# Equating distributional derivatives

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 $$<\Lambda,\phi>=\int_{|x|R}\frac{\phi(x)}{x}$$, $$\forall \phi \in S(\mathbb{R})$$. 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 $$:=<\Lambda,P(\phi)>$$ $$\forall \phi \in S(\mathbb{R})$$.

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

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: $$<(-2 \pi i H +c)',\phi>=<-2 \pi i H ,\phi'>=$$, 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 $$==\lim_{x\rightarrow +\infty} \phi(x) - \lim_{x\rightarrow -\infty} \phi(x) =0$$ by definition of $$S(\mathbb{R})$$.

• Isn't it $<(-2 \pi i H +c)',\phi>=<-2 \pi i H ,\color{red}-\phi'>$ with a minus sign (in red)? – 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.