L¹(ℝ)中函数f与g(x)=e^(2iπx)的卷积计算及相关定理

Let's break this down step by step, starting with key definitions and background, then applying them to compute the convolution.

Key Definitions to Recall #

• $L^1(\mathbb{R})$ Space: This consists of all measurable functions $f$ where the integral of their absolute value is finite:
  $$\int |f(t)|dt < \infty$$
• Convolution: For two functions $f$ and $g$, their convolution at a point $x$ is defined as:
  $$(f*g)(x) = \int f(x-\tau)g(\tau)d\tau$$
• Convolution Commutativity: A handy property here is that $fg = gf$ — we'll use this to simplify our integral later.
• Euler's Formula: The complex exponential can be expressed using trigonometric functions, though we might not need this directly here:
  $$e^{i2\pi x} = \cos(2\pi x )+ i\sin(2\pi x)$$

Relevant Theorem (Theorem 1) #

Let $f \in L^1(\mathbb{R})$ and $g\in C^p(\mathbb{R})$. If all $k$-th derivatives of $g$ (for $k=0,1,\dots,p$) are bounded, then:

1. $f*g \in C^p(\mathbb{R})$;
2. For $k=1,\dots,p$, $(fg)^{(k)} = fg^{(k)}$.

Quick note: Since $g(x) = e^{2i\pi x}$ is infinitely differentiable (all its derivatives are just scalar multiples of itself, hence bounded), this theorem tells us $f*g$ is infinitely smooth — a good sanity check for our result.

Computing the Convolution #

First, use the commutativity of convolution to rewrite the integral in a more manageable form:
$$\begin{aligned}(fg)(x) &= (gf)(x) = \int g(x-\tau)f(\tau)d\tau \&= \int e^{i2\pi(x-\tau)}f(\tau)d\tau\end{aligned}$$

Next, factor out the term that doesn't depend on the integral variable $\tau$ ($e^{i2\pi x}$ is constant with respect to $\tau$):
$$\begin{aligned}(f*g)(x) &= e^{i2\pi x} \int e^{-i2\pi \tau}f(\tau)d\tau\end{aligned}$$

That integral $\int e^{-i2\pi \tau}f(\tau)d\tau$ is exactly the Fourier transform of $f$ evaluated at $\xi=1$, usually written as $\hat{f}(1)$. So we can simplify the result to a neat expression:
$$(f*g)(x) = \hat{f}(1) \cdot e^{i2\pi x}$$

In short, convolving an $L^1$ function with this complex exponential just scales the exponential by the value of $f$'s Fourier transform at 1.

内容的提问来源于stack exchange，提问作者Tiger Blood