Find derivative of $\lfloor{x}\rfloor$ in distribution

Question

Find derivative in distribution of $f(x)=\lfloor{x}\rfloor=E(x)$

$$E(x)≤x≤E(x+1)$$

• Answer is :

$$\lfloor{x}\rfloor '=\displaystyle\sum_{k=-\infty}^{\infty}\delta_{k}$$

I don't have any idea about how to.

Can you assist?

I'm too thankful

Answer 1

Let $f(x)=\lfloor x\rfloor$. For any test function $\phi(x)$, we have

$$\begin{align} \langle f',\phi\rangle&=-\langle f,\phi'\rangle\\\\ &=-\int_{-\infty}^\infty \lfloor x\rfloor \phi'(x)\,dx\\\\ &=-\sum_{n=-\infty}^\infty \int_{n}^{n+1} n\phi'(x)\,dx\\\\ &=-\sum_{n=-\infty}^\infty n (\phi(n+1)-\phi(n))\\\\ &=-\sum_{n=-\infty}^\infty n \phi(n+1)+\sum_{n=-\infty}^\infty n\phi(n)\\\\ &-\sum_{n=-\infty}^\infty (n-1) \phi(n)+\sum_{n=-\infty}^\infty n\phi(n)\\\\ &=\sum_{n=-\infty}^\infty \phi(n)\\\\ &=\sum_{n=-\infty}^\infty \langle \delta_n ,\phi\rangle\\\\ &= \langle \sum_{n=-\infty}^\infty\delta_n ,\phi\rangle \end{align}$$

Hence, in distribution $$\lfloor x\rfloor' = \sum_{n=-\infty}^\infty \delta(x-n)$$as was to be shown!