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We know the topology of the space $\mathscr{S_n}$ of rapidly decreasing functions is strictly weaker than the topology of the test function space $\mathcal{D}\mathbb{(R^n)}$. So I wish to construct a specific sequence ${ϕ_m}$ in $\mathcal{D}\mathbb{(R^n)}$ such that $ϕ_m → 0$ in the topology of $\mathscr{S_n}$ but not in the topology of $\mathcal{D}\mathbb{(R^n)}$. Are there any suggestions?

Abdelmalek Abdesselam
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Mathman
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    Take a bump function $\rho$, see https://en.wikipedia.org/wiki/Bump_function , and let $\phi_m(x)=\frac{1}{m}\rho(x/m)$. The point is that to converge in $\mathscr{D}$ a sequence $\phi_m$ must have a common compact support. If you really want to understand this you need to learn the topology of $\mathscr{D}$. See, e.g., https://math.stackexchange.com/questions/3510982/doubt-in-understanding-space-d-omega/3511753#3511753 – Abdelmalek Abdesselam May 02 '21 at 17:03
  • @AbdelmalekAbdesselam Thanks! But how do I show $\phi_m(x)=\frac{1}{m}\rho(x/m)$ converges to 0 in $\mathscr{S_n}$? Why does convergence in $\mathscr{S_n}$ not require a common compact support? – Mathman May 03 '21 at 02:06

1 Answers1

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To answer the question in the comments:

First see:

Doubt in understanding Space $D(\Omega)$

for what it means to define the topology of a vector space using a collection of seminorms.

Now let $\mathscr{S}(\mathbb{R}^n)$ be the Schwartz space of smooth functions of rapid decay on $\mathbb{R}^n$. The topology is defined by the countable family of seminorms $$ ||f||_{\alpha,k}=\sup_{x\in\mathbb{R}^n}\langle x\rangle^k|\partial^{\alpha}f(x)| $$ indexed by $k\in\mathbb{N}_0$ and $\alpha\in\mathbb{N}_0^n$.

Here $\partial^\alpha\phi_m(x)=\frac{1}{m^{|\alpha|+1}}\partial^\alpha \rho(x/m)$.

Suppose the support of $\rho$ is contained in $[-M,M]^n$, then

$$ ||\phi_m||_{\alpha,k}\le \frac{1}{m^{|\alpha|+1}}\times (1+nM^2)^{\frac{k}{2}} \times \sup_{x\in [-M,M]^n}|\partial^\alpha\rho(x)| $$ which goes to zero when $m\rightarrow \infty$. Since this true for all $k$ and $\alpha$, this means the sequence $\phi_m$ converges to zero in Schwartz space.

Abdelmalek Abdesselam
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