Consider arbitrary elements $1 \leqslant p < \infty$ and $0 < \lambda \leqslant n$. Furthermore, during this post I considered the usual Lebesgue measure over $\mathbb R^n$ and I denote the Lebesgue measure of a set by $|\cdot|$.
My goal is to show that every essentialy bounded function $f$ with compact support satisfies the following two conditions:
$$ f \in L^p_{\text{loc}}(\mathbb R^n) \quad \text{ and } \quad \sup_{x \in \mathbb R^n, \, r > 0} r^{-\lambda} \int_{B(x,r)} |f(y)|^p \, dy < \infty. $$
My attempt. I was able to show that $f \in L^p_{\text{loc}}(\mathbb R^n)$ in the following way: for every compact set $K \subset \mathbb R^n$ we have that
$$ \int_K |f(y)|^p \, dy \leqslant \| f \|_{L^\infty(\mathbb R^n)}^p \int_K 1 \, dy = \| f \|_{L^\infty(\mathbb R^n)}^p |K| < \infty, $$
since $f$ is essentially bounded by hypothesis and since compact sets have finite Lebesgue measure.
Therefore, all that's left to prove is that $f$ also satisfies the second property. Note that to show that $f$ is locally $p-$integrable everything I needed to use was the essential boundedness of $f$, but such simplicity doesn't occur for the second property. Let us see why this is the case: simple calculations tell us that
$$ \sup_{x \in \mathbb R^n, \, r > 0} r^{-\lambda} \int_{B(x,r)} |f(y)|^p \, dy \leqslant \sup_{r > 0} r^{-\lambda} \| f \|_{L^\infty(\mathbb R^n)}^p |B(x,r)| = \sup_{r > 0} r^{n-\lambda} \| f \|_{L^\infty(\mathbb R^n)}^p |B(0,1)|, $$
and the right-hand side supremum is not necessarily finite for $0 < \lambda \leqslant n$. So I believe that I shall use the fact that $f$ has compact support to prove this property but I am having a hard time trying to do so.
Thanks for any help in advance.
Best Answer
So what you essentially want to show that $L^\infty_c(\mathbb{R}^n)\hookrightarrow L^{p,\lambda}(\mathbb{R}^n)$ for $0 < \lambda \leq n$, where the latter space is the Morrey space. Your calculations already show that $L^\infty_c(\mathbb{R}^n)\hookrightarrow L^{p,n}(\mathbb{R}^n)$. Next, we prove that $L^{p,n}(\mathbb{R}^n) \cap L^\infty_c(\mathbb{R}^n)\hookrightarrow L^{p,\lambda}(\mathbb{R}^n)$ for all $0 < \lambda < n$.
Let $R>0$ such that $\text{supp}\;f\subset B(0,R)$. Then for all $r\in (0,R)$ $$ \int_{B(x,r)}\vert f \vert^p\;dx = r^{n} r^{-n}\int_{B(x,r)}\vert f \vert^p \;dx \leq r^{\lambda} R^{n-\lambda} r^{-n}\int_{B(x,r)}\vert f \vert^p \;dx.$$ Thus, $$ \sup_{x\in\mathbb{R}^n, r > 0} r^{-\lambda}\int_{B(x,r)}\vert f \vert^p \;dx \leq R^{n-\lambda}\sup_{x\in\mathbb{R}^n, r > 0} r^{-n}\int_{B(x,r)}\vert f \vert^p \;dx, $$ which concludes the proof.
More generally, you can show that $$ L^{q,\mu}(\Omega) \hookrightarrow L^{p,\lambda}(\Omega)$$ for $1\leq p \leq q <\infty$, $\lambda, \mu \geq 0$ with $\frac{\lambda - n}{p} \leq \frac{\mu-n}{q}$, and $\Omega\subset\mathbb{R}^n$ is a bounded domain.