Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In a paper (dealing with spectra of certain Schrodinger operators) I found the following assumption for a function $f\in C^\infty(\mathbb R^n;\mathbb R)$:

there exists a constant $C>0$ and a compact set $K\subset\mathbb R^n$ such that for $x\in\mathbb R^n\setminus K$

(i) $|\nabla f(x)|\geq \frac 1C$

(ii) $|\text{Hess}f(x)|\leq C|\nabla f(x)|^2$

I'm bit confused about this and don't find it intuitive. I understand that $f(x)= |x|^\alpha$ is ok for $\alpha\geq 1$ and it is not ok for $\alpha<1$ because (i) doesn't hold. I would like to understand it better by finding (preferably simple,onedimensional) examples of functions that

a) satisfy (i) but not (ii) (or viceversa)

b) satisfy (i)' $|\nabla f(x)|\rightarrow \infty$ for $|x|\rightarrow\infty$ instead of (i) but not (ii)

c)show the relation (if any) of (i) and (ii) with convexity/concavity properties.

d) show the relation (if any) to $e^{-f} \in L^2$

e) (i) and (ii) hold but not the following (ii)': there exists $\rho>0$ such that $\sup_{y\in B_{\rho}(x)}|\text{Hess}f(y)|\leq C|\nabla f(x)|^2$ with $B_\rho(x)$ the ball of radius $\rho$ centered in $x$.

In general any kind of intuition and comments are appreciated, and of course also partial answers are more than welcome.

Many thanks

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.