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I am currently working through this lecture notes and on page 164, there it is said

The space of $\mathcal{D}(\mathbb{R}^n)$ of smooth complex-valued functions with compact support is contained in the Schwartz space $\mathcal{S}(\mathbb{R}^n)$. If $f_k \to f$ in $\mathcal{D}$, then $f_k \to f$ in $\mathcal{S}$, so $\mathcal{D}$ is continuously embedded in $\mathcal{S}$. Furthermore, if $f\in \mathcal{S}$, and $\eta \in C_c^{\infty}(\mathbb{R}^n)$ is a cutoff function with $\eta_k(x) = \eta(x/k)$, then $\eta_k f \to f$ in $\mathcal{S}$ as $k \to \infty$, so $\mathcal{D}$ is dense in $\mathcal{S}$.

I don't understand the arguments in this paragraph, for a subset $\mathcal D$ of $\mathcal S$ to be dense in $\mathcal S$ for every element $s$ of $\mathcal S$ I need to find a sequence in $\mathcal D$ which converges to $s$, but there just stands that $\eta_k f \to f$ in $\mathcal{S}$, but what i need is a sequence in $\mathcal{D}$ not in $\mathcal{S}$, so why does it follow from this that $\mathcal{D}$ is dense in $\mathcal{S}$?

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As far as I understand, $\eta_k f \in \mathcal{D}$. –  Siminore Jul 2 '12 at 13:47
    
is this enough to conclude that $f \in \mathcal{D}$? –  Stefan Jul 2 '12 at 13:52
    
Why do you need that? You must prove that any $f \in \mathcal{S}$ is the limit of a sequence from $\mathcal{D}$. –  Siminore Jul 2 '12 at 13:54
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Wait a moment. $\mathcal{D}$ is here endowed with the same topology as $\mathcal{S}$. Be careful: although $\mathcal{D}$ might have a different topology, we are thinking of it as a subset of $\mathcal{S}$, and hence it inherits the topology of $\mathcal{S}$. –  Siminore Jul 2 '12 at 14:18
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In a nutshell: read the statement "$\eta_k f\to f$ in $\mathcal S$" as "$\eta_k f\to f$ in [the topology of] $\mathcal S$". –  user31373 Jul 2 '12 at 17:52
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up vote 3 down vote accepted

The sequence $\{\eta_k\}$ is contained in $\mathcal D(\Bbb R^n)$ and we can check that the product of a function in the Schwartz space with a test function is a test function (it's smooth because the product of two smooth functions is smooth, and it has a compact support because it's contained in the support of the test function).

We have $\mathcal D(\Bbb R^n)\subset \mathcal S(\Bbb R^n)$, and what we want to see is that it's a dense subset for the topology of $\mathcal S(\Bbb R^n)$. So what we have to show is that $$\sup_{x\in \Bbb R^n}|x^p\partial^{\alpha}(\eta_kf-f)(x)|\to 0$$ for all integer $p$ and all $\alpha\in\Bbb N^n$.

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guess this holds for almost every kind of convergence cause $$\eta_k(x) = 1$$ iff $|x| \le k$, so if i just make $k$ big enough, the term $\eta_k f - f$ vanishes, is it that trivial? –  Stefan Jul 3 '12 at 17:25
    
It vanishes for a fixed point. The most difficult is to show that it converges to $0$ uniformly. –  Davide Giraudo Jul 3 '12 at 17:26
    
but because for every $k$ there is some $x$ such that $\eta_k(x) = 0$ it is always the case that $x^p \partial^{\alpha}(\eta_k f - f)(x) = x^p \partial^{\alpha}f > 0$. –  Stefan Jul 4 '12 at 9:27
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