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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 $<fLambda,phi>:=<Lambda,fphi>$ $forall phi in S(mathbb{R})$

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 $0$):

$widehat{Lambda}'=-iwidehat{xLambda}=-iwidehat{1}=-2 pi idelta_0$.

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 $<c',phi>=<c,phi'>=lim_{xrightarrow +infty} phi(x) - lim_{xrightarrow -infty} phi(x) =0$ by definition of $S(mathbb{R})$.

Thanks in advance!

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

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, $langle (-2ipi H+c)’,,phirangle=langle -2ipi H +c,, phi’rangle neq langle -2ipi H,phi’+crangle$, among others because $langle H,,1rangle$ is not defined.