A kind of Gagliardo-Nirenberg inequality proof Could any one give a proof for this inequility here? I just know its some kind of Gagliardo-Nirenberg inequility, but where does the second term come from? Thx~
$$
\int_{B_r}|u|^q\le C\left(\int_{B_r}|\nabla u|^2\right)^a \left(\int_{B_r}|u|^2\right)^{\frac{q}{2}-a}+\frac{C}{r^{2a}} \left(\int_{B_r}|u|^2\right)^{\frac{q}{2}}
$$
$$
2\le q\le 6, a=\frac{3}{4}(q-2)
$$
Here $B_r$ is a ball of radius $r$ and $C$ is a constant independent of $r$. If complete detailed proof could be provided, that will help me a lot!!
 A: By context, I infer that you are working in 3 dimensions.
Step 1: reduce to case $r = 1$
Consider the mappings $S_r$ that map
$$ S_r u(x) := r^{-3/q} u(r^{-1} x) $$
We have that $S_r: L^q(B_1) \to L^q(B_r)$ is an isometric bijection.
Note that
$$ \|\nabla S_r u \|_{L^2(B_r)} = r^{-3/q - 1 + 3/2} \|\nabla u\|_{L^2(B_1)} $$
and
$$ \| S_r u\|_{L^2(B_r)} = r^{3/2 - 3/q} \|u\|_{L^2(B_1)} $$
A direct computation shows that if we know
$$ \|u\|_{L^q(B_1)}^q \leq C \|\nabla u\|_{L^2(B_1)}^{2a} \|u\|_{L^2(B_1)}^{q-2a} + C \|u\|_{L^2(B_1)}^{q} $$
Then for $u\in H^1(B_r)$, applying the inequality to $S_r^{-1} u\in H^1(B_1)$ and using the scaling identity above we see that the desired inequality holds for all $r$.
Step 2: Extension and GNS
The unit ball has $C^1$ boundary and is a Sobolev extension domain. In particular, there exists a bounded linear operator $E: H^1(B_1) \to H^1_0(B_2)$.
Gagliardo-Nirenberg-Sobolev inequality says that
$$ \|Eu\|_{L^6(B_2)} \leq C \|\nabla Eu\|_{L^2(B_2)} \leq C \|Eu\|_{H^1(B_2)}   $$
So (for a potentially different $C$)
$$ \|u\|_{L^6(B_1)} \leq \|Eu\|_{L^6(B_2)} \leq C\|u\|_{H^1(B_1)}$$
Step 3: interpolate
Apply the standard interpolation inequality
$$ \|u\|_{L^q} \leq \|u\|_{L^2}^{1-\theta} \|u\|_{L^6}^{\theta} $$
for an appropriate $\theta\in [0,1]$ and $q\in [2,6]$, you get
$$ \|u\|_{L^q(B_1)}^q \leq C \|u\|_{L^2(B_1)}^{1-a} \|u\|_{H^1(B_1)}^{a} $$
Finally expand
$$ \|u\|_{H^1}^a \leq C'' \|u\|_{L^2}^a + C'' \|\nabla u\|_{H^1}^{a} $$
and you are done.
