1
$\begingroup$

From Wikipedia, I get the following fourier transform pairs:

$$\frac{d}{dx}f(x) \rightarrow i2\pi\xi \hat{f}(\xi)$$ and $$\frac{sgn(x)}{2} \rightarrow \frac{1}{i2\pi\xi}$$

When the right hand side of these equations are multiplied, I get $\hat{f}(\xi)$. Translating this to left hand side, I expect the convolution of $\frac{d}{dx}f(x)$ and $\frac{sgn(x)}{2}$ to be equal to $f(x)$ (because the RHS is $\hat{f}(\xi)$).

However, if I expand $$\frac{d}{dx}f(x) * \frac{sgn(x)}{2} = \frac{1}{2}\bigg(\int\limits_{-\infty}^{x}\frac{d}{dy}(f(y)) dy - \int\limits_{x}^{\infty}\frac{d}{dy}(f(y)) dy)\Bigg)$$ this equals $f(x) + \frac{f(-\infty)\ +\ f(\infty)}{2}$

Questions

  • Are the starting fourier transform pairs correct or is there some missing factor? OR
  • Is there an error in the steps that I have used for convolution? OR
  • Is there some property due to which the extra factor of $\frac{f(-\infty)\ +\ f(\infty)}{2}$ become equal to 0?
$\endgroup$
7
  • $\begingroup$ This answer explains why $f(-\infty)$ and $f(\infty)$ must be zero under the assumption that $f'$ is integrable. $\endgroup$
    – user169852
    Aug 8, 2016 at 18:25
  • $\begingroup$ @Bungo: I am afraid that the linked answer is not true, even though given by a high reputation user. One may imagine a function as a succession of disjoint "bumps", each one higher than the preceding one, but also much narrower, such that the added areas of these bumps should be a finite number. Such a function would be in $L^1$, but would not vanish at infinity. (Take bump "n" to be of base $[n - 4^{-n}, n + 4^{-n}]$ and height $2^n$, and extend the function by $0$ between bumps.) $\endgroup$
    – Alex M.
    Aug 8, 2016 at 18:32
  • $\begingroup$ @AlexM. But the hypothesis (explicitly in the linked question, implicitly here) is that not only $f$ but also $f'$ is in $L^1$. Moreover, since $f$ is differentiable, it is continuous. Therefore, $f(x) = \int_0^x f'(t) dt + f(0)$ for every $x$, by the fundamental theorem of calculus. Since the limit of the RHS exists as $x \to \infty$, so must the limit of the LHS exist. So this would seem to exclude your bump function as a possibility. Apologies if I'm missing something. $\endgroup$
    – user169852
    Aug 8, 2016 at 18:45
  • $\begingroup$ @Bungo: I apologize, but I do not see how one deduces the vanishing of $f$ at infinity even with the added assumption that $f'$ is in $L^1$. The answer that you link to has a proof that I find mistaken (the argument with $|f| \ge \frac L 2$ for large enough $x$) that never uses $f' \in L^1$. $\endgroup$
    – Alex M.
    Aug 8, 2016 at 18:56
  • 1
    $\begingroup$ @Bungo: Oh, yes, this is obvious. This shows that, in my example with bumps, the derivative is not integrable. Probably the suprema of $f'$ on the bumps form a sequence that, even when twisted with the length of a very narrow base ($4^{-n}$ in the case above), still yield a divergent series. I've learned something new, thank you for this discussion! $\endgroup$
    – Alex M.
    Aug 8, 2016 at 19:19

3 Answers 3

Reset to default
2
$\begingroup$

There is no mistake in your reasoning. You are just forgetting that, in order for your statements to make sense, you must specify the space of functions on which you choose to define the Fourier transform.

An often used choice is to define it on the space of rapidly decreasing functions (the "Schwartz functions"), thus turning the Fourier transform into an automorphism of this space. These have the property, among others, of vanishing at infinity, so for them $f(-\infty) = f(\infty) = 0$. The problem is that $\operatorname{sgn}$ is not such a function, tending to $1$ at $\infty$.

To cure this, another choice is to use tempered distributions. I guess that this is what you want. In this context your equality will be true.

There are several other choices (most of them including the space $L^1$), but tempered distributions are the most convenient setting.

$\endgroup$
6
  • $\begingroup$ I don't see any assumption here that $f$ is a Schwartz function. Certainly the Fourier transform can be inverted on a larger class than $\mathcal S$. The link I provided in the comment above explains why $f(\infty) = f(-\infty) = 0$ under the assumption that $f'$ is integrable. $\endgroup$
    – user169852
    Aug 8, 2016 at 18:27
  • $\begingroup$ @Bungo: Only on this space is the Fourier transform an automorphism (in fact, this is what motivated its construction and study). $\endgroup$
    – Alex M.
    Aug 8, 2016 at 18:28
  • $\begingroup$ @Bungo: I agree with you, I have changed my answer. $\endgroup$
    – Alex M.
    Aug 8, 2016 at 18:56
  • $\begingroup$ This seems reasonable and general enough (+1). The point I was making with the comments above is that a sufficient set of conditions seems to be: $f$ is differentiable (hence continuous), and both $f$ and $f'$ are in $L^1$. These are certainly less restrictive conditions than $f \in \mathcal S$. :-) and more importantly, they seem to be more or less implicit in this question. $\endgroup$
    – user169852
    Aug 8, 2016 at 19:05
  • $\begingroup$ @Bungo: While the Fourier transform may be safely used in the less restrictive hypotheses that you mention, I do not know whether nice properties of it still hold (such as $\hat {f'} = \xi \hat f$ and $\widehat {f * g} = \hat f \hat g$, that are used in the question). I could check it, but I am too lazy to do it, so I prefered to use an already created framework that is sooo convenient! $\endgroup$
    – Alex M.
    Aug 8, 2016 at 19:24
0
$\begingroup$

The transforms you have given above assume that the function $f$ decays to $0$ at $\pm \infty$, which would send your troubled term to $0$. Indeed, to show that $$ \int_{-\infty}^{\infty} f'(x) e^{-2\pi i x \xi} \,dx = 2 \pi i \xi \hat{f}(\xi) $$ you perform an integration-by-parts on the left-hand integral, moving the derivative from $f$ to the Fourier kernel. However, when you do this (or any) integration-by-parts, you need to worry about the boundary terms, which are zero in the right-hand term.

$\endgroup$
0
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\,\mathcal{#1}} \newcommand{\mrm}[1]{\,\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Notation:
1. $\ds{\mrm{f}\pars{x} = \int_{-\infty}^{\infty}\hat{\mrm{f}}\pars{k}\expo{-\ic kx} \,{\dd k \over 2\pi}\,,\qquad\hat{\mrm{f}}\pars{x} = \int_{-\infty}^{\infty}\mrm{f}\pars{x}\expo{\ic kx}\,\dd k}$.

2. $\ds{\Theta: \mbox{Heaviside Step Function}}$. $\ds{\mrm{sgn}\pars{x} = \Theta\pars{x} - \Theta\pars{-x}}$.

\begin{align} \Theta\pars{x} & = -\int_{-\infty}^{\infty}{\expo{-\ic kx} \over k + \ic 0^{+}} \,{\dd k \over 2\pi\ic} \quad\imp\quad\hat{\Theta}\pars{k} = {\ic \over k + \ic 0^{+}} \end{align}

\begin{align} \mrm{sgn}\pars{x} & = -\int_{-\infty}^{\infty}{\expo{-\ic kx} \over k + \ic 0^{+}} + \int_{-\infty}^{\infty}{\expo{\ic kx} \over k + \ic 0^{+}} \,{\dd k \over 2\pi\ic} = -\int_{-\infty}^{\infty}{\expo{-\ic kx} \over k + \ic 0^{+}} - \int_{\infty}^{-\infty}{\expo{-\ic kx} \over -k + \ic 0^{+}} \,{\dd k \over 2\pi\ic} \\[5mm] & = \int_{-\infty}^{\infty}\pars{-\,{1 \over k + \ic 0^{+}} - {1 \over k - \ic 0^{+}}}\expo{-\ic kx} \,{\dd k \over 2\pi\ic} \\[5mm] \imp\quad & \hat{\mrm{sgn}}\pars{k} = {\ic \over k + \ic 0^{+}} + {\ic \over k - \ic 0^{+}} \end{align}

$\ds{\pm\ic 0^{+}}$ are understood 'under the integral sign'. For instance: $$ \int_{-\infty}^{\infty}{\phi\pars{k} \over k + \ic 0^{+}}\,\dd k \quad\mbox{means}\quad \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty}{\phi\pars{k} \over k + \ic\epsilon}\,\dd k $$ and it serves to the purpose of a shortcut in an 'operational side'.

By the way; $$ \totald{}{x}\bracks{\mrm{f}\pars{x}\,{\mrm{sgn}\pars{x} \over 2}} = \half\,\mrm{f}'\pars{x}\mrm{sgn}\pars{x} + \mrm{f}\pars{0}\delta\pars{x} $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .