Let $\Omega$ be a bounded domain ($|\Omega| < \infty$, where $|\cdot|$ represents the Lebesgue measure), $1 \leqslant p \leqslant q < \infty$, $0 \leqslant \lambda, \mu \leqslant n$ such that
$$ \frac{\lambda - n}{p} \leqslant \frac{\mu-n}{q}, $$
where $n \in \mathbb{N}$ is arbitrary. My goal is to show that
$$ \left( r^{-\lambda}\int_{B(x,r) \cap \Omega} |f(y)|^p \, dy\right)^{1/p} \leqslant c \left( r^{-\mu} \int_{B(x,r) \cap \Omega} |f(y)|^q \, dy \right)^{1/q},$$
for every $ r > 0$ and $x \in \mathbb{R^n}$, where $c$ is some positive constant ($c > 0$).
My attempt. It seems to me that the only way to approach this problem is to apply Hölder's Inequality with exponent $q/p$. Doing so, one obtains that
$$ \int_{B(x,r) \cap \Omega} |f(y)|^p \, dy \leqslant \left( \int_{B(x,r) \cap \Omega} |f(y)|^q \, dy \right)^{p/q} \, |B(x,r) \cap \Omega|^{1-p/q}. $$
Now, I try to use the fact that $\Omega$ is a bounded domain in order to find the constant $c$. With this in mind, I use the bound $|B(x,r) \cap \Omega| \leqslant |\Omega|$ to conclude that
$$ \int_{B(x,r) \cap \Omega} |f(y)|^p \, dy \leqslant \left[ |\Omega|^{1/p - 1/q} \left(\int_{B(x,r) \cap \Omega}|f(y)|^q \, dy \right)^{1/q} \right]^p $$
or, equivalently,
$$ \left( r^{-\lambda} \int_{B(x,r) \cap \Omega}|f(y)|^p \, dy \right)^{1/p} \leqslant r^{-\lambda/p} |\Omega|^{1/p - 1/q} \left( \int_{B(x,r) \cap \Omega} |f(y)|^q \right)^{1/q}. $$
Now, if I could guarantee that $r^{-\lambda/p} \leqslant r^{-\mu/q},$ the proof would be complete and I would conclude that the constant $c$ is just $|\Omega|^{1/p - 1/q}$. The problem is that I don't think we can guarantee that $r^{-\lambda/p} \leqslant r^{-\mu/q}$. Let me elaborate:
By hypothesis, we know that
$$ \frac{\lambda - n}{p} \leqslant \frac{\mu - n}{q}. $$
Solving this for $-\lambda/p$, we have that
$$ 0\geqslant-\frac{\lambda}{p} \geqslant -\frac{\mu}{q} + \frac{n}{q} - \frac{n}{p},$$
which doesn't allow me to conclude that $-\lambda/p \leqslant -\mu/q$, since $n/q - n/p \leqslant 0$.
With this in mind, I am posting this question in order to find any possible mistake that I've done or to find another approach to this problem.
I am mainly looking for hints and not full answers.
Thanks for any help in advance.
We need to assume that $$ \frac{\mu-n}q - \frac{\lambda-n}p \ge 0 \ge \frac{\mu}q - \frac{\lambda}p. $$ The first inequality is an assumption in the question. The second is necessary to prove the claim. At the end there is a counterexample to the claim if $\frac{\mu}q - \frac{\lambda}p>0$.
By this assumption, there is $s\in [0,1]$ such that $$ s(\frac{\mu-n}q - \frac{\lambda-n}p) + (1-s)( \frac{\mu}q - \frac{\lambda}p) =0 $$ or equivalently $$ s( \frac np- \frac{n}q) + \frac{\mu}q - \frac{\lambda}p =0. $$
Using H"older inequality $$ r^{-\lambda} \int_{B_r \cap \Omega} |f|^p \le r^{-\lambda} \ \left(\int_{B_r\cap \Omega} |f|^q \right)^{p/q} |B_r\cap \Omega|^{1-\frac pq} . $$ We continue with $$ |B_r\cap \Omega| \le |B_r|^s |\Omega|^{1-s} \le c\ r^{ns} . $$ Then we get $$ r^{-\lambda} \int_{B_r \cap \Omega} |f|^p \le c\ r^{-\lambda} \ \left(\int_{B_r\cap \Omega} |f|^q \right)^{p/q} r^{ns(1-\frac pq)} $$ so that $$ \left(r^{-\lambda} \int_{B_r} |f|^p \right)^{1/p} \le c\ r^{-\frac\lambda p + s(\frac np - \frac nq) + \frac\mu q}\left(r^{-\mu} \int_{B_r} |f|^q \right)^{1/q}. $$ The exponents add up to zero, hence we get the desired inequality (without restrictions on $r$): $$ \left(r^{-\lambda} \int_{B_r} |f|^p \right)^{1/p} \le c\ \left(r^{-\mu} \int_{B_r} |f|^q \right)^{1/q} \quad \forall r>0. $$
If you do not like additional assumptions, there is bad news: if $\frac{\mu}q - \frac{\lambda}p>0$ then the only $L^q(\mathbb R^n)$-function satisfying the claim for all $r>0$ is $f=0$.
Suppose the desired estimate $$ \left(r^{-\lambda} \int_{B_r} |f|^p \right)^{1/p} \le c\ \left(r^{-\mu} \int_{B_r} |f|^q \right)^{1/q}. $$ is valid for all $r>0$. Rearranging gives $$ \left( \int_{B_r} |f|^p \right)^{1/p} \le c \ r^{-\frac\mu q + \frac\lambda p } \left( \int_{B_r} |f|^q \right)^{1/q}. $$ For $r\to\infty$, the terms in brackets converge to the $L^p(\mathbb R^n)$- and $L^q(\mathbb R^n)$ norm of $f$ (extend $f$ by zero to $\mathbb R^n$). Hence if $f\in L^q(\mathbb R^n)$ then the right-hand side converges to zero, hence $f=0$.