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  • $f(x)$ is a convex function and twice differentiable. The domain is $\mathbb{R}$.
  • suppose $f'(x_{0})=0$ , and $x_{0}$ is unique.
  • My question is: what is $inf(\frac{x-x_{0}}{f'(x)})$,$x\neq x_{0}$?

I'm aware that this function has a limit when x approaching $x_{0}$ and $f''(x_{0})$ is not $0$ (despite that the limit may not be important in this question).

  • $\lim_{x\rightarrow x_{0}}\frac{x-x_{0}}{f'(x)}=\frac{1}{f''(x_{0})}$

But can we always get a infimum of this set? I tried Taylor decomposition， apparently the sign of the Lagrange residue is uncertain (even it is third order differentiable).

  • $f'(x_{0})=f'(x)+f''(x)(x_{0}-x)+\frac{f'''(\varepsilon )}{2}(x_{0}-x)^2$

the question is can we always get a infimum of $\frac{x-x_{0}}{f'(x)}$? if yes, what is it? if not, what condition can make this true.

Ocean
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  • Welcome to Math.SE! Since $f$ is twice-differentiable, try writing out $f''(x_0)$ using the limit definition of the derivative. – David M. Jan 26 '19 at 04:40
  • thank you very much. I've edited my question. can you be more specific? – Ocean Jan 26 '19 at 05:01
  • In your edit you say $f''(x_0)$ is not zero--what if $f(x)=x^4$ (so $x_0=0$)? – David M. Jan 26 '19 at 05:02
  • Also, every set of real numbers which is bounded from below has an infimum. It's not too hard to show this set is bounded below by zero, so the infimum always exists. – David M. Jan 26 '19 at 05:06
  • Then we might add a condition to this question: $f''(x_{0})\neq0$. – Ocean Jan 26 '19 at 05:08
  • Okay! I would add that to the question (and maybe clarify what you mean by “get an infimum”—do you mean the infimum exists, or there exists an $x$ that attains the infimum?) – David M. Jan 26 '19 at 05:10
  • I think it is bounded by $1/2$ when $f(x)=x^2$. – Ocean Jan 26 '19 at 05:11
  • yes, I want to find this infimum. – Ocean Jan 26 '19 at 05:30

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