Proving integrability given local Lp integrability

Question

This is a qualifying exam problem, and therefore I am interested in tips/advice/approach and not the actual solution.

Problem.

Define $$E_r:=\{(x_1,x_2)\in\mathbb{R}^2:r\le x_1^4+x_2^2\le 2r\}.$$ Fix $f\in L_{loc}^{3/2}(\mathbb{R}^2)$ with $$\int_{E_r}|f|^{3/2}\le r^{-a}$$ for some $a>3/8$ and for every $r\ge 1$. Show that $f\in L^1(\mathbb{R}^2)$.

My instincts tell me there is a clever application of Holder's inequality applied to $\int |f|$ along with partitioning the plane $\mathbb{R}^2$ into pieces $S_n$ for which $\sum_{n}\int_{S_n}|f|<+\infty$ thereby proving the integrability claim.

My main question (which is a solution) can probably be answered on my own if I knew the answers to the following two:

(1) what is the relationship between $3/2$ and $3/8$ in this problem?

(2) How does one work with the funky domain $E_r$ in this problem?

Edit(1). To elaborate on my attempt, we have \begin{align} \int_{E_r}|f|&\le (\int|f|^{3/2})^{2/3}\mu(E_r)^{1/3}\\ &\le r^{-2a/3}\mu(E_r)^{1/3}\\ &< r^{-1/4}\mu(E_r)^{1/3} \end{align} (3) Is it true that $\mu(E_r)\le r^{-3(3/4+\epsilon)}$?

Answer 1

Extensive revision: I have decided to rewrite entirely my answer since the question is interesting and deserves more than few sketchy (and sometimes buggy) hints. Now lets start again by giving hints: to start with a little notation, define $\mu(E)$ as the Lebesgue measure of the set $E\subset\mathbb{R}^2$.

1. $f\in L^{3/2}_\mathrm{loc}(\mathbb{R}^2)$ implies $f\in L^1_\mathrm{loc}(\mathbb{R}^2)$: as a matter of fact, by Hölder's inequality $$ \Vert f\Vert_{L^1(K)}\leq \mu(K)^{1/3}\Vert f\Vert_{L^{3/2}(K)} \quad \text{for all compact sets }K\subset\mathbb{R}^2\tag{1}\label{1} $$ See the Wikipedia article on Locally integrable functions for a comprehensive explanation.
2. After having obtained inequality \eqref{1}, the next step is trying to find a countable covering of $\mathbb{R}^2$ made of pairwise disjoint sets $\{S_n\}_{n\in\mathbb{N}}$ build by using the family $\{E_r\}_{r\geq1}$. If we succeed in this search, this would enable us to use \eqref{1} on each set of the covering and find a condition for the finiteness of $\Vert f\Vert_{L^1(\mathbb{R}^2)}$ since formally $$ \Vert f\Vert_{L^1(\mathbb{R}^2)}=\int_{\mathbb{R}^2}|f|\mathrm{d}\mu=\sum_{n=1}^\infty \int_{S_n}|f|\mathrm{d}\mu=\sum_{n=1}^\infty\Vert f\Vert_{L^1(S_n)}\tag{2}\label{2} $$ Claim: the sought for covering $\{S_n\}_{n\in\mathbb{N}}$ of $\mathbb{R}^2$ can be build from the sets $$ S_n\triangleq E_{n+1}\setminus E_n=\{(x_1,x_2)\in\mathbb{R}^2:2n\le x_1^4+x_2^2\le 2(n+1)\}\quad \forall n\in\mathbb{N} $$ or the sets $$ S_n\triangleq E_n\setminus E_{n+1}=\{(x_1,x_2)\in\mathbb{R}^2:n\le x_1^4+x_2^2\le n+1\}\quad \forall n\in\mathbb{N} $$ (It is up to yo to see what would be the better choice)
3. After choosing the proper covering $\{S_n\}_{n\in\mathbb{N}}$ next step is to calculate the area $\mu(S_n)$ for all $n\in\mathbb{N}$: in this way, by using \eqref{1} and the estimate $$ \Vert f\Vert_{L^{3/2}(E_n)}=\left|\int_{E_r}|f|^{3/2}\mathrm{d}\mu\right|^{2/3}\le r^{-\frac{2a}{3}} $$ we can obtain an estimate $$ \Vert f\Vert_{L^1(s_n)}\leq \mu(S_n)^{1/3}\Vert f\Vert_{L^{3/2}(S_n)}\le n^{\,p(a)} \quad \forall n\in\mathbb{N}\tag{3}\label{3} $$ where $p(a)$ is a function which surely depends on $a$ and arises from the estimation of $\Vert f\Vert_{L^{3/2}(S_n)}$ and $\mu(S_n)$ as a function of $n$.
   Hint on the calculation of $\mu(E_r)$. The estimation of $\mu(E_r)$ is a key step in the quantitative evaluation of \eqref{3}, since estimates of $\mu(S_n)$ and of $\Vert f\Vert_{L^{3/2}(S_n)}$ depend on it: in order to do this, the first thing to note is that $E_r$ is homeomorphic to a ring domain through the mapping $(x_1,x_2)\mapsto(\rho,\theta)$ defined as $$ x_1= \begin{cases} \rho^\frac{1}{2}|\sin\theta|^\frac{1}{2} & 0\leq\theta<\pi\\ -\rho^\frac{1}{2}|\sin\theta|^\frac{1}{2} & \pi\leq\theta<2\pi \end{cases}\quad x_2=\rho\cos\theta $$ where $\rho\in[r,2r]$ and $\theta\in[0,2\pi]$. Then, by calculating the modulus of the Jacobian determinant of the mapping, we have that $$ \mu(E_r)=\int_{E_r}\mathrm{d}x_1\mathrm{d}x_2=\int\limits_r^{2r}\mathrm{d}\rho\int\limits_0^{2\pi}|J(\rho,\theta)|\mathrm{d}\theta $$ The integral at the last term of the inequality is not easy to calculate: however it is easyly estimable as a function of $r$ by the Fubini-Tonelli theorem, and this leads to the sought for estimate \eqref{3}.
4. By using estimate \eqref{3} we have that $$ \Vert f\Vert_{L^1(\mathbb{R}^2)}=\sum_{n=1}^\infty\Vert f\Vert_{L^1(s_n)}\le\sum_{n=1}^\infty\mu(S_n)^{1/3}\Vert f\Vert_{L^{3/2}(S_n)}\le \sum_{n=1}^\infty n^{\,p(a)}\tag{4}\label{4} $$ Therefore, from this inequality we have that $$ p(a)<-1\implies f\in L^1(\mathbb{R}^2)\tag{5}\label{5} $$

Now, after those hint we are able to give a direct answer to the OP questions.

Answer to question 1. The exponent $3/2$ and $3/8$ are related by the condition $p(a)<-1$ of condition \eqref{5} for convergence of the series $\sum s_n$.

Answer to question 2. The funky domains $E_r$ should be used to construct a partition of $\mathbb{R}^2$ as a family of disjoint sets, such as one of the families $\{S_n\}_{n\in\mathbb{N}}$.

Answer to question 3. No. Intuitively, since the area of the "funky" region $E_r$, $r>0$ grows, its measure cannot vanish. More precisely, following the hint for the calculation of $\mu(E_r)$ above, we have that $$ |J(\rho,\theta)|=\frac{1}{2}\rho^\frac{1}{2}\left||\sin\theta|^\frac{1}{2}\sin\theta+|\sin\theta|^{-\frac{1}{2}}\cos^2\theta\right|.\\ $$ This implies that $$ \begin{align} \mu(E_r)=&\int\limits_r^{2r}\mathrm{d}\rho\int\limits_0^{2\pi}|J(\rho,\theta)|\mathrm{d}\theta\\ =&\frac{1}{2}\int\limits_r^{2r}\rho^\frac{1}{2} \mathrm{d}\rho\int\limits_0^{2\pi}\left||\sin\theta|^\frac{1}{2}\sin\theta+|\sin\theta|^{-\frac{1}{2}}\cos^2\theta\right|\mathrm{d}\theta\\ =&\frac{C_\theta}{2}\int\limits_r^{2r}\rho^\frac{1}{2} \mathrm{d}\rho = \frac{C_\theta}{3}\left[\rho^\frac{3}{2}\right]_r^{2r}=\frac{C_\theta}{3}\left[(2r)^\frac{3}{2}-r^\frac{3}{2}\right]\\ =&C_\theta\frac{\sqrt[2]{8}-1}{3}r^\frac{3}{2} \end{align} $$ where $C_\theta$ is the (constant) value of the trigonometric integral above, i.e. $$ C_\theta=\int\limits_0^{2\pi}\left||\sin\theta|^\frac{1}{2}\sin\theta+|\sin\theta|^{-\frac{1}{2}}\cos^2\theta\right|\mathrm{d}\theta $$ This implies $\mu(E_r)=O\left(r^\frac{3}{2}\right)$ as $r\to\infty$.