About some differential identities of Vandermonde Determinant I have three identities to prove:

I proved (1) and (2), except (3) 
I know that f(x) can be factored out (xi-xj) terms(exactly n(n-1)/2 factors)
, but this is all I know.
Any help would be appreciated. 
 A: The Vandermonde determinant is a sum of monomials whose exponent vectors are permutations of the vector $(0,1,2,\dotsc,n-1)$, with alternating signs. That is, the monomial $x_1^0 x_2^1 x_3^2 \dotsm x_n^{n-1}$ has coefficient $+1$, and for each permutation $x_1^{\pi(0)} \dotsm x_n^{\pi(n-1)}$, for $\pi \in S_n$, the coefficient is the sign of the permutation $\pi$.
Now, for each $i$, the second derivative $\partial^2 f/\partial x_i^2$ involves monomials whose exponent vectors have $n-1$ distinct numbers from $0$ to $n-1$, adding up to $\frac{n(n-1)}{2}-2$, with exactly one repetition. These are the only possible terms that can appear in the expression in equation (3), $\sum_i \partial^2 f/\partial x_i^2$.
So consider a possible term $x_1^{a_1} \dotsm x_n^{a_n}$, where $a_1,\dotsc,a_n$ are in $\{0,\dotsc,n-1\}$, sum to $\frac{n(n-1)}{2}-2$, and have exactly one repetition. Say $a_i = a_j$, $i<j$.
Consider the two terms in $f$ given by $x_1^{a_1} \dotsm x_i^{a_i+2} \dotsm x_j^{a_j} \dotsm x_n^{a_n}$, and $x_1^{a_1} \dotsm x_i^{a_i} \dotsm x_j^{a_j+2} \dotsm x_n^{a_n}$. They appear in $f$ with opposite signs because the pairs $(a_i+2,a_j)$ and $(a_i,a_j+2)$ have been transposed (and all other powers stay the same between these two terms, and $a_i=a_j$).
The first term contributes $\pm a_i(a_i-1) x_1^{a_1} \dotsm x_n^{a_n}$ to $\partial^2 f/\partial x_i^2$. The second term contributes $\mp a_j(a_j-1) x_1^{a_1} \dotsm x_n^{a_n}$ to $\partial^2 f/\partial x_j^2$. In the sum in (3), these contributions cancel out.
A: Hint: Write it as an exponential and then take its derivative.
Also, each identity has a Lie group interpretation, when you think of the operators as representing a Lie algebra:
(1) Invariance under translation.
(2) Homogeneous under scaling.
(3) Invariance under Gaussian convolution.
A: I have a short proof for (3). Try to show that $$\Delta f(X) := \sum_{j = 1}^{n} \frac{\partial^2 f(X)}{\partial x_j^2}$$ equals zero when $x_1  = x_2$. 
This will imply that $\Delta f(X)$ has a factor of $x_1 - x_2$. Similarly $\Delta f(X)$ has a factor of $x_i - x_j$ for every pair of $i < j$. This means that $\Delta f(X)$, as a polynomial, has a factor of $f(X)$. Since $\Delta f(X)$ is 2 degrees lower than $f(X)$, it can only equal to zero. 
