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I've often seen it repeated that for any convergent sequence of test functions $\phi_i$ in $C_0^\infty(\Omega)$, there must exist a compact set $K$ such that for all $i$, the support of $\phi_i$ is in $K$. I'm having trouble proving this, and in fact it seems false to me.

Let $K_n$ be an increasing sequence of compact sets whose union is $\Omega$, then define $\phi_i$ to be some smooth function which is zero on $K_i$, but has a little bump of height $1$ somewhere in $\Omega\backslash K_i$. Does this sequence not converge to $0$ in the test function topology?

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    $\begingroup$ That the test functions must be supported in the same compact set is the definition of the test function (pseudo)topology. You can certainly come up with a sequence of test functions that fails this condition. So, I am not sure what it is you are trying to disprove or prove. $\endgroup$
    – user8960
    Commented Oct 25, 2016 at 22:16
  • $\begingroup$ @user8960 I thought the test function topology was the one induced by the semi-norms $||\cdot||_{K_n, N}$, the supremum of the at-most-N'th order partial derivatives of $\phi$ on the compact $K_n$. $\endgroup$
    – Jack M
    Commented Oct 25, 2016 at 22:21
  • $\begingroup$ You can drop this condition whenever you consider a compactly supported (set of) distributions. But if your distribution isn't, it can grow very fast as $x \to \infty$, so dropping the compact condition will be obviously a problem when trying to bound $\langle T,\varphi_n \rangle$. And note that in the context of tempered distributions, we replace the compact set condition by a growth rate condition and everything works well. $\endgroup$
    – reuns
    Commented Oct 25, 2016 at 22:31
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    $\begingroup$ Start by proving that for any sequence of test functions such that $\bigcup_n supp(\varphi_n) = \mathbb{R}$ there exists a distribution $T$ such that $\langle T,\varphi_n \rangle $ diverges $\endgroup$
    – reuns
    Commented Oct 25, 2016 at 22:36
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – user8960
    Commented Oct 25, 2016 at 22:50

2 Answers 2

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The topology induced by the seminorms $|| . ||_{K_n,N}$ is the topology f uniform convergence on compact sets (with all its derivatives). The "commonly-used" topology on the space f test functions is strictly finer.

A distribution is continuous with respect to the topology induced by the seminorms if (and only if!) the distribution has compact support.

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Take $\Omega := \mathbb{R}$ and $$ \phi_k(x) := \begin{cases} C_k \cdot \exp\left(\frac{1}{x^2 - k^2}\right), & \text{if } |x|< k, \\ 0, &\text{elsewhere,} \end{cases} \qquad \text{where } \frac{1}{C_k} := \int_{-k}^{k} \phi_k(x) dx. $$ Then, $\phi_k \in \mathcal{C}_{\text{c}}^{\infty}(\mathbb{R})$ for all $k \in \mathbb{N}$ and $\text{supp}(\phi_k) \subset [-k, k]$, but we can't find a compact interval $K$ so that $\text{supp}(\phi_k) \subset K$ for all $k \in \mathbb{N}$. And we observe $$ \lim_{k \to \infty} \int_{\mathbb{R}} \phi_k(x) dx = 1 \neq 0 = \int_{\mathbb{R}} \lim_{k \to \infty} \phi_k(x) dx. $$

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