Find the distribution $T$ defined in $\mathcal S'(\mathbb R)$

Question

Let $\mathcal S'$be the space of all tempered distributions.

Find a distribution $T \in\mathcal S'(\mathbb R)$ such that $xT = \sin^2(x) $, and moreover $T(\phi) = \pi$ for $\phi(x) = \exp(−x^2)$.

Answer 1

The concept of the product of distribution and a smooth function is a well-established one in the theory of distributions. One can regard $\frac 1 x$ as a distribution (the simplest way is to use the fact that the function $\log |x|$ is locally integrable and so a distribution). We then define $\frac 1 x$ simply to be the distributional derivative of the latter---this is equivalent to the alternative one using principal values. We now define the distribution $T_0$ to be the product of this distribution with the smooth function $\sin^2 x$. This takes care of the first condition. For the second one, we use the known fact that any solution $T$ of the above equation has the form $T_0 + c \delta_0$ for some constant $c$. The latter can be computed using the second condition (note that the function $\frac 1x \exp(-x^2) \sin^2 x$ is odd and so its integral over the reals is $0$). All the above facts about distributions can be found in the usual texts (e.g., the monograph of L. Schwartz for the approach using duality theory and functional analysis or, more user-friendly for those not comfortable with this level of abstraction, in "Introduction to the theory of distributions" by Campos Ferreira which is based on the approach of J. Sebastiao e Silva and uses methods which are at the level of a one-variable analysis course).