I am trying to prove that if a function $f:\mathbb{R} \to \mathbb{R}$ is bounded and the second derivative $f''$ is bounded, then the first derivative $f'$ is also bounded. My hint is to use Taylor's theorem.

I know for any $x$ and $y$, $f(y) = f(x) + f'(x)(y-x) + \frac{1}{2} f''(c)(y-x)^2,$ is the Taylor approximation where $c$ is some point in between $x$ and $y$. From this I get

$$f'(x)(y-x) = f(y) - f(x) - \frac{1}{2}f''(c)(y-x)^2.$$

If $|f(x)| < C$ and $|f''(x)| < D$ are bounds, then

$$|f'(x)| < \frac{|f(x)| + |f(y)|}{|y-x|} + \frac{1}{2} |f''(c)| |y-x| < \frac{2C}{|x-y|}+ \frac{D}{2}|x-y|.$$

I seem to be stuck at this point because both terms can be unbounded depending on $x$ and $y$.

Thank you for any help.

  • $\begingroup$ See: math.stackexchange.com/questions/1905598/… $\endgroup$ – John11 Mar 6 '17 at 21:23
  • $\begingroup$ @SAS: You can look at link provided by John11 offering a different approach, but there is an easy way to continue with what you started. $\endgroup$ – RRL Mar 6 '17 at 21:58


Your inequality is true for every $y \in \mathbb{R}$ and

$$\min_{0 < z < \infty} \left(\frac{2C}{z} + \frac{Dz}{2}\right)= 2 \sqrt{CD}$$

  • $\begingroup$ I get it now. Thanks. $\endgroup$ – SAS Mar 6 '17 at 23:26

Here is a solution with Tyler's expansion: $$ \forall x\in\mathbb{R},\forall h>0,\\\ f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(\eta_1)\text{ ,where $\eta_1\in(x,x+h)$}\\ f(x-h)=f(x)-hf'(x)+\frac{h^2}{2}f''(\eta_2)\text{ ,where $\eta_2\in(x-h,x)$}\\ \Rightarrow f(x+h)-f(x-h)=2hf'(x)+\frac{h^2}{2}[f''(\eta_1)+f''(\eta_2)]\\ \text{Hence, }|2hf'(x)|\leq|f(x+h)-f(x-h)|+|\frac{h^2}{2}[f''(\eta_1)+f''(\eta_2)]|\leq\\ 2\sup|f|+h^2\sup |f''| $$ We deduce that: $$2\sup|f'|\leq\frac{2\sup|f|}{h}+h\sup|f''|$$ Since $\sup|f^{(k)}|$ is a constant. We can let $$h=\sqrt{\dfrac{2\sup|f|}{\sup|f''|}}$$ We deduce that $$\boxed{\sup |f'| \le \sqrt{2}\cdot \sqrt{\sup |f| \cdot \sup|f''|}}$$ The first derivative is bounded since the function and its second derivative are bounded.


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