Multiplication in $\mathcal D'(R)$.

Question

I was reading in my text book that multiplication of elements of $\mathcal D'(R)$ is not a continuous function although it is defined and can be seen from the convergence of $\sin(nx)$ to $0$ as $n$ goes to infinity in $\mathcal D'(R)$. And not in the case of $\sin^2(nx)$ as $n$ goes to infinity. Please explain me what does he mean. Thank you in advance.

Answer 1

In the sense of distributions, $\sin nx \to 0$ as $n\to\infty$, since

$$ \lim_{n\to\infty}\int_{-\infty}^\infty \phi(x)\,\sin nx\,dx = 0$$

for every test function $\phi$. (This follows for example from Riemann-Lebesgue's lemma.)

On the other hand $\sin^2 nx = \frac12 - \frac12\cos 2nx$ so $$ \lim_{n\to\infty}\int_{-\infty}^\infty \phi(x)\,\sin^2 nx\,dx = \frac12 \int_{-\infty}^\infty \phi(x)\,dx,$$ which means that $\sin^2 nx$ tends to $1/2$ in the sense of distributions as $n \to \infty$.

Hence, the operation $f \mapsto f^2$ can't be continuous on $\mathcal{D}'$. If it were, it would map a sequence of distributions tending to $0$ to another sequence of distributions tending to $0$.

(Multiplication of two arbitrary distributions can't even be defined in general: what would $\delta^2$ be?)