we are given that $|f''(x)|\le C$ and $|f(x)|\le A$. Therefore, using these bounds in $(5)$ yields

$$|f'(x)|\le \frac{A}{h}+\frac{Ch}{2} \tag 6$$

for all $h>0$. The maximum value of the right-hand side of $(6)$ occurs at $h=\sqrt{\frac{2A}{C}}$ and is equal to

$$\frac{A}{\sqrt{\frac{2A}{C}}}+\frac{C\sqrt{\frac{2A}{C}}}{2}=\sqrt{2AC}$$

Therefore, we have

$$|f'(x)|\le \sqrt{2AC}$$

as was to be shown!

Mark Viola
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