Question about Embedding I have two spaces $$H^1((-\infty,+\infty))=\lbrace u, u\in AC, u'\in L^2\rbrace$$ 
with the norm $||u||^2=\int_{-\infty}^{+\infty} u'^2+\int_{-\infty}^{+\infty} u^2$
and $C_p((-\infty,+\infty))=\lbrace u, u\in C((-\infty,+\infty)),\sup_{t\in\mathbb{R}} p(t)|u(t)|<+\infty\rbrace$$
where $p$ is a weight's fuction it can be $L^1$.
Please how to prove that $H^1((-\infty,+\infty))\subset C_p((-\infty,+\infty))$ ?
 I must find the correct space for the fuction $p$ 
Please help me thank you.
 A: I would reformulate (or correct) the question as: characterise the linear space 
$$
\{p~|~p(t)u(t)\in L^{\infty}(\mathbb{R}), \forall u\in W^{1,2}(\mathbb{R})\}.
$$
I have no idea about any familiar Sobolev space it is equal to. But certainly it contains as subset any function $p$ satisfying the decaying condition $|p(t)|\leq C |t|^{-1/2}, \forall~|t|\geq1$ and bounded.
I hope you can see this by writing
$$
u(t)=u(0)+\int_0^t u'(\tau)~d\tau.
$$
I am also curious about the answer. When you get it from your professor, will you post it here? Thanks!
Edit
I just found that $u\in H^1(\mathbb{R})$ must imply that $u\in L^{\infty}(\mathbb{R}).$ So the above condition on $p$ is indeed too strong.
A: If I am not mistaken,the condition given by Hui Zhang turns out to be optimal. For a sketch of the proof, consider the following:
Fix 
$$
g\in M:=\left\{ p\,\middle|\, p\cdot u\in L^{\infty}\left(\mathbb{R}\right)\:\forall u\in H^{1}\left(\mathbb{R}\right)\right\} .
$$
Using the closed graph theorem, it is then easy to show that the map
$$
\Gamma:H^{1}\left(\mathbb{R}\right)\rightarrow L^{\infty}\left(\mathbb{R}\right),u\mapsto p\cdot u
$$
is bounded, where we use the norm $\left\Vert f\right\Vert _{H^{1}}=\left\Vert f\right\Vert _{2}+\left\Vert f'\right\Vert _{2}$
in $H^{1}\left(\mathbb{R}\right)$.
Now let $t\in\left[1,\infty\right]$ be fixed. Consider
$$
f_{t}\left(x\right):=\begin{cases}
0, & x\leq0,\\
e^{x}-1, & x\in\left[0,t\right],\\
\left(e^{t}-1\right)\cdot\left(\left(t+1\right)-x\right), & x\in\left[t,t+1\right],\\
0, & x\geq t+1.
\end{cases}
$$
Then $f\in H^{1}\left(\mathbb{R}\right)$ with (weak) derivative
$$
f'_{t}\left(x\right)=\begin{cases}
0, & x\leq0,\\
e^{x}, & x\in\left[0,t\right],\\
1-e^{t}, & x\in\left[t,t+1\right],\\
0, & x\geq t+1.
\end{cases}
$$
Using this, one can calculate
$$
\left\Vert f_{t}\right\Vert _{2}+\left\Vert f'_{t}\right\Vert _{2}\lesssim\sqrt{t+1}\cdot e^{t}.
$$
We now derive
$$
e^{t}\cdot\left|p\left(t\right)\right|\lesssim\left|p\left(t\right)\right|\cdot\left(e^{t}-1\right)=\left|p\left(t\right)\cdot f_{t}\left(t\right)\right|\overset{\circledast}{\leq}\left\Vert p\cdot u\right\Vert _{\infty}=\left\Vert \Gamma\left(u\right)\right\Vert _{\infty}\leq\left\Vert \Gamma\right\Vert \cdot\left\Vert f\right\Vert _{H^{1}}\lesssim\left\Vert \Gamma\right\Vert \cdot e^{t}\cdot\sqrt{t+1}.
$$
By canceling $e^{t}$ on both sides, you obtain the required estimate.
One minor technical detail: In the step marked with $\circledast$,
the estimate is strictly speaking only valid for almost every $t\in\mathbb{R}$
(i.e. for all $t\in\mathbb{R}\setminus N_{t}$, where $N_{t}\subset\mathbb{R}$
is a null set that unfortunately depends upon $f_{t}$ and hence on
$t$ itself). There are two possibilities:
You can ignore this technical problem, or you can use some trickery
with Lebesgue-points to show that the set $N$ can be chosen so that
it does not(!) depend upon $t$ (using the continuity of the $f_{t}$).
EDIT: An analogous proof should be possible for $t \leq -1$. It is easy to see that $p$ has to be essentially bounded on $[-1,1]$. This completes the proof.
