Compute $\lim_{|h|\rightarrow \infty} \int_{\mathbb{R}^n} |f(y+h)+f(y)|^p dy$

Question

Suppose $f\in L^p(\mathbb{R}^n), 0<p<\infty$, and compute $\displaystyle \text{lim}_{|h|\rightarrow \infty} \int_{\mathbb{R}^n} |f(y+h)+f(y)|^p dy$

I have no idea from where to start since I do not know anything about $f$. Can anyone please give me a hint to start? (Please do not downvote this question, I cannot show research effort since I don't have any idea. I only need a hint, not a complete answer)

Answer 1

If $p\in(1,\infty)$":

Show it for $g$ that is continuous and has compact support . Then use the fact that $C_c(\mathbb R^n)\subset L^p(\mathbb R^n)$ is a dense subset and use triangle inequality like that $\|f_h-f\|_p\leq \|f_h-g_h\|_p+\|g_h-g\|_p+\|g-f\|_p$.

$\|f_h-g_h\|_p,\|g-f\|_p<\epsilon$ for $\epsilon >0$.

Take $lim_{|h|}inf\|g_h-g\|_p\geq \|g(min(D_g)+g\|_p$ and $lim_{|h|}sup\|g_h-g\|_p\leq \|g(max(D_g)+g\|_p$

Answer 2

For $1\le p\lt\infty$, $$ \|f(x)\|_p=\lim_{M\to\infty}\left\|[|x|\le M]\,f(x)\vphantom{\sum}\right\|_p\tag{1} $$ where $[\cdots]$ are Iverson brackets. Since $$ \|f(x)\|_p^p=\left\|[|x|\le M]\,f(x)\vphantom{\sum}\right\|_p^p+\left\|[|x|\gt M]\,f(x)\vphantom{\sum}\right\|_p^p\tag{2} $$ there is some $M_\epsilon$ so that for $M\ge M_\epsilon$ $$ \left\|[|x|\gt M]\,f(x)\vphantom{\sum}\right\|_p\le\epsilon\tag{3} $$ Now if $h\gt2M_\epsilon$, the supports of $[|x+h|\le M_\epsilon]f(x+h)$ and $[|x|\le M_\epsilon]f(x)$ are disjoint; therefore, $$ \begin{align} &\left\|[|x+h|\le M_\epsilon]f(x+h)+[|x|\le M_\epsilon]f(x)\vphantom{\sum}\right\|_p^p\\ &=\left\|[|x+h|\le M_\epsilon]f(x+h)\vphantom{\sum}\right\|_p^p+\left\|[|x|\le M_\epsilon]f(x)\vphantom{\sum}\right\|_p^p\\ &=2\left\|[|x|\le M_\epsilon]f(x)\vphantom{\sum}\right\|_p^p\tag{4} \end{align} $$ The triangle inequality, along with $(3)$, yields $$ \begin{align} 2^{1/p}\left\|[|x|\le M_\epsilon]f(x)\vphantom{\sum}\right\|_p &\ge2^{1/p}\left(\|f\|_p-\left\|[|x|\gt M_\epsilon]f(x)\vphantom{\sum}\right\|_p\right)\\ &\ge2^{1/p}\|f\|_p-2^{1/p}\epsilon\\[3pt] &\ge2^{1/p}\|f\|_p-2\epsilon\tag{5} \end{align} $$ The triangle inequality and $(3)$ say that $$ \begin{align} &\left\|[|x+h|\gt M_\epsilon]f(x+h)+[|x|\gt M_\epsilon]f(x)\vphantom{\sum}\right\|_p\\ &\le\left\|[|x+h|\gt M_\epsilon]f(x+h)\vphantom{\sum}\right\|_p+\left\|[|x|\gt M_\epsilon]f(x)\vphantom{\sum}\right\|_p\\ &=2\left\|[|x|\gt M_\epsilon]f(x)\vphantom{\sum}\right\|_p\\ &\le2\epsilon\tag{6} \end{align} $$ $(4)$, $(5)$, $(6)$, and the triangle inequality give that $$ 2^{1/p}\|f\|_p-4\epsilon\le\left\|f(x+h)+f(x)\vphantom{\sum}\right\|_p\le2^{1/p}\|f\|_p+2\epsilon\tag{7} $$ Therefore, $$ \lim_{h\to\infty}\left\|f(x+h)+f(x)\vphantom{\sum}\right\|_p^p=2\|f\|_p^p\tag{8} $$