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Context. Let $\Omega \subset \mathbb R^n$ denote an arbitrary non-empty open set and let $C_0^\infty(\Omega)$ denote the usual space of infinitely differentiable functions that have compact support in $\Omega$. We say that a sequence $(\varphi_j)_{j \in \mathbb N} \subset C_0 ^\infty(\Omega)$ converges to $\varphi \in C_0^\infty(\Omega)$ if there exists a compact set $K \subset \Omega$ such that

$$ \text{supp } \varphi_j \subset K, \text{ for all } j \in \mathbb N $$

and

$$ D^\alpha \varphi_j \to D^\alpha \varphi \text{ uniformly on } K \text{ as } j \to \infty, \text{ for all } \alpha \in \mathbb N_0^n. $$

Here, $D^\alpha$ stands for the usual differentiation operator (I shall not get into much details about this since I believe it won't be needed).

Question. Given a function $f \in L^1_{\operatorname{loc}}(\Omega)$ (the usual space of locally integrable functions), define the operator $T_f : C_0^\infty(\Omega) \to \mathbb C$ by

$$ \langle T_f, \varphi \rangle := \int_{\Omega} f(x) \, \varphi(x) \, dx. $$

Show that $T_f$ is continuous from $C_0^\infty(\Omega)$ to $\mathbb C$.

My attempt. Consider a sequence $(\varphi_j)_{j \in \mathbb N}$ that converges to $\varphi$ in $C_0^\infty(\Omega)$. Our goal is to show that the sequence $(\langle T_f, \varphi_j \rangle)_{j \in \mathbb N}$ converges to $\langle T_f, \varphi \rangle$ in $\mathbb C$. We begin by noting that, due to our hypothesis, there exists a compact set $K \subset \Omega$ such that $$ \text{supp } \varphi_j \subset K, \text{for all } j \in \mathbb N. $$ Therefore, it is clear that $\varphi_j(x) = 0$, for all $x \notin K$ and $j \in \mathbb N$. Hence, we have that

$$ \langle T_f, \varphi_j \rangle = \int_{\Omega} f(x) \varphi_j(x) \, dx = \int_K f(x) \varphi_j(x) \, dx = \int_\Omega f(x) \varphi_j(x) \chi_K(x) \, dx, $$ for all $j \in \mathbb N$. Now, since each $\varphi_j$ belongs to $C_0^\infty(\Omega)$, each $\varphi_j$ is also bounded, which allows us to define

$$ M = \sup_{\substack{x \in \Omega \\[.1cm] j \in \mathbb N}} |\varphi_j(x)| < \infty. $$

Therefore, it follows that

$$ |f(x) \varphi_j(x) \chi_K(x)| \leqslant M |f(x) \chi_K(x)|, \text{ for all } j \in \mathbb N \text{ and } x \in \Omega. $$

Noting that the latter function is integrable (in fact, f is locally integrable and $K$ is a compact set), this means that we can apply the Lebesgue Dominated Convergence Theorem. Doing so, it follows that

$$ \lim_{j \to \infty} \langle T_f, \varphi_j \rangle = \lim_{j \to \infty} \int_\Omega f(x) \varphi_j(x) \chi_K(x) \, dx = \int_K f(x) \lim_{j \to \infty} \varphi_j(x) \, dx. $$

Now, again using our hypothesis, we know that $(\varphi_j)_{j \in \mathbb N}$ converges uniformly to $\varphi$ on $K$, as $j \to \infty$. Therefore, it is clear that $(\varphi_j)_{j \in \mathbb N}$ also converges pointwise almost everywhere on $K$ to $\varphi$, as $j \to \infty$. Hence, resuming our previous reasoning, we have

$$ \lim_{j \to \infty} \langle T_f , \varphi_j \rangle = \int_K f(x) \lim_{j \to \infty} \varphi_j(x) \, dx = \int_K f(x) \varphi(x) \, dx. $$

This is almost what I want to reach, but I should now justify the equality

$$ \color{red}{\int_K f(x) \varphi(x) \, dx = \int_\Omega f(x) \varphi(x) \, dx,} $$

which is something I don't know how to do. Intuitively, this tells me that I should prove that the support of $\varphi$ is also contained in $K$ somehow, but I don't know how to reach this conclusion.

Thanks for any help in advance.

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2 Answers 2

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I think your definition is wrong. It should include that the support of $\varphi$ is also contained in $K$. Otherwise you could change $\varphi$ outside of $K$ as you like and the limit would also be nonunique.

Edit: The definition from your text does not make any sense. Take the zero sequence $\varphi_j \equiv 0$. Since the support is empty, we can choose $K = \emptyset$. Now, your requirement on uniform convergence on $K$ is satisfied for every $\varphi$, since $K$ is empty. Consequently, the zero sequence converges to arbitrary functions.

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  • $\begingroup$ Do you have any reference for this definition you proposed? Because the text I am following doesn't say anything about the support of $\varphi$. $\endgroup$ Commented Nov 13, 2024 at 17:38
  • $\begingroup$ I have looked into the (German) functional analysis book by Dobrowolski. But this should be in any text on distributions. Your text seems to be blatantly wrong, see edit. $\endgroup$ Commented Nov 14, 2024 at 7:23
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The notation $C_0^{\infty}(\Omega)$ is no good, because it suggests smooth functions which go to zero as one approaches the boundary of $\Omega$, and may not necessarily have compact support. Better is $C_{\rm c}^{\infty}(\Omega)$, but I will use Laurent Schwartz's notation $\mathscr{D}(\Omega)$ instead.

For $K$ a compact subset of $\Omega$, I will write $C^{\infty}_{K}(\Omega)$ for the subspace of $\mathscr{D}(\Omega)$ made of test functions $\varphi$ such that ${\rm supp}(\varphi)\subseteq K$. Let also $\iota_K:C^{\infty}_{K}(\Omega)\rightarrow\mathscr{D}(\Omega)$ be the inclusion map, i.e., such that $\iota_K(\varphi)=\varphi$ for all $\varphi\in C^{\infty}_{K}(\Omega)$.

Now, define the map $\rho:\mathscr{D}(\Omega)\rightarrow [0,\infty)$ given by $$ \rho(\varphi):=|\langle T_f,\varphi\rangle|=\left| \int_{\Omega} f(x)\varphi(x)\ d^nx \right|\ . $$ This map is well defined and it is a seminorm. Namely it satisfies $$ \rho(\varphi_1+\varphi_2)\le \rho(\varphi_1)+\rho(\varphi_2) $$ and $$ \rho(\lambda\varphi)=|\lambda|\rho(\varphi)\ . $$

For $K$ a compact subset, $\rho\circ\iota_K$ is a seminorm on $C^{\infty}_{K}(\Omega)$. Moreover, it is a continuous seminorm on that space, because for any $\varphi\in C^{\infty}_{K}(\Omega)$, $$ \rho(\varphi)=\left| \int_{\Omega} f(x)\varphi(x)\ d^nx \right|\le C \sup_{x\in K}|\varphi(x)| $$ with $$ C:=\int_K |f|<\infty\ . $$ The last finiteness property easily follows from local integrability of $f$, and the use of a finite open cover of $K$.

The above argument shows that $\rho$ is an admissible seminorm on $\mathscr{D}(\Omega)$, i.e., for every compact subset $K$ of $\Omega$, it restricts to a continuous seminorm on $C^{\infty}_{K}(\Omega)$. By definition of the topology of $\mathscr{D}(\Omega)$, this means that $\rho$ is a continuous seminorm on $\mathscr{D}(\Omega)$. Since the linear map $T_f$ is such that $$ |\langle T_f,\varphi\rangle|\le\rho(\varphi) $$ for all $\varphi\in\mathscr{D}(\Omega)$, this map is continuous and preserves convergence of sequences.

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