Are the three statements the same?

Question

Let $\mathcal{S}(\mathbb{R}^n)$ be the Schwartz function on $\mathbb{R}^n$. Consider two statements which have the same proof. $$ f\in \mathcal{S}(\mathbb{R}^n)\,\,\Longrightarrow\,\,f\in L^p(\mathbb{R}^n)$$ and $$\mathcal{S}(\mathbb{R}^n)\subset L^p(\mathbb{R}^n).$$

My question is, if we know the fact $\mathcal{S}(\mathbb{R}^n)$ is a subspace of $L^p$, does this imply $\mathcal{S}(\mathbb{R}^n)$ is dense in $L^p(\mathbb{R}^n)?$

In other words, the three statements are equivalent. \begin{align*} f\in \mathcal{S}(\mathbb{R}^n)\,\,&\Longrightarrow\,\,f\in L^p(\mathbb{R}^n),\\ \mathcal{S}(\mathbb{R}^n)&\subset L^p(\mathbb{R}^n),\\ \mathcal{S}(\mathbb{R}^n)\quad&\text{is dense}\quad\text{in}\quad L^p(\mathbb{R}^n). \end{align*}

Could anyone correct my mistakes? How to prove $\mathcal{S}(\mathbb{R}^n)$ is dense in $L^p(\mathbb{R}^n)?$

Answer 1

The statement

$\mathcal{S}(\mathbb{R}^n)$ is dense in $L^p(\mathbb{R}^n)$

cannot be infered from

$f \in \mathcal{S}(\mathbb{R}^n) \Rightarrow f \in L^p(\mathbb{R}^n)$.

What is said in the notes that you're refering to is :

$\mathcal{S}(\mathbb{R}^n)$ is dense in $L^p(\mathbb{R}^n)$ because $\mathcal{D}(\mathbb{R}^n) \subseteq \mathcal{S}(\mathbb{R}^n)$ and it has been established in Lemma 1.5 that $\mathcal{D}(\mathbb{R}^n) $ is dense in $L^p(\mathbb{R}^n)$.

Additional comment : What does "$X$ is dense in $Y$ mean ? It means that every neighbourhood of every $y \in Y$ contain an element of $x$, i.e. fix a $y \in Y$ at random and select any neighbourhood of this $y \in Y$ then there ought to be a $x \in X$ in this neighbourhood.

Whenever you can prove that $X$ is dense in $Y$ and that $X \subseteq Z \subseteq Y$ than you can conclude that $Z$ is dense in $Y$ too.

Why ? Well fix $y \in Y$ and fix a neighbourhood of $y$. This neighbourhood contains a certain $x \in X$ because $X$ is dense in $Y$ by hypothesis. But $x \in X \subseteq Z$. Therefore you can see $x$ as an element of $Z$ and conclude that every neighbourhood of any $y \in Y$ contain an element of $Z$.

Example : I'll answer the question you asked in a comment here. Define $$X := \big\{ (q_1, q_2) :~~ q_1, q_2 \in \mathbb{Q}\big\}\\ Z := \big\{ (q, r) :~~ q \in \mathbb{Q}, r \in \mathbb{R}\big\} \\ Y := \big\{ (r_1, r_2) :~~ r_1, r_2 \in \mathbb{R}\big\}= \mathbb{R}^2.$$

Clearly $X \subseteq Z \subseteq Y$. Also you can easily prove that $X$ is dense in $Y$.

Now, can we readily conclude that $Z$ is dense in $Y$ (that is without having to provide a proof) ? Yes ! And here is why :

Let $(r_1, r_2)$ be any element of $Y= \mathbb{R}^2$ and consider a neighbourhood of $(r_1, r_2)$. This neighbourhood contains an element $(q_1, q_2)$ of $X$ (because we know that $X$ is dense in $Y$). Hence this neighbourhood contains an element of $Z$, namely $(q_1, q_2)$ itself. This concludes the proof that $Z$ is dense in $Y$.