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 kind of Gagliardo-Nirenberg inequality proof
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.


