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Let $\phi \colon \mathbb{R}^n \to \mathbb{R}$ be continuous with compact support. Furthermore, suppose that $\phi$ is spherically symmetric (i.e., assume $\phi(Tx) = \phi(x)$ for every orthogonal transformation $T\colon\mathbb{R}^n \to \mathbb{R}^n$).

Next, for $(x,t) \in \mathbb{R}^n \times (0, \infty)$, let $G(x,t) = (4\pi t)^{-\frac{n}{2}}e^{-\frac{|x|^2}{4t}}$ be the fundamental solution to the heat equation. Prove that the convolution $f(x,t) = \displaystyle \int_{\mathbb{R}^n} G(y, t)\phi(x-y)dy$ is also spherically symmetric in the variable $x \in \mathbb{R}^n$. In other words, we want to show that, for any $t > 0$, and any orthogonal $T\colon\mathbb{R}^n \to \mathbb{R}^n$, we have $f(Tx, t) = f(x, t)$.

This question appears on an old PDE qual I am studying, and I'm stumped so far. One thing I have noticed is that the function $G$ is spherically symmetric in the variable $x$. So, I'm wondering if there is a more general result I could apply, something like: "the convolution of two integrable, spherically symmetric functions on $\mathbb{R}^n$ is again spherically symmetric."

I also know that the convolution is a commutative operation, so I could just as well define $f$ by $f(x,t) = \displaystyle \int_{\mathbb{R}^n} G(x-y, t)\phi(y)dy$. But I am failing to see how that would make the proof easier.

Hints or solutions are greatly appreciated.

share|cite|improve this question
Use the change of variables, $y = Ty_1$, then the volume element in $y$ coordinates is $dV_y = \det(T) dV_{y_1}$. Since $T$ is orthogonal, we have $\det(T) = \pm 1$. – user17762 Dec 24 '12 at 6:27
Or use the uniqueness argument as for the Laplace equation. – user53153 Dec 24 '12 at 6:41
By the way, this is uncomfortably close to being a duplicate of… You could have just edited the previous question to clarify. – user53153 Dec 24 '12 at 6:43
@PavelM: I agree that editing… would have been smarter than posting this new question. In the future, I'll be more careful about the duplication issue. – JZShapiro Dec 24 '12 at 6:55

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