(old qual question in analysis) If $A_\lambda=\lbrace (x,y): \lambda \le x^4+y^2\le 2\lambda \rbrace$ and $f$ is locally in $L^{(3/2)}(\mathbb{R}^2)$ and there is an $a>3/8$, such that $$\int_{A_\lambda}\vert f(x)\vert ^{(3/2)}dx\le \frac{1}{\lambda ^a}$$ for $\lambda \ge 1$, then $f\in L^1(\mathbb{R})^2$.
Here's what the region looks like for $\lambda=1:$
Let $S_k$ denote the square of sidelength $k$ centered at the origin. $$\int_{\mathbb{R}^2} \vert f\vert dx\times dy= \int_{S_1} \vert f\vert + \sum_k \int_{S_{2^{k+1}}-S_{2^k}}\vert f \vert $$
where the first integral is bounded above by $(\int_S \vert f \vert ^{3/2} )^{2/3}\cdot 2^{-1/2}$
If I am on to something, how to I estimate the sum? I'm not sure how to use the regon $A_{\lambda}$.
It suffices to show that $$\sum_{k=0}^\infty \int_{A_{2^k}} |f|<\infty \tag{1}$$ because $\bigcup_{k=0}^\infty A_{2^k}$ covers the plane except for a subset of finite measure (and on a subset of finite measure $L^{3/2}\subset L^1$). Naturally, Hölder's inequality comes into play: $$ \int_{A_{2^k}} |f| \le \left(\int_{A_{2^k}} |f|^{3/2}\right)^{2/3} \left(\int_{A_{2^k}} 1 \right)^{1/3} \tag{2} $$ Here $\int_{A_{2^k}} |f|^{3/2} \le 2^{-ak}$ by assumption. Estimate the measure of $A_\lambda$ from above, by comparing with a rectangle: the bound is $O(\lambda^{3/4})$. So, $$ \int_{A_{2^k}} |f| \le 2^{-2ak/3} 2^{k/4} \tag{3} $$ and the rest is arithmetics of exponents.