We fix the space $\mathcal{D}=\mathcal{C}^\infty_0(\mathbb{R}^n)$ as space of testfunctions. Let $(f_n)$ be a sequence of distributions with $\lim_{n\to\infty} f_n(\varphi)$ existing for all $\varphi\in\mathcal{D}$. Define $f(\varphi):=\lim_{n\to\infty} f_n(\varphi)$. Then the claim is that $f$ is also a distribution. My question: For proving that this holds, how to see that $f$ is continuous w.r.t convergence in $\mathcal{D}$? (We define a sequence $(\varphi_n)$ in $\mathcal{D}$ to converge to some $\varphi\in\mathcal{D}$ iff there is some compact set $K\subset\mathbb{R}^n$, s.t. $\mathrm{supp}(\varphi_n)\subset K$ for each $n\in\mathbb{N}$ and $\|D^k\varphi_n-D^k\varphi\|_\mathrm{sup}\to 0$ as $n\to\infty$ for each $k\in\mathbb{N}_0^n$.) Thanks!


For $K \subset \mathbb{R}^n$ compact, consider the space

$$\mathcal{D}_K := \{ \varphi \in \mathcal{D} : \operatorname{supp} \varphi \subset K\}$$

endowed with the seminorms

$$\lVert\varphi\rVert_k = \sup \{ \lvert D^k\varphi(x)\rvert : x \in \mathbb{R}^n\}$$

for $k \in \mathbb{N}^n$.

It is straightforward to show that $\mathcal{D}_K$ is then a Fréchet space, and for compact $K \subset L$, $\mathcal{D}_K$ is a closed subspace of $\mathcal{D}_L$. Denote the inclusion maps by $j^K_L \colon \mathcal{D}_K \to \mathcal{D}_L$ and $\iota^K \colon \mathcal{D}_K \to \mathcal{D}$.

As sets (or as vector spaces without considering the topology), we have

$$\mathcal{D} = \bigcup_{\substack{K\subset\mathbb{R}^n \\ K \text{compact}}} \mathcal{D}_K,$$

or for any sequence $K_m$ of compact subsets of $\mathbb{R}^n$ such that $K_m \subset \overset{\Large\circ}{K}_{m+1}$ for all $m$ and $\bigcup\limits_{m=0}^\infty K_m = \mathbb{R}^n$

$$\mathcal{D} = \bigcup_{m=0}^\infty \mathcal{D}_{K_m}.$$

The topology one endows $\mathcal{D}$ with - the locally convex inductive limit topology with respect to the inclusions $\iota^K$ - is however strictly finer than the topology induced by the family $\{ \lVert\cdot\rVert_k : k \in \mathbb{N}^n\}$ of seminorms inducing the topology on the $\mathcal{D}_K$.

But, and this is the crucial property, a linear map $T\colon \mathcal{D} \to E$, where $E$ is a locally convex topological vector space, is continuous if and only if $T\circ \iota^K \colon \mathcal{D}_K \to E$ is continuous for all compact $K \subset \mathbb{R}^n$.

If, as seems likely, the topology on $\mathcal{D}$ is not explicitly introduced, and the continuity property of distributions is defined in terms of convergent sequences, we have the same fact, since the convergence of a sequence $\bigl(\varphi_m\bigr)_{m\in\mathbb{N}}$ to $\varphi$ in $\mathcal{D}$ is defined as

  1. There is a compact $K\subset \mathbb{R}^n$ such that $\varphi_m \in \mathcal{D}_K$ for all $m\in \mathbb{N}$ and $\varphi \in \mathcal{D}_K$, and
  2. $\varphi_m$ converges to $\varphi$ in the space $\mathcal{D}_K$.

Thus, to see that the pointwise limit $f$ of the sequence $\bigl(f_m\bigr)_{m \in \mathbb{N}}$ of distributions is again a distribution, we need to see that $f \circ \iota^K$ is continuous for every compact $K\subset \mathbb{R}^n$.

Fix an arbitrary compact $K\subset \mathbb{R}^n$, and write $g = f\circ \iota^K$ and $g_m = f_m \circ \iota^K$ for $m\in \mathbb{N}$. By the above, we have $g_m \in \mathcal{D}_K'$, i.e. $g_m$ is a continuous linear functional on $\mathcal{D}_K$, and $g$ is the pointwise limit of the $g_m$,

$$\bigl(\forall \varphi \in \mathcal{D}_K\bigr)\Bigl(\lim_{m\to\infty} g_m(\varphi) = g(\varphi)\Bigr).$$

Now, $\mathcal{D}_K$ is a Fréchet space, and hence the pointwise limit of a sequence of continuous linear functionals on $\mathcal{D}_K$ is again a continuous linear functional. That is, in greater generality, theorem 2.8. in Rudin's "Functional Analysis", a consequence of the Banach-Steinhaus theorem (theorem 2.6. ibid). Nevertheless, let us add a proof here:

For $m \in \mathbb{N}$, let $A_m = \{\varphi \in \mathcal{D}_K : \lvert g_m(\varphi)\rvert \leqslant 1\}$. Since $g_m$ is continuous, $A_m$ is a closed neighbourhood of $0$ in $\mathcal{D}_K$, and since $g_m$ is linear, $A_m$ is convex and balanced. Therefore

$$A := \bigcap_{m=0}^\infty A_m$$

is a closed convex and balanced subset of $\mathcal{D}_K$. Since convergent sequences are bounded, $A$ is absorbing, that is, for every $\varphi \in \mathcal{D}_K$ there is a $\delta > 0$ such that $\lambda\cdot\varphi \in A$ for all scalars $\lambda$ with $\lvert \lambda\rvert \leqslant \delta$. In short, $A$ is a barrel (a closed, convex, balanced, and absorbing set) in $\mathcal{D}_K$. By Baire's theorem, and since $\varphi \mapsto \lambda\cdot \varphi$ is a homeomorphism for every fixed $\lambda \neq 0$, $A$ has nonempty interior ($\mathcal{D}_K = \bigcup_{r=1}^\infty r\cdot A$ because $A$ is absorbing), and since $A$ is convex and balanced, it follows that $A$ is a neighbourhood of $0$ in $\mathcal{D}_K$. (The last shows that Fréchet spaces are barrelled spaces, locally convex spaces in which each barrel is a neighbourhood of $0$. The argument shows that the conclusion that a pointwise limit of a sequence of continuous linear maps is again continuous holds for barrelled spaces.) Since $g$ is the pointwise limit of the $g_m$, it follows that

$$A \subset g^{-1}\bigl(\{ z : \lvert z\rvert \leqslant 1\}\bigr),$$

so $g^{-1}\bigl(\{ z : \lvert z\rvert \leqslant 1\}\bigr)$ is a neighbourhood of $0$, and hence $g$ is continuous.

So, as needed, $f\circ \iota^K$ is continuous for every compact $K\subset \mathbb{R}^n$, and that is, as mentioned above, equivalent to $f$ being continuous, i.e. $f$ being a distribution.


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