Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that $g:\mathbb{R}^{2n}\rightarrow\mathbb{R}$ is a function such that:


holds for all $f:\mathbb{R}^{2n}\rightarrow\mathbb{R}$. Here $x$ and $y$ denote points in $\mathbb{R}^{n}$.

The properties that interest me are:

  1. Does it follow that $g$ satisfies $\left[\nabla_{x}^{2}g\left(x,y\right)\right]_{x=y}=0$ ?

  2. Does it follow that $g$ satisfies $\left[\nabla_{x}g\left(x,y\right)\right]_{x=y}=0$ ?

Note: I am only considering smooth functions $f$ and $g$ here.

share|improve this question
Please, feel free to retitle and/or retag this question if you come up with something more appropriate. –  becko Mar 20 '13 at 3:35
Also, this is not a homework exercise. I only placed the tag [mathematical-physics] because this problem shows up in some quantum mechanics calculations that I was doing. –  becko Mar 22 '13 at 0:28
$\nabla_x$ means $\frac{\partial}{\partial x_1}+...+\frac{\partial}{\partial x_n}$? –  Occupy Gezi Mar 22 '13 at 1:16
@AnilBaseski Yes. And $\nabla_x^2 = \nabla_x \cdot \nabla_x$ is the Laplacian with respect to $x_1,...,x_n$. –  becko Mar 22 '13 at 1:20
Why do you think they must be zero? They don't include differantials $d\ x_i$? –  Occupy Gezi Mar 22 '13 at 1:26

1 Answer 1

up vote 2 down vote accepted

Since this is holding for every $f$, let's take $f = 1$. It answers the first question.

Then take $f(x,y) = x$, it answers the second question.

share|improve this answer
So that answers the first question: $\left[\nabla_{x}^{2}g\left(x,y\right)\right]_{x=y}=0$. As for the second question, you cannot take $f(x,y)=x$, because $x$ is a vector in $\mathbb{R}^{n}$ and $f(x,y)$ is supposed to be a scalar. –  becko Mar 22 '13 at 13:50
To fix it, just take $f(x,y)=a_1x_1+...+a_nx_n$. Then vary $(a_1,...,a_n)$, to conclude that $\nabla_x g(x,y)=0$. –  Tomás Mar 25 '13 at 12:39
@Tomás +1 Thanks. –  becko Mar 26 '13 at 19:14

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.