📂Fourier Analysis

Riemann-Lebesgue Lemma

Theorem¹

Let’s assume $f \in$ $L^{1}$ is given. Then, the following equation holds.

$$\lim \limits_{n \to \pm \infty} \hat{f}(\xi) = 0$$

Here, $\hat{f}$ is the Fourier transform of $f$.

Proof

step 1 Proofs for the step function $f$, and generalization in step 2. Note that $f$ in step 1 and step 2 are not the same.

• case 1

  Assume $f$ is the following step function.

  $$f(x) = \sum \limits_{j=1}^n c_{j} \chi_{j}(x)$$

  $c_{j}$ is a constant and $\chi_{j}(x)=\chi_{ [-x_{j}-a_{j} ,\ x_{j}+a_{j}] }(x)$. Using $\mathcal{F} \left[ f(x-a) \right] ( \xi ) = e^{-ia\xi}\hat{f}(\xi)$, $\chi_{[-x_{j} - a_{j},\ x_{j} + a_{j}]}(x)=\chi_{[-a_{j} ,\ a_{j}]}(x-x_{j})$, and $\mathcal{F} \left[ \chi_{[-a,a]}(x) \right] = \dfrac{2 \sin(a\xi) }{\xi}$ here,

  $$\begin{align*} \mathcal{F}\left[ \chi_{[-x_{j}-a_{j} ,\ x_{j}+a_{j}]} (x)\right] (\xi) &= \mathcal{F}\left[ \chi_{ [-a_{j} ,\ a_{j}]} (x-x_{j})\right] (\xi) \\ &= e^{-i\xi x_{j}} \mathcal{F} \left[ \chi_{[-a_{j},\ a_{j}]}(x)\right] (\xi) \\ &= e^{-i\xi x_{j}}\dfrac{2\sin (a_{j}\xi) }{\xi} \end{align*}$$

  Therefore, the Fourier transform of $f$ is

  $$\hat{f} (\xi) = \sum \limits_{j=1}^n2 c_{j}e^{-i\xi x_{j}} \dfrac{ \sin (a_{j} \xi) }{\xi} \rightarrow 0 \quad \mathrm{as}\ \ \xi \rightarrow \pm \infty$$

• step 2

  Assume $f \in L^{1}$. Then, there exists a sequence of step functions $\left\{ f_{n} \right\}$ that satisfy the following equation.

  $$\int |f_{n}(x) -f(x)| dx \rightarrow 0$$

  Using this,

  $$\begin{align*} \sup \limits_{\xi}| \hat{f_{n}}(\xi) - \hat{f}(\xi)| &= \sup \limits_{\xi} \left| \int \big( f_{n}(x)-f(x) \big) e^{-i\xi x} dx \right| \\ &\le \sup \limits_{\xi} \int \left| f_{n}(x)-f(x) \right| dx \rightarrow 0 \end{align*}$$

  Therefore, $\hat{f_{n}}(\xi)$ uniformly converges to $\hat{f}(\xi)$. And from the result of step 1,

  $$\hat{f}(\xi) \rightarrow 0 \quad \mathrm{as} \ \ \xi \rightarrow \pm \infty$$

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