Suppose $\int _a^b\vert f\vert^p<\infty$ for some $p\ge 1$ and for all $a,b\in \mathbb{R}$, and for some $a>p-1$ $$\int_{2\vert y-x\vert \le x}\vert f(y)\vert ^pdy\le \vert x\vert^{-a}$$ when $\vert x \vert \ge 1$. I'd like a hint on how to show $f\in L^1(\mathbb{R}).$
The region described in the limits of integral is
I rewrite the integral as $\int_{2\vert y-x\vert \le x}\vert f(y)\vert ^pdy=\int_{-2x}^{-x/2}\vert f \vert^p + \int_{x/2}^{2x}\vert f\vert^p$ but I'm not sure what is required to start describing $\int \vert f \vert$.
Most problems I have seen on $L^p$ spaces require Holder's inequality, but here I am not even given that $f\in L^p$. May I have a hint from anyone with an idea on this?
$$ \int_{-\infty}^\infty|f(y)\,dy=\int_{-1}^1|f(y)|\,dy+\sum_{k=0}^\infty\int_{3^k}^{3^{k+1}}|f(y)|\,dy+\sum_{k=0}^\infty\int_{-3^k}^{-3^{k+1}}|f(y)|\,dy. $$ For the first integral $$ \int_{-1}^1|f(y)|\,dy\le\Bigl(\int_{-1}^1|f(y)|^p\,dy\Bigr)^{1/p}2^{1-1/p}<\infty. $$ I will estimate the first sum, the other being similar. We have $$ [3^k,3^{k+1}]=\{y:2\,|y-2\cdot3^k|\le2\cdot3^k\}. $$ Thus $$ \int_{3^k}^{3^{k+1}}|f(y)|\,dy\le\Bigl(\int_{3^k}^{3^{k+1}}|f(y)|^p\,dy\Bigr)^{1/p}(2\cdot3^k)^{1-1/p}\le C\,3^{-\bigl(\tfrac{a+1}{p}-1\bigr)k}. $$ Since $a>p-1$ we have $(a+1)/p-1>0$ and the sum converges.