If $f \in L^p(R)$, then $\lim_{y \to \infty}\|f(x+y)+f(x)\|_p=2^{1/p}\|f\|_p$ If $f \in L^p(R)$, then $\lim_{y \to \infty} \|f(x+y)+f(x)\|_p = 2^{1/p}\|f\|_p$ 
I am not sure how to proceed. To me, it seems like a density argument problem, and I can show this is true for continuous functions with compact support. However, I do not know how to extend it to all $L^p$ functions. Given a continuous function with compact support, we simply take $y$ big enough, so that the support of $f(x+y)$ disjoint with support of $f(x)$ so that they do not intersect at all. Thus for such large $y$ we know that $|f(x+y)+f(x)|^p=2|f(x)|^p$ for each $x$. Hence the limit follows. Now is it possible to extend it to any integrable function be density? How would one does that?

You can extend to general functions as follows. Given $\epsilon > 0$, let $g$ be continuous with compact support such that $f = g + h$ with $\|h\|_p < \epsilon$. Then 
$$\|f(x+y) + f(x)\|_p = \|g(x+y) + g(x) + h(x+y) + h(x)\|_p$$
By the triangle inequality you have
$$\|g(x+y) + g(x) + h(x+y) + h(x)\|_p \leq \|g(x+y) + g(x)\|_p + \|h(x+y)\|_p + \|h(x)\|_p$$
$$< \|g(x+y) + g(x)\|_p + 2\epsilon$$
By the triangle inequality in another form you have
$$\|g(x+y) + g(x) + h(x+y) + h(x)\|_p \geq \|g(x+y) + g(x)\|_p  - \|h(x+y) + h(x)\|_p$$
$$\geq \|g(x+y) + g(x)\|_p - \|h(x+y)\|_p -\| h(x)\|_p$$
$$> \|g(x+y) + g(x)\|_p - 2\epsilon$$
Hence you have
$$\|g(x+y) + g(x)\|_p - 2\epsilon < \|f(x+y) + f(x)\|_p < 
\|g(x+y) + g(x)\|_p + 2\epsilon$$
Now try using the result for $g(x)$ to get the full result. 
