The $L_p$-norm of sum of a function and its translation at infinity I met this problem in real analysis:
Let $1 \leq p < \infty$ and assume $f \in L^p(\mathbb{R}^n)$. Prove that 
$$
\lim_{|y| \to \infty} \| f(x+y)+f(x) \|_p = 2^{\frac{1}{p}} \| f\|_p
$$
I know that we need to prove the following:
$$
\lim_{|y| \to \infty}\int_{\mathbb{R^n}} [f(x+y)+f(x)]^p - 2(f(x))^p dx= 0
$$
I feel that, from intuition, the result makes sense to me. Because, when $y$ goes to infinity, the influence of "overlapping" of $f(x+y)$ and $f(x)$ when raised to $p$-th power is "wiped out", thus we have double $(f(x))^p$. But I have difficulty proving it. Could anyone give me some hint? Thank you so much!
 A: Some assembly required:
Let $f_k = f \cdot 1_{[-k,k]^n}$ and $(\sigma_y f)(x) = f(x+y)$, it is not hard to show that $\|f_k + \sigma_y f_k\| \to \sqrt[p]{2} \|f_k\|$.
$f+\sigma_y f  = f - f_k + f_k + \sigma_y f_k - \sigma_y f_k + \sigma_y f$ and so
$| \|f+\sigma_y f\| - \| f_k + \sigma_y f_k\|| \le \| f-f_k\| + \|\sigma_y(f-f_k)\| = 2 \|f-f_k\|$.
A: Let $C_c(\mathbb R^n)$ the set of continous functions with compact support. We know that $C_c(\mathbb R^n)$ is dense in $L^p(\mathbb R^n)$. Assume that $f \in C_c(\mathbb R^n)$ and let $M > 0$ such that $\text{supp} (f) \subset B(0,M)$. For $|y| > M$ $B(-y,M) \cap B(0,M) = \emptyset$ and so $$\int_{\mathbb R^n} |f(x+y) + f(x)|^p \mathrm d x = \int_{B(0,M)} |f(x) + f(x+y)|^p \mathrm d x + \int_{B(-y,M)} |f(x) + f(x+y)|^p \mathrm d x + \int_{\left\|x\right\| \ge M,~ \left\|x+y\right\| \ge M} |f(x) + f(x+y)|^p \mathrm d x = \int_{B(0,M)} |f(x)|^p \mathrm d x + \int_{B(-y,M)} |f(x+y)|^p \mathrm d x = 2\int_{B(0,M)} |f(x)|^p \mathrm d x = 2\int_{\mathbb R^n} |f(x)|^p \mathrm d x$$ This proves that the statement we are looking for holds for $C_c(\mathbb R^n)$ i.e. : $$\forall f \in C_c(\mathbb R^n),~\lim_{|y|\rightarrow \infty} \left\|f(\cdot+y) + f\right\|_p = 2^{\frac1p} \left\|f\right\|_p$$. Now we will try to prove it for $L^p(\mathbb R^n)$ using density. Indeed, let $f \in L^p(\mathbb R^n)$, $\epsilon > 0$ and $f_\epsilon \in C_c(\mathbb R^n)$ such that $\left\|f-f_\epsilon\right\|_p^p \le \frac\epsilon5$ $$\left|\left\|f(\cdot + y) + f\right\|_p - 2^{\frac1p}\left\|f\right\|_p\right| \le \left|\left\|f_\epsilon(\cdot + y) + f_\epsilon\right\|_p - 2^{\frac1p}\left\|f_\epsilon\right\|_p\right| + \left|\left\|f(\cdot + y) + f\right\|_p - \left\|f_\epsilon(\cdot + y) + f_\epsilon\right\|_p\right| + \left|2^{\frac1p}\left\|f\right\|_p- 2^{\frac1p}\left\|f_\epsilon\right\|_p\right| $$
Or $$\left|\left\|f(\cdot + y) + f\right\|_p - \left\|f_\epsilon(\cdot + y) + f_\epsilon\right\|_p\right| \le \left\|f(\cdot + y) + f - f_\epsilon(\cdot + y) + f_\epsilon\right\|_p \le \left\|f(\cdot + y) - f_\epsilon(\cdot + y)\right\|_p + \left\|f-f_\epsilon\right\| \le 2 \left\|f-f_\epsilon\right\|$$ So $$\left|\left\|f(\cdot + y) + f\right\|_p - 2^{\frac1p}\left\|f\right\|_p\right| \le \left|\left\|f_\epsilon(\cdot + y) + f_\epsilon\right\|_p - 2^{\frac1p}\left\|f_\epsilon\right\|_p\right| + 4\left\|f-f_\epsilon\right\|_p$$
