I am looking at the following theorem:

  • $C_C^{\infty}(\mathbb{R}^n) \subset S(\mathbb{R}^n)$ and the embedding is continuous.
  • $C_C^{\infty}(\mathbb{R}^n)$ is dense in $S(\mathbb{R}^n)$
  • $S(\mathbb{R}^n) \subset L_1(\mathbb{R}^n)$ and the embedding is continuous.

Note that a function $\phi \in C^{\infty}(\mathbb{R}^n)$ for which $||\phi||_{\alpha, \beta}=\sup_x |x^{\alpha} D^{\beta} \phi|<+\infty \ \ \ \forall \alpha, \beta$ belongs to $S(\mathbb{R}^n)$.

  • Let $\phi \in C_C^{\infty}(\mathbb{R}^n)$. We have that $supp (D^{\beta} \phi) \subset supp(\phi)$ since $D^{\beta} \phi \neq 0 \Rightarrow \phi \neq 0$, right? From this we have that $D^{\beta}\phi \in C_C^{\infty}(\mathbb{R}^n)$, thus also $x^{\alpha} D^{\beta} \phi \in C_C^{\infty}(\mathbb{R}^n), \forall \alpha$. And so we deduce that $\sup_x |x^{\alpha} D^{\beta} \phi|<+\infty \Rightarrow \phi \in S(\mathbb{R}^n)$.

    Right? But how can we show that the embedding is continuous?

  • Let $\phi \in S(\mathbb{R}^n)$. Let $\rho(x)$ such that $\int_{\mathbb{R}^n} \rho(x) dx=1, \ \ \ supp{\rho } \subset \{ ||x|| \leq 1\}$ . Consider $\phi_j(x)=\rho{\left( \frac{x}{j}\right)} \phi(x) \in C_C^{\infty}(\mathbb{R}^n)$.

    So it suffices to show that $\phi_j(x) \to\phi(x)$. How could we show this?

    • Let $\phi \in S(\mathbb{R}^n)$. Then $ \sup_x |x^{\alpha} D^{\beta}\phi |<+\infty$ . For $\alpha=0, \beta=0$ we get that $\sup_x|\phi|<+\infty$. So $\int |\phi(x)| dx<+\infty$ and so $\phi \in L_1(\mathbb{R}^n)$. Is this right? How can we show that the embedding is continuous?

2 Answers 2


While $C_c^\infty(\mathbb R^n)$ fails to be first countable, it is still true that for any linear map $T: C^\infty_c(\mathbb R^n) \to X$ where $X$ is a topological vector space whose topology is induced by a family seminorms (i.e., a locally convex TVS), we have that $T$ is continuous iff $T$ is sequentially continuous. It's immediate that $C^\infty_c(\mathbb R^n) \subset S(\mathbb R^n)$; to prove that the embedding is continuous to need to prove that the inclusion map $\iota: C^\infty_c(\mathbb R^n)\to S(\mathbb R^n)$ is continuous. I brought up the first point, because it is easiest to prove the inclusion map is sequentially continuous. To do this, take a sequence $\{f_k\}$ in $C^\infty_c(\mathbb R^n)$ converging to $f \in C^\infty_c(\mathbb R^n)$ and prove that this implies that $f_k \to f$ in $S(\mathbb R^n)$. Depending on how you define the topology on $C^\infty_c(\mathbb R^n)$, this is actually fairly trivial.

On the second point, you're on the right track. I'm not sure it is important for $\rho$ to integrate to $1$. Instead, I would take a sequence $\{ \rho_k\}$ in $C^\infty_c(\mathbb R^n)$ with $\rho_k = 1$ on $[-k,k]^n$ and $\rho = 0$ outside of $[-(k+1),k+1]^n$. Then for any $\phi \in S(\mathbb R^n)$, we should have $\phi \cdot \rho_k \to \phi$ in $S(\mathbb R^n)$ and each $\phi \cdot \rho_k$ is in $C^\infty_c(\mathbb R^n)$.

For the last part, again it is clear that $S(\mathbb R^n) \subset L^1(\mathbb R^n)$. To prove the embedding is continuous, you need to prove that if $f_k \to f$ in $S(\mathbb R^n)$ then $f_k \to f$ in $L^1(\mathbb R^n)$. This should be fairly clear because the Schwartz topology is very strong.

I can type up a solution to the 3rd one as an example, if necessary.

EDIT: To answer questions from the comments. To show that $f_k \to f$ in $S(\mathbb R^n)$, you need to show that for every multi-indices $\alpha, \beta$ we have $x^\alpha D^\beta f_k \to x^\beta \partial^\alpha f$ uniformly on $\mathbb R^n$.

The reason $\phi \cdot \rho_k \to \phi$ in $S(\mathbb R^n)$ is because $\rho_k$ doesn't change $\phi$ on $[-k,k]^n$ and outside of $[-k,k]^n$ (as $k$ gets large) $\phi$ and all of its derivatives decay rapidly to zero.

I think your justification is fine. No, you need to prove something stronger: you need to prove that if $x^\alpha D^\beta f_k \to x^\beta \partial^\alpha f$ uniformly on $\mathbb R^n$ for all multi-indices $\alpha, \beta$, then $\int_{\mathbb R^n} \lvert f_k(x) - f(x) \rvert dx \to 0$. Proving that $\int_{\mathbb R^n} \lvert f_k(x) - f(x) \rvert dx < +\infty$ is not strong enough.

  • $\begingroup$ For the first part, could you explain to me how we can show that $f_k \to f$ in $S(\mathbb{R}^n)$ ? $$$$ For the second part, why does it hold that $\phi \cdot \rho_k \to \phi$ in $S(\mathbb{R}^n)$ ? $$$$ For the last part, is my justification why $S(\mathbb{R}^n) \subset L^1(\mathbb{R}^n)$ right? In order to show that the embedding is continuous, to we have to show that if $\sup_x |x^{\alpha} D^{\beta}(f_k-f)|<+\infty$ then $\int_{\mathbb{R}^n} (f_k-f) dx<+\infty$ ? Do we use the Cauchy-Schwarz inequality? If so , how? $\endgroup$
    – Evinda
    May 26, 2016 at 21:04
  • $\begingroup$ To show that $f_k \to f$ in $S(\mathbb R^n)$, you need to show that for every multi-indices $\alpha, \beta$ we have $x^\beta \partial^\alpha f_k \to x^\beta \partial^\alpha f$ uniformly on $\mathbb R^n$. $\endgroup$
    – User8128
    May 27, 2016 at 14:24
  • $\begingroup$ I realized I wouldn't be able to answer these in a comment so I edited my answer. $\endgroup$
    – User8128
    May 27, 2016 at 14:30

You must know continuous immersion in locally convex spaces, in this case we can also speak of Fréchet spaces for $\mathcal{S}(\mathbb{R}^n)$ or $L^1(\mathbb{R}^n)$, instead $C_{c}^\infty(\mathbb{R}^n)$ is a LCS. Precisely theorem (1) of this question: Continuous inclusions in locally convex spaces

The problem is that you have to consider the appropriate topologies.

  • $\begingroup$ Could you explain it further to me what we have to do? I haven't seen a similar proof. @JohnMartin $\endgroup$
    – Evinda
    May 27, 2016 at 9:09
  • $\begingroup$ @Evinda you understand what is a continued inclusion between locally convex spaces? $\endgroup$
    – user288972
    May 27, 2016 at 11:07
  • $\begingroup$ To be honest, I haven't heard of locally convex spaces before. @JohnMartin $\endgroup$
    – Evinda
    May 27, 2016 at 11:10
  • $\begingroup$ But you know how the topology defined in $\mathcal{S}(\mathbb{R}^n)$? the continuous inclusion between LCS are characterized as a particular case of Theorem 1, of the previous link. If you understand this, I can answer you. @Evinda $\endgroup$
    – user288972
    May 27, 2016 at 11:13
  • $\begingroup$ I have just seen the definition of $S(\mathbb{R}^n)$. $\endgroup$
    – Evinda
    May 27, 2016 at 11:20

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