Prove :If $u(x)\in \displaystyle{H^{l}(R)}$, then $u(x)\in \displaystyle C^{l-1}(R)$ I come across this problem in my functional analysis book.Prove:
If $u(x)\in \displaystyle{H^{l}(R)}$,   then 
$u(x)\in \displaystyle C^{l-1}(R)$,$\displaystyle\lim_{x\to|\infty|}D^{\alpha}u=0  ,\alpha=0,1,...l$
where $\displaystyle\lim_{x\to|\infty|}u(x)=0,$$H^{l}(R)$ is a Soblev space $W^{l,p}(R)$ with $p=2$,The norm of $C^{l-1}(R)$ is
$||u||=\displaystyle \max_{|a|\leq l-1}\max_{x\in R}|D^{\alpha}u|$.

I want to use Poincare Inequility, but I don't know how to do that.
 A: Hint 1: it suffices to do the case $\ell = 1$, since $D^\alpha u \in C^0$ implies that $D^{\alpha - 1}u \in C^1$. 
Hint 2: Consider Hölder's inequality (or Cauchy-Schwarz if you will) for $-\infty < a < b < \infty$
$$ \int_a^b |f(x)| \mathrm{d}x = \int_a^b \mathbb{1} |f(x)| \mathrm{d}x \leq \left(\int_a^b \mathbb{1}^2\mathrm{d}x \right)^{1/2}\left(\int_a^b |f(x)|^2 \mathrm{d}x\right)^{1/2} \tag{*}$$
this implies that $f\in L^2(R)\implies f\in L^1_{\mathrm{loc}}(R)$ so that the function $F(x) = \int_0^x f(x) \mathrm{d}x$ is absolutely continuous by the fundamental theorem of calculus for Lebesgue integrals, and $F'(x) = f(x)$ almost everywhere. This in particular implies that $f\in H^1(R) \implies f \in C^0(R)$. 
Hint 3: For the decay statement, there must be a typo: you can only prove that $\lim_{|x|\to\infty} D^\alpha u = 0$ for $\alpha < \ell$. For $\alpha = \ell$ it is false. (Let 
$$ g(x) = \sum_{n = 1}^{\infty} n \chi_{[n,n+\frac{1}{2n^4}]}(x) $$
it is evidently $L^2$ but $\lim_{|x|\to\infty} g(x)$ does not exist.)
To actually prove the statement for $\alpha < \ell$, it suffices to consider $\alpha = 0$ and $\ell = 1$ again as before. Now you can use equation (*) to show that if $f \in H^1(R)$, the total variation of $f$ in the interval $[n,n+1]$ must goes to 0 as $|n|\to\infty$. From that $f\in L^2(R)$ you then get that $\liminf_{|x|\to\infty} |f| = 0$. Put the two together you get that the limit itself must be 0. 
