In my book I have seen that if I have a disitribution $T$ in $D'$ (continuous linear functional on the space $D$) which can be defined for every $C^{\infty}$ function
(no necessarily compact supported ) then $T$ must have compact support since otherwise we can construct $\phi\in C^{\infty}$ such that $\langle T,\phi\rangle$ is not finite.
Does anyone have idea how to construct such function?
thank you
Best Answer
Suppose that the support of $T$ is not compact. Then this support intersects with infinite number of intervals of the form $[n,n+1]$, $n\in\Bbb Z$. Denote the indices of intervals that intersect with the support of $T$ by $n_k$, $k\in \Bbb Z$.
Now, $\forall k\in \Bbb Z$ there exist a test function $\phi_k$ with support in $[n_k,n_k+1]$ such that $(T,\phi_k)=1$ (this quickly follows from our definition of $n_k$ and from linearity of $T$).
The function $\Phi(x)=\sum_{k\in\Bbb Z}\phi_k(x)$ is $C^\infty(\Bbb R)$. Yet $$(T,\Phi) = \sum_{k\in\Bbb Z}(T,\phi_k(x))=\sum_{k\in\Bbb Z}1=+\infty.$$