# Why does a convergent sequence of test functions have to be supported in a single compact set?

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?

• 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. Commented Oct 25, 2016 at 22:16
• @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 $K_n$. Commented Oct 25, 2016 at 22:21
• 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. Commented Oct 25, 2016 at 22:31
• 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 Commented Oct 25, 2016 at 22:36
• en.wikipedia.org/wiki/… Commented Oct 25, 2016 at 22:50

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.
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 $$\lim_{k \to \infty} \int_{\mathbb{R}} \phi_k(x) dx = 1 \neq 0 = \int_{\mathbb{R}} \lim_{k \to \infty} \phi_k(x) dx.$$