📂Fourier Analysis

Convolution of Multivariable Functions

Definition

Let’s say we have $f,g:\mathbb{R}^{n}\to \mathbb{C}$ and $\mathbf{x},\mathbf{y} \in \mathbb{R}^{n}$. Then the convolution of these two multivariable functions is as follows:

$$f \ast g(\mathbf{x})=\int f(\mathbf{y})g(\mathbf{x}-\mathbf{y})d\mathbf{y}$$

In this case, the integral mentioned above is the integral of a multivariable function.

Properties

The convolution of multivariable functions also satisfies the same desirable properties as the convolution of single-variable functions.

(a) Commutative Law

$$f \ast g = g \ast f$$

(b) Distributive Law

$$f \ast (g+h)=f \ast g + f \ast h$$

(c) Associative Law

$$f\ast (g\ast h)=(f \ast g)\ast h$$

(d) Associative Law of Scalar Multiplication

$$a(f \ast g)=(af \ast g)=(f\ast ag)$$

(e) Differentiation

$$\partial_{j}(f \ast g)=(\partial _{j}f) \ast g=f \ast (\partial _{j}g)\quad \text{where } \partial_{j}=\frac{ \partial }{ \partial_{j}}$$

Explanation

Naturally, convolution convergence theorem and convolution norm convergence theorem are also satisfied.

Let’s say we have $g\in L^{1}$ and $\int g(\mathbf{x})d\mathbf{x}=1$. And let’s consider $g_{\epsilon (\mathbf{x})}=\epsilon^{-n}g(\epsilon^{-1}\mathbf{x})$.

• (i) If $f$ is bounded or if $g$ is $0$ outside a certain closed interval, then $f \ast g$ is well-defined, and if $f$ is continuous in $\mathbf{x}$, the following holds:

  $$\lim \limits_{\epsilon \to 0} f \ast g_{\epsilon}(\mathbf{x})=f(\mathbf{x})$$

  If $f$ is continuous on a closed and bounded $D$, then the convergence above is uniform convergence on $D$.

• (ii) If we have $f\in L^{2}$, then the following holds:

  $$\lim \limits_{\epsilon \to 0} \left\| f \ast g_{\epsilon}-f \right\| =0$$