5 added 13 characters in body; edited tags

I understand that this question would be trivial for experts, sorry for that, I just need to clarify things.

So let $S(\mathbb{R}^n)$ denote the Schwartz space on $\mathbb{R}^n$ and $W_p$, $W_q$ are the Sobolev spaces, or in the other words completions of $S(\mathbb{R}^n)$ with respect to $p$ and $q$ - Sobolev norms.

Let us assume that $\phi:$ $W_q\rightarrow W_p$ is a map , continuously extending the identity map on $S$ assuming that $p < q$.

Question: why is $\phi$ an embedding? It is only clear that $\phi$ is embedding of $S$ ( as it is identity on it).

Similar question for Sobolev embedding theorem into $C^k$.

P.S. To clarify my question here is what I mean by the Sobolev space. For a real number $p\in\mathbb{R}$ the $p$-Sobolev norm on $S$ is given by ${|f|}^2_p=\int{(1+|\xi|)}^{2p}{|\hat{f}(\xi)|}^2d\xi$, where $\hat{f}(\xi)$ is a Fourier transform of $f(x)\in S$. So $W_p$ is a completion of $S$ with respect to this norm.

P.P.S. To make things little more clear I am mostly interested in case of non-integer or negative $p$ when the Sobolev $p$-norm can not be defined through the weak derivatives.

4 added 180 characters in body

I understand that this question would be trivial for experts, sorry for that, I just need to clarify things.

So let $S(\mathbb{R}^n)$ denote the Schwartz space on $\mathbb{R}^n$ and $W_p$, $W_q$ are the Sobolev spaces, or in the other words completions of $S(\mathbb{R}^n)$ with respect to $p$ and $q$ - Sobolev norms.

Let us assume that $\phi:$ $W_q\rightarrow W_p$ is a map , extending the identity map on $S$ assuming that $p < q$.

Question: why is $\phi$ an embedding? It is only clear that $\phi$ is embedding of $S$ ( as it is identity on it).

Similar question for Sobolev embedding theorem into $C^k$.

P.S. To clarify my question here is what I mean by the Sobolev space. For a real number $p\in\mathbb{R}$ the $p$-Sobolev norm on $S$ is given by ${|f|}^2_p=\int{(1+|\xi|)}^{2p}{|\hat{f}(\xi)|}^2d\xi$, where $\hat{f}(\xi)$ is a Fourier transform of $f(x)\in S$. So $W_p$ is a completion of $S$ with respect to this norm.

P.P.S. To make things little more clear I am mostly interested in case of non-integer or negative $p$ when the Sobolev $p$-norm can not be defined through the weak derivatives.

3 added 325 characters in body

I understand that this question would be trivial for experts, sorry for that, I just need to clarify things.

So let $S(\mathbb{R}^n)$ denote the Schwartz space on $\mathbb{R}^n$ and $W_p$, $W_q$ are the Sobolev spaces, or in the other words completions of $S(\mathbb{R}^n)$ with respect to $p$ and $q$ - Sobolev norms.

Let us assume that $\phi:$ $W_q\rightarrow W_p$ is a map , extending the identity map on $S$ assuming that $p < q$.

Question: why is $\phi$ an embedding? It is only clear that $\phi$ is embedding of $S$ ( as it is identity on it).

Similar question for Sobolev embedding theorem into $C^k$.

P.S. To clarify my question here is what I mean by the Sobolev space. For a real number $p\in\mathbb{R}$ the $p$-Sobolev norm on $S$ is given by ${|f|}^2_p=\int{(1+|\xi|)}^{2p}{|\hat{f}(\xi)|}^2d\xi$, where $\hat{f}(\xi)$ is a Fourier transform of $f(x)\in S$. So $W_p$ is a completion of $S$ with respect to this norm.

2 fixed latex
1