📂Fourier Analysis

Fourier Inversion Theorem

Buildup

The process of deriving the Fourier transform also derived the definition of the inverse transform. However, this was simply explained to aid understanding, and the transformation formula was not accurately derived. The Fourier inverse transformation is as follows:

$$\begin{equation} f(x) =\dfrac{1}{2\pi} \int \hat{f}(\xi) e^{i\xi x}d\xi \end{equation}$$

This equation implies that from $f$, we can obtain $\hat{f}$ and from $\hat{f}$, we can retrieve $f$ again. This might seem obvious or trivial, but it’s not at all. For instance, consider differentiation and integration. When we differentiate a polynomial, we lose the constant term, which cannot be retrieved through integration. However, the Fourier transform preserves information. From $f$, we obtain the Fourier transform $\hat{f}$, and by applying the inverse transform to this $\hat{f}$, we can obtain $f$ exactly as it was.

$$\begin{equation} \hat{f} (\xi) := \int f(x) e^{-i\xi x}dx \end{equation}$$

When defining the Fourier transform as above, $(1)$ becomes its inverse as demonstrated below.

Fourier Inverse Transform Theorem

Let’s define the Fourier transform of $f$ as $\hat{f}$ in $(2)$. Assume $f$ is integrable and piecewise continuous. At points of discontinuity, $f$ is defined as follows:

$$f(x)=\dfrac{1}{2} \big[ f(x-)+f(x+)\big]$$

Then, the following equation holds:

$$f(x)=\lim \limits_{\epsilon \rightarrow 0}\dfrac{1}{2\pi}\int\hat{f}(\xi)e^{i\xi x}e^{-\epsilon^2\xi^2/2}d\xi$$

Moreover, if we assume that $\hat{f} \in$ is $L^1$, then $f$ is continuous and

$$f(x) =\dfrac{1}{2\pi}\int\hat{f}(\xi) e^{i\xi x}d\xi = \mathcal{F}^{-1}[\hat{f}] (x)$$

Corollary

If we assume $\hat{f}=\hat{g}$, then $f=g$ follows.

Proof

$$f=\mathcal{F}^{-1}[\hat{f}]=\mathcal{F}^{-1}[\hat{g}]$$

Therefore,

$$f=\mathcal{F}^{-1}[\hat{g}]=\mathcal{F}^{-1}\mathcal{F}[g]=g$$

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Explanation

Upon observing this corollary, one might think, ‘Isn’t this obvious?’ but it’s not at all. Consider the operation of differentiation. Suppose $f(x)=x^2+1$, $g(x)=x^2+4$. In this case, the fact that $f^{\prime}(x)=2x=g^{\prime}(x)$ does not guarantee $f(x)=g(x)$.

Proof

Strategy:

To solve the problem easily, a cutoff function will be used. A cutoff function has the effect of cutting off outside a certain range when multiplied and taken to a limit, hence the name. It is used to maintain the original value near the origin and to converge to $0$ as it moves away from the origin. It’s a concept similar to a mollifier. If this explanation is difficult to understand, it’s okay to skip it. Either way, such a cutoff function will be multiplied into the equation to derive the inverse transformation.

$$\eta (\xi)=\dfrac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}, \quad \eta_\epsilon (\xi)=\dfrac{1}{\epsilon}\eta \left( \frac{\xi}{\epsilon} \right)=\dfrac{1}{\epsilon\sqrt{2\pi}}e^{-\frac{x^2}{2\epsilon^2}}$$

Multiplying this cutoff function yields

$$\dfrac{1}{2\pi}\int\hat{f} (\xi) e^{i\xi x}e^{-\epsilon^2\xi^2/2}d\xi$$

This equation, by the definition of the Fourier transform, is as follows.

$$\dfrac{1}{2\pi}\int \int f(y) e^{-i\xi y}dy e^{i\xi x}e^{-\epsilon^2\xi^2/2}d\xi$$

Using the definition of the Fourier transform and convolution, among others, the equation is transformed as follows:

$$\begin{align*} \dfrac{1}{2\pi}\int \int f(y) e^{-i\xi y}dy e^{i\xi x}e^{-\epsilon^2\xi^2/2}d\xi &= \dfrac{1}{2\pi}\int {\color{blue}\int} f(y) {\color{blue}e^{-i\xi (y-x)}e^{-\epsilon^2\xi^2/2}}dy {\color{blue}d\xi} \\ &= \dfrac{1}{2\pi}\int {\color{blue}\mathcal{F} \left[ e^{-\epsilon^2 \xi^2 /2}\right] (y-x)} f(y) dy \\ &= \dfrac{1}{2\pi}\int {\color{blue}\sqrt{\dfrac{2\pi}{\epsilon^2}}e^{-\frac{(y-x)^2}{2\epsilon^2}}} f(y) dy \\ &= \int \dfrac{1}{\epsilon \sqrt{2\pi}} e^{-\frac{(x-y)^2}{2\epsilon^2}} f(y) dy \\ &= \int \eta_\epsilon (x-y)f(y) dy \\ &= f \ast \eta_\epsilon (x) \end{align*}$$

The third equality used the formula below.

Fourier Transform of a Gaussian Function

The Fourier transform of a Gaussian function $f(x)=e^{-Ax^2}$ is as follows:

$$\mathcal{F}[f] (\xi) = \mathcal{F} \left[ e^{-Ax^2} \right] (\xi)=\sqrt{\frac{\pi}{A}}e^{-\frac{\xi ^2}{4A}}$$

Then, assuming $f$ is piecewise continuous, the mollification of $f$ converges as follows.

$$\begin{align*} \lim \limits_{\epsilon \rightarrow 0} \dfrac{1}{2\pi}\int\hat{f} (\xi) e^{i\xi x}e^{-\epsilon^2\xi^2/2}d\xi &= \lim \limits_{\epsilon \rightarrow 0} f \ast \eta_\epsilon (x) \\ &= \dfrac{1}{2} \big[ f(x-)+f(x+)\big] \\ &= f(x) \end{align*}$$

Furthermore, assuming $\hat{f} \in L^1$,

$$\left| e^{i\xi x}e^{-\epsilon^2|\xi|^2 /2} \hat{f}(\xi) \right| \le \left| \hat{f}(\xi) \right|$$

thus, by the Dominated Convergence Theorem,

$$\begin{align*} f(x) &= \lim \limits_{\epsilon \rightarrow 0} \dfrac{1}{2\pi}\int\hat{f} (\xi) e^{i\xi x}e^{-\epsilon^2\xi^2/2}d\xi \\ &= \dfrac{1}{2\pi}\int \lim \limits_{\epsilon \rightarrow 0} \hat{f} (\xi) e^{i\xi x}e^{-\epsilon^2\xi^2/2}d\xi \\ &= \dfrac{1}{2\pi}\int \hat{f} (\xi) e^{i\xi x}d\xi \end{align*}$$

Therefore,

$$f(x)=\dfrac{1}{2\pi}\int \hat{f} (\xi) e^{i\xi x}d\xi$$

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