Derivative of Compact Supported Distribution is Compactly Supported

Question

If you have a smooth, time-dependent distribution $u$ that maps the real numbers to compactly-supported distributions, then must its derivative map reals to compactly-supported distributions? By derivative, I refer to $$\left\langle u'(t),\phi\right\rangle=\lim_{h\to 0}\left\langle\frac{u(t+h)-u(t)}{h},\phi\right\rangle.$$

Answer 1

Let $u(t) = t \chi_{(-1/t, 1/t)}$ for $t\neq 0,$ and $u(0)=0$. Then for $\varphi \in C_c^\infty(\mathbb{R})$ we have $$ \langle u'(t), \varphi \rangle = \int_{-1/t}^{1/t} \varphi(x) \, dx - \frac{1}{t}(\varphi(1/t) - \varphi(-1/t)). $$ For $t$ small enough both $\varphi(1/t)$ and $\varphi(-1/t)$ will vanish since $1/t$ and $-1/t$ will be outside of the support of $\varphi.$ Thus $$\langle u'(0), \varphi \rangle = \int_{-\infty}^{\infty} \varphi(x) \, dx = \langle 1, \varphi \rangle,$$ i.e. $u'(0) \not\in \mathcal{E}'(\mathbb{R}).$

Answer 2

With $T_h u$ the distribution defined by $\langle T_h u, \phi\rangle = \langle u,\phi(.-h)\rangle$ It is obvious that $\frac{T_h u-u}{h}$ is compactly supported.

The only point is to show that $\lim_{h \to 0}\frac{T_h u-u}{h}$ converges in the sense of distributions, which is quite obvious too : $$\lim_{h \to 0}\langle \frac{T_h u-u}{h},\phi\rangle=\lim_{h \to 0}\langle u,\frac{\phi(.-h) -\phi}{h}\rangle=\langle u,-\phi'\rangle$$ where the last step follows from $\frac{\phi(.-h) -\phi}{h}\to -\phi'$ in the test function topology and the definition of distributions as linear functionals continuous for the test function topology.