If $f \in L^p(R)$, then $\lim_{y \to \infty}\|f(x+y)+f(x)\|_p=2^{1/p}\|f\|_p$

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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?

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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.