José A. Cañizo — Research · Publications · Teaching · Other

Convolution of symmetric, radially decreasing functions

A function $f \: \R^d \to \R$ is radially symmetric if $f(x) = F(|x|)$ for some $F \: [0,+\infty) \to \R$. We say a radially symmetric function is radially increasing if its corresponding $F$ is increasing, and similarly for radially decreasing.

I want to give a simple proof of the fact that the convolution of nonnegative, radially symmetric, and radially decreasing functions is again radially decreasing. We used this in a recent paper on the heat equation, and we found that it is almost directly a consequence of Lemma 2.2 (i) in [Lieb1983] (thanks to Rupert Frank for the reference!). In Lieb’s paper the proof is not explicitly given (though the main arguments are given in another lemma), and I was not able to find a full proof anywhere. Since it is not an obvious proof, I want to record it here in case it is useful.

Here is the precise result with its proof:

Lemma 1. Let $d \geq 1$ and $f, g \: \R^d \to [0,+\infty)$ be nonnegative, radially symmetric functions for which the convolution $f * g$ is well defined for all $x \in \R^d$. Then $f * g$ is radially symmetric and furthermore:

  1. If both $f$ and $g$ are radially nonincreasing, then $f*g$ is radially nonincreasing.
  2. If $f$ is radially nondecreasing and $g$ is radially nonincreasing, then $f * g$ is radially nondecreasing.

Proof. The fact that $f*g$ is radially symmetric is a basic property of convolution which can be checked by a change of variables $y \mapsto Qy$, with $Q$ any given orthogonal transformation of $\R^d$. In order to check the other properties, call (for $R \geq 0$ and $x \in \R^d$) \begin{equation*} 𝟙_R (x) := \begin{cases} 1 &\text{if $|x| < R$,} \\ 0 &\text{if $|x| \geq R$.} \end{cases} \end{equation*} It is clear that for any $R,S \geq 0$ we have that $𝟙_R * 𝟙_S$ is radially nonincreasing, since the volume of the intersection of $B_R(0)$ and $B_S(x)$ is radially nonincreasing in $x$. Now, if $f$ and $g$ are radially nonincreasing we can use the layercake decomposition to write \begin{equation*} f(x) = \int_0^\infty 𝟙_R(x) \d \mu(R), \qquad g(x) = \int_0^\infty 𝟙_S(x) \d \nu(S) \quad\text{almost everywhere in $\R^d$}, \end{equation*} where $\mu$ and $\nu$ are minus the distributional derivatives of $f$ and $g$ in the radial direction. Notice that $\mu$ and $\nu$ are finite nonnegative measures on $[0,+\infty)$. Then, for all $x \in \R^d$, \begin{equation*} f * g (x) = \ird f(x-y) g(y) \d y = \int_0^\infty \int_0^\infty \ird 𝟙_R(x-y) 𝟙_S(y) \d y \d \nu(S) \d \mu(R), \end{equation*} which is clearly radially nonincreasing in $x$ since the inner integral is. On the other hand, if $f$ is radially nondecreasing we call \begin{equation*} 𝟙_R^{\mathrm{c}} (x) := 1 - 𝟙_R(x) = \begin{cases} 0 &\text{if $|x| < R$,} \\ 1 &\text{if $|x| \geq R$.} \end{cases} \end{equation*} Since $𝟙_R * 𝟙_S$ is radially nonincreasing, $𝟙_R^{\mathrm{c}} * 𝟙_S = (1 - 𝟙_R) * 𝟙_S$ is radially nondecreasing. We can write, again with almost everywhere equality, \begin{equation*} f(x) = \int_0^\infty 𝟙_R^{\mathrm{c}} (x) \d \mu(R), \end{equation*} where the measure $\mu$ is now the distributional derivative of $f$ in the radial direction, again a nonnegative measure. As before, \begin{equation*} f * g (x) = \ird f(x-y) g(y) \d y = \int_0^\infty \int_0^\infty \ird 𝟙_R^{\mathrm{c}}(x-y) 𝟙_S(y) \d y \d \nu(S) \d \mu(R), \end{equation*} which is radially nondecreasing, since the inner integral is. ∎

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