Few questions about the following sequence . If i have a sequence $u_n=u(x+n)$ , $u_n\in C_c^\infty R$,  is it bounded in $W^{1,p} (R)$ ? and is it true that for no $q\ge1$ does there exist a subsequence converging strongly in $L^q(R)$ . 
As per my lecture notes the claim is that $\int_R u_n\phi\to 0$ and $\int_R u_n'\phi\to 0$ as $n\to \infty$ . 
I am not able to clarify these statements . Any explanations, hints etc would really help me . 
 A: *

*Yes. Since for any integrable function $f$ and any $y\in\mathbb{R}$ you have that $\int_{\mathbb{R}}f(x) \mathrm{d}x = \int_{\mathbb{R}} f(x + y)\mathrm{d}x$, that is, the Lebesgue measure is translation invariant, you have that $\|u_n\|_{W^{1,p}} = \|u\|_{W^{1,p}}$ for any $n$. Hence the sequence is bounded. 

*Yes. If $u \in C^\infty_c(R)$, we have that there exists some $M$ such that $u(x) = 0$ whenever $|x| > M$. Therefore for $k > 2M$ we have that $\|u_{n+k} - u_{n}\|_{L^q} = 2\|u\|_{L^q} > 0$ (using that $u_{n+k}$ and $u_n$ have disjoint supports). This implies that for any subsequence, Cauchy's criterion must fail, and the subsequence cannot converge. 

*If $\phi \in L^1$, we have that $\lim_{R\to \infty} \int_{-R}^R |\phi| \mathrm{d}x = \int_{\mathbb{R}}|\phi|\mathrm{d}x$. This in particular requires that $\lim_{R\to\infty} \int_{(-\infty,-R)\cup(R,\infty)} |\phi|\mathrm{d}x = 0$. Using that $u_n$ has compact support, if we choose $n > M+R$, where $M$ is defined as before, we have that by examining the support of $u_n$
$$ \int u_n\phi \mathrm{d}x = \int_{-M-n}^{M-n}u_n(x)\phi(x)\mathrm{d}x $$
taking absolute values we have
$$ \left|\int u_n\phi\mathrm{d}x \right| \leq \| u\|_{L^\infty} \int_{-M-n}^{M-n} |\phi(x)|\mathrm{d}x \leq \|u\|_{L^\infty} \int_{-\infty}^{-R} |\phi(x)|\mathrm{d}x $$
Now you just need to take the limit on both sides. 

Edit to address comment:


*

*By definition $\phi \in L^1$ means that $\int_{-\infty}^\infty |\phi| \mathrm{d}x < \infty$. In the Riemann sense of the integral you have to define it as an improper integral. This means that $\lim_{a\to\infty}\lim_{b\to -\infty}\int_b^a |\phi|\mathrm{d}x < \infty$, as well as when the order of the limits is switched, and the two limits must be equal. By using the definition of convergence, this means that for any $\epsilon > 0$ and sufficiently large $|a|$ and $|b|$ where $ab < 0$ you have that
$$ \left|\int_{-\infty}^\infty |\phi|\mathrm{d}x - \int_b^a |\phi|\mathrm{d}x \right| < \epsilon $$
where $\int_{-\infty}^\infty$ is interpreted as the improper integral (defined via the converging limit). The difference can be written as $\int_{-\infty}^b + \int_a^\infty$, again interpreted as an improper integral. 
In other words, if you take integrals in the Riemann sense the implication $\lim_{R\to \infty} \int_{(-\infty,-R) \cup(R,\infty)} |\phi| \mathrm{d}x$ follows from the definition of "convergence" and the definition of "improper integral". 
If you choose to interpret the integral in the Lebesgue sense, using that the Lebesgue measure is countably additive over disjoint measurable sets, we can write
$$ \int_{-\infty}^\infty |\phi(x)|\mathrm{d}x = \int_{(-1,1]} |\phi(x)|\mathrm{d}x + \sum_{n = 1}^\infty \int_{(-n-1,-n] \cup (n,n+1]} |\phi(x)|\mathrm{d}x $$
and by definition of the convergence of an infinite sum, the desired expression follows. 
Another way of saying the statement is in words that "any integrable function can be approximated (in the $L^1$ norm) by a sequence of functions with compact support", which is a direct consequence of Littlewood's first principle. 

*For the last step the taking of the limit can be dealt with in the following way. For $n > M+ 1$, define $R(n) = n - M$. Then $n > M + R(n)$. So we can get the expression as before
$$ \left| \int u_n \phi\mathrm{d}x\right| \leq \|u\|_{L^\infty} \int_{-\infty}^{-R(n)} |\phi(x)|\mathrm{d}x $$
Plugging in $R(n) = n-M$ we take the limit in $n$ on both sides. The left hand side is your desired expression. The right hand side has only one term that depends on $n$, which is $\int_{-\infty}^{-n + M} |\phi(x)|\mathrm{d}x$, which tends to $0$ as $n$ tends to $\infty$ by the previous argument. 
