# Properties of $g$ satisfying $f(x,x)[\nabla_{x}^{2}g(x,y)]_{x=y}+2[\nabla_{x}f(x,y)]_{x=y}\cdot[\nabla_{x}g(x,y)]_{x=y}=0$ for all $f$

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

$$f\left(x,x\right)\left[\nabla_{x}^{2}g\left(x,y\right)\right]_{x=y}+2\left[\nabla_{x}f\left(x,y\right)\right]_{x=y}\cdot\left[\nabla_{x}g\left(x,y\right)\right]_{x=y}=0$$

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.

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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
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.
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