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Problem (Laplacian Invariance by Rotations). Let $U \subset \mathbb{R}^{m}$ be an open set. For any function $f: U \to \mathbb{R}$ twice differentiable, the Laplacian of $f$ is the function $\triangle f: U \to \mathbb{R}$, defined by $$\displaystyle \triangle f = \frac{\partial^{2} f}{\partial x^{2}_{1}} + \cdots + \frac{\partial^{2} f}{\partial x^{2}_{m}}.$$ Prove that if $T: \mathbb{R}^{m} \to \mathbb{R}^{n}$ is an orthogonal linear transformation, then $$\triangle (f \circ T) = (\triangle f)\circ T: V \to \mathbb{R},$$ where $V = T^{-1}(U)$.

I tried to solve this using the Hessian matrice and writing the partial derivatives as inner product, but I didn't get something promising. I appreciate any hint!

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Let $e_1, \ldots, e_m$ be the standard $\mathbb R^m$ basis. Then we have $$Te_j = \sum_{i}a_{ij}e_i$$ where $a_{ij} = \langle Te_j , e_i \rangle$. Let $T_i(x)$ be the $i$-th coordinate of $T(x)$.

So we have $$T_i(x) = \sum_{j} a_{i,j}x_j.$$

Lets denote $\partial / \partial x_i$ by $\partial_i$. We have $$\partial_i [f(T(x))] = \sum_j \partial_i T_j(x)\partial_j(f)(T(x)) $$ and since $ \partial_i T_j(x) = a_{ij}$ we obtain $$\partial_i [f(T(x))] = \sum_j a_{ij}\partial_j(f)(T(x)).$$

Let's now compute the second derivative: $$\partial_i^2 [f(T(x))] = \sum_j a_{ij}\partial_i[\partial_j(f)(T(x))].$$

By the above calculation, putting $\partial_j(f)(T(x))$ in the place of $f(T(x))$, we obtain $$\partial_i[\partial_j(f)(T(x))] = \sum_k a_{ik}\partial_k\partial_j(f)(T(x))$$ and therefore $$\partial_i^2 [f(T(x))] = \sum_{j,k} a_{ij}a_{ik}\partial_k\partial_j(f)(T(x)).$$

Since $T$ is ortonormal we have $$\sum_{i}a_{ij}a_{ik} =\langle Te_j,Te_k\rangle = \langle e_j, e_k \rangle = \delta_{jk}.$$

Therefore \begin{align*} \Delta[f\circ T(x)] &= \sum_{i}\partial_i^2 [f(T(x))] =\sum_i\sum_{j,k}a_{ij}a_{ik}\partial_k\partial_j(f)(T(x)) \\&= \sum_{j,k} \partial_k\partial_j(f)(T(x)) \sum_i a_{ij}a_{ik} = \sum_{j,k} \partial_k\partial_j(f)(T(x)) \delta_{jk} \\&= \sum_k \partial_k^2(f)(T(x)) = \Delta(f) \circ T(x). \end{align*}

And that ends the proof.

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  • $\begingroup$ A more elegant proof can be done using the Fourier transform and that it commutes with ortogonal transformations. $\endgroup$ – Hugocito Jun 10 '18 at 0:51
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    $\begingroup$ Nice proof! Thank you for the support! $\endgroup$ – Lucas Corrêa Jun 10 '18 at 4:07

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