📂Fourier Analysis

Inverse Fourier Transform Theorem for Smooth Functions

Theorem

Assuming that $f$ is integrable on $\mathbb{R}$ and piecewise smooth, then the following equation holds:

$$\lim \limits_{r\to \infty} \frac{1}{2\pi} \int_{-r}^{r}e^{i\xi x} \hat{f}(\xi)d\xi= \frac{1}{2}\big[f(x-)+f(x+) \big],\quad \forall x\in \mathbb{R}$$

Here, $f(x+)$ and $f(x-)$ are respectively the right-hand limit and left-hand limit of $f$ at $x$.

Description

The inverse Fourier transform theorem used a cutoff function instead of requiring a relatively weak condition for $f$. The theorem above is another form of the inverse Fourier transform theorem. The resulting equation is as powerful as the condition is strong. The inverse Fourier transform theorem only required that $f$ be continuous at best, but here, it even demands smoothness. It could be considered that the following cutoff function was used.

$$\eta (x)=\begin{cases} 1&-r \le x \le r \\ 0 & \text{otherwise} \end{cases}$$

Proof

First, let’s calculate the following equation:

$$\begin{align*} \int_{-r}^{r}e^{i\xi (x-y)}d\xi &= \left. \frac{e^{i\xi (x-y)}}{i(x-y)}\right]_{\xi=-r}^{r} \\ &= \frac{e^{ir(x-y)}-e^{-ir(x-y)}}{i(x-y)} \\ &=\frac{2i \sin r(x-y)}{i(x-y)}=\frac{2\sin r(x-y)}{(x-y)} \end{align*}$$

Using this, we obtain the following equation:

$$\begin{align*} \frac{1}{2\pi}\int_{-r}^{r}e^{i\xi x}\hat{f}(\xi)d\xi &= \frac{1}{2\pi}\int_{-r}^{r} e^{i\xi x}\int _{-\infty} ^{\infty}f(y)e^{-i\xi y}dyd\xi \\ &= \frac{1}{2\pi}\int _{-\infty} ^{\infty}\int_{-r}^{r} e^{i\xi (x-y)}d\xi f(y)dy \\ &= \frac{1}{\pi}\int _{-\infty} ^{\infty}\frac{\sin r(x-y)}{(x-y)}f(y)dy \end{align*}$$

Substituting with $x-y=y$, we get:

$$\begin{align} \frac{1}{\pi}\int _{-\infty} ^{\infty}\frac{\sin r(x-y)}{(x-y)}f(y)dy &= -\frac{1}{\pi}\int _{\infty} ^{-\infty}\frac{\sin (ry)}{y}f(x-y)dy \nonumber \\ &= \frac{1}{\pi}\int _{-\infty} ^{\infty}\frac{\sin (ry)}{y}f(x-y)dy \label{eq1} \end{align}$$

It is well known that the ideal integral of the sinc function is as follows:

$$\begin{equation} \int_{-\infty}^{0} \frac{\sin x}{x}dx=\int_{-\infty}^{0} \frac{\sin (ax)}{x}dx=\frac{\pi}{2}=\int_{0}^{\infty} \frac{\sin x}{x}dx=\int_{0}^{\infty} \frac{\sin (ax)}{x}dx \label{eq2} \end{equation}$$

Using the two equations $(1)$ and $(2)$, we can obtain the following equation:

$$\begin{align} &\frac{1}{2\pi}\int_{-r}^{r}e^{i\xi x}\hat{f}(\xi)d\xi-\frac{1}{2}\big[ f(x+)+f(x-)\big] \nonumber \\ =&\ \frac{1}{\pi}\int _{-\infty} ^{\infty}\frac{\sin (ry)}{y}f(x-y)dy -\frac{1}{\pi}\int_{-\infty}^{0}\frac{\sin (ry)}{y}dyf(x+)-\frac{1}{\pi}\int_{0}^{\infty}\frac{\sin (ry)}{y}dyf(x-) \nonumber \\ =&\ \frac{1}{\pi}\int _{-\infty} ^{0}\frac{\sin (ry)}{y}\big[f(x-y)-f(x+) \big]dy +\frac{1}{\pi}\int _{0} ^{\infty}\frac{\sin (ry)}{y}\big[f(x-y)-f(x-) \big]dy \label{eq3} \end{align}$$

Since the solution method is the same, it suffices to consider only the second term of the last line of the above equation. Let’s divide the integration interval as follows for $K\ge 1$.

$$\begin{align} &\frac{1}{\pi}\int _{0} ^{\infty}\frac{\sin (ry)}{y}\big[f(x-y)-f(x-) \big]dy \nonumber \\ = &\frac{1}{\pi}\int _{0} ^{K}\frac{\sin (ry)}{y}\big[f(x-y)-f(x-) \big]dy+\frac{1}{\pi}\int _{K} ^{\infty}\frac{\sin (ry)}{y}\big[f(x-y)-f(x-) \big]dy \label{eq4} \end{align}$$

Let’s look at the second term of $(4)$ in two parts. Since $x \ge 1$ with respect to $\left| \frac{\sin x}{x} \right| \le 1$, we obtain the following equation:

$$\begin{align*} \left| \frac{1}{\pi}\int _{K} ^{\infty}\frac{\sin (ry)}{y}f(x-y)dy\right| &\le \frac{1}{\pi}\int _{K} ^{\infty}\left|\frac{\sin (ry)}{y}f(x-y)\right|dy \\ &\le \frac{1}{\pi}\int _{K} ^{\infty}\left|f(x-y)\right|dy \end{align*}$$

Since by assumption $f$ is integrable, $\lim \limits_{x \to \infty}f(x)$ does not diverge. Thus, the above integration approaches $0$ when $K \to \infty$. The remaining part can be calculated as follows:

$$\begin{align*} \frac{1}{\pi}\int _{K} ^{\infty}\frac{\sin (ry)}{y}f(x-)dy &= \frac{f(x-)}{\pi}\int _{K} ^{\infty}\frac{\sin (ry)}{y}dy \\ &= \frac{f(x-)}{\pi}\int _{rK} ^{\infty}\frac{\sin (y)}{y}dy \end{align*}$$

Since $\frac{\sin x}{x}$ is also integrable, similar to above, irrespective of the value of $r$, the integration value approaches $0$ when $K \to \infty$. Therefore, by assuming $K$ is adequately large, we can make the second term of $(4)$ as small as we want. Now, it’s time to calculate the first term of $(3)$, but first, let’s assume the function $g$ as follows.

$$g(y) =\begin{cases} \frac{f(x-y)-f(x-)}{y}, &0<y<K \\ 0, & \text{otherwise}\end{cases}$$

Then, we obtain the following equation:

$$\begin{align*} \int \frac{e^{iry}-e^{-iry}}{2i}g(y)dy &=\frac{1}{2i}\left[\int g(y)e^{iry}dy-\int g(y)e^{-iry}dy \right] \\ &= \frac{1}{2i}\big[ \hat{g}(r)-\hat{g}(-r) \big] \end{align*}$$

Assuming $f$ is a smooth function, $g$ is smooth everywhere except at $y=0$. And the function value $g(y)$ approaches $f^{\prime}(x-)$ as $y$ decreases close to $0$, so $g$ is bounded at $[0,K]$ and integrable. Therefore, by the Riemann-Lebesgue lemma, $\lim \limits_{r\to\infty}\hat{g}(\pm r)=0$. Combining all, we obtain the following equation for the second term of $(3)$.

$$\lim \limits_{r\to \infty}\frac{1}{\pi}\int _{0} ^{\infty}\frac{\sin (ry)}{y}\big[f(x-y)-f(x-) \big]dy =0$$

Similarly, for the first term of $(3)$, we obtain the following equation:

$$\lim \limits_{r\to \infty}\frac{1}{\pi}\int _{-\infty} ^{0}\frac{\sin (ry)}{y}\big[f(x-y)-f(x-) \big]dy =0$$

Therefore,

$$\lim \limits_{r \to \infty} \frac{1}{2\pi}\int_{-r}^{r}e^{i\xi x}\hat{f}(\xi)d\xi-\frac{1}{2}\big[ f(x+)+f(x-)\big]=0$$

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