I'm trying to get unstuck from some problems I encountered while studying the Fourier transform on tempered distributions. I'll discuss them using the exercise that originated them.

Let $\Lambda=\text{p.v.}(1/x)$ be the tempered distribution defined by $<\Lambda,\phi>=\int_{|x|<R}\frac{\phi(x)-\phi(0)}{x}+\int_{|x|>R}\frac{\phi(x)}{x}$, $\forall \phi \in S(\mathbb{R})$. One can show the definition doesn't depend on $R>0$.

I'm trying to understand the steps to compute $\Lambda '$, the derivative in the sense of tempered distributions of $\Lambda$.

I will write some notation provided by my book:

  • let $P$ be a polynomial, then for $\Lambda \in S'(\mathbb{R}^n)$ one can define the $P(\Lambda)$ to be the distribution such that $<P(\Lambda),\phi>:=<\Lambda,P(\phi)>$ $\forall \phi \in S(\mathbb{R})$.

  • let $f \in S(\mathbb{R}^n)$, then $f \Lambda $ is the distribution such that $<f\Lambda,\phi>:=<\Lambda,f\phi>$ $\forall \phi \in S(\mathbb{R})$

Now, one can show that $x \Lambda =1$, where $1$ stands for the tempered distribution such that $<1,\phi>=\int_{\mathbb{R}} \phi(x)$.

Applying the theorem for the (distributional) derivative of a Fourier transformed tempered function and because $\hat{1}=2 \pi \delta_0$ ($\hat{} $ denotes the transformed function, $\delta_0$ is the Dirac delta centered at $0$):

$\widehat{\Lambda}'=-i\widehat{x\Lambda}=-i\widehat{1}=-2 \pi i\delta_0$.

Let $H$ be the tempered distribution associated to the Heaviside function. Then:

$\widehat{\Lambda}'=-2 \pi i H'=(-2 \pi i H)'$

From this my book concludes $\widehat{\Lambda}=-2 \pi i H +c$ for some constant $c$ (*).

Now, according to the provided notation: $<(-2 \pi i H +c)',\phi>=<-2 \pi i H ,\phi'>=<H,-2 \pi i \phi' +c >$, which is not well defined. Is the problem in the definition of notation or is it in my understanding of the topic? Furthermore how can (*) be justified? My intution is that it is because $<c',\phi>=<c,\phi'>=\lim_{x\rightarrow +\infty} \phi(x) - \lim_{x\rightarrow -\infty} \phi(x) =0$ by definition of $S(\mathbb{R})$.

Thanks in advance!

  • $\begingroup$ Isn't it $<(-2 \pi i H +c)',\phi>=<-2 \pi i H ,\color{red}-\phi'>$ with a minus sign (in red)? $\endgroup$ – Jean Marie Dec 12 '18 at 23:12

There is a much easier characterization: since $x\Lambda=1$, $(x\Lambda)’=0$ thus $x\Lambda’=-\Lambda$.

Regardless, the first thing you want to prove is that if $T$ is a distribution such that $T’=0$, then $T$ is constant. Indeed, some Schwartz function is the derivative of a Schwartz function iff it has integral $0$ (the direct sense is basically what you wrote at the end of your post), $T$ is a linear form that vanishes where $1$ vanishes, so $T$ is a multiple of $1$ (this is general linear algebra).

On the other hand, $\langle (-2i\pi H+c)’,\,\phi\rangle=\langle -2i\pi H +c,\, \phi’\rangle \neq \langle -2i\pi H,\phi’+c\rangle$, among others because $\langle H,\,1\rangle$ is not defined.


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