Here I would like to prove a result that I assume is known but I am having difficulty proving. I will give the set up. This problem is really coming from functions with are 'doubly radial' or 'multi-radial'. Take $ 3 \le m \le n$ and integer and consider functions $u=u(s,t)$, which are smooth for $1<s^2+t^2<4$ and assume the function is even with respect to $s$ and $ t$ and zero on the boundary of the annulus. Since the function is even and smooth we can work on the first quadrant and assume $u$ satisfies a Neumann BC on the intersection of the annulus and the axis.


Let $ 1 <p< \frac{2n}{n-2}$; i want to show the existence of a $C$ (indepandent of $u$) such that
$$ \left( \int\_{ \tilde{\Omega}} | u(s,t)|^p s^{m-1} t^{n-1} ds dt \right)^\frac{2}{p} \le C \int\_{\tilde{\Omega}} | \nabla\_{s,t} u(s,t)|^2 s^{m-1} t^{n-1} ds dt.$$ Here $ \tilde{\Omega}$ is the annulus or the annulus intersect the first quadrant. So inuitively this should be true since
if we stay away from the axis this is really just a two dimensional problem and we have good imbeddings. So it appears if it fails then there must be some concentration near one of the axis and then the inequality starts to look like one in dimension $n$ or $m$ (for radial functions) and one has a worse imbedding in dimension $n$.


Anyhow if I attempt to write out a 'proof' I fail. Any comments on this would be appreciated.



Sobolev imbedding result; proof
This first part is an **extended comment**: I think you have your scaling wrong.


Let $u\_k = \phi(\lambda\_k(s-s\_k), \mu\_n(t-t\_k))$, where $\phi$ has compact support in the ball of radius 1.


We assume that $(s\_k, t\_k) \to (1.5,0)$ from within $\tilde{\Omega}$.
That $\mu\_k > 2/t\_k$, and $\lambda\_k > 2$, so $u\_k$ has compact support in $\tilde{\Omega}$.


For large $k$, the LHS is approximately
$$ \int |u\_k(s,t)|^p s^{m-1} t^{n-1} ~ds ~dt \approx \int |\phi(\lambda\_k(s - s\_k), \mu\_k(t-t\_k))|^p t^{n-1} ~ds ~dt $$
Our assumption on $\mu\_k$ implies that the support is concentrated near $|t-t\_k| < t\_k/2$, so we can further replace by
$$ \approx t\_k^{n-1} \int |\phi(\lambda\_k s, \mu\_k t)|^p ~ds ~dt = t\_k^{n-1} \lambda\_k^{-1} \mu\_k^{-1} \|\phi\|\_{L^p}^p$$


The RHS on the other hand is
$$ \int |\nabla u\_k|^2 s^{m-1} t^{n-1} ~ds ~dt \approx $$
$$ \int \left[ \lambda\_k^2 |\partial\_s \phi(\lambda\_k (s-s\_k), \mu\_k(t-t\_k))|^2 + \mu\_k^2 |\partial\_t \phi(\lambda\_k(s - s\_k), \mu\_k(t-t\_k))|^2 \right] t^{n-1} ~ds ~dt $$
The support property for $t$ gives
$$ \approx t\_k^{n-1} (\lambda\_k \mu\_k^{-1} + \lambda\_k^{-1} \mu\_k) \|\nabla \phi\|\_{L^2}^2 $$


For the desired inequality to hold you need


$$ t\_k^{(n-1)(1/2 - 1/p)} (\lambda\_k^{1/p+1/2} \mu\_k^{1/p-1/2} + \lambda\_k^{1/p-1/2} \mu\_k^{1/p+1/2} ) \overset{?}{\gtrsim} 1$$


Let $\lambda\_k = \kappa \mu\_k$ for $\kappa > 1$, this requires
$$ t\_k^{(n-1)(1/2 - 1/p)} \mu^{2/p} (\kappa^{1/p+1/2} + \kappa^{1/p-1/2}) \overset{?}{\gtrsim} 1 $$
Choose $\mu\_k = 3/t\_k$, we need
$$ t\_k^{(n-1)(1/2 - 1/p) - 2/p} \overset{?}{\gtrsim} 1 $$
and since $t\_k\to 0$ we need
$$ 2/p \geq (n-1)(1/2-1/p) $$
which requires
$$ 1/p \geq \frac{n-1}{2(n+1)} $$
or
$$ p \leq \frac{2(n+1)}{n-1} $$


**This makes sense** as in the limit where the function is concentrated near one of the axes, you should be looking at a function defined on $\mathbb{R}\times\mathbb{R}^n$, radial in the second variable. So the Sobolev scaling should be computed based on the total dimension $(n+1)$, not on $n$.




---


Now the proof should be fairly simple.


First, for convenience, use polar coordinates so you have $s = r\cos\theta$ and $t = r\sin\theta$. You are integrating with the measure


$$ r^{m+n-1} ~dr~ \cos(\theta)^{m-1} \sin(\theta)^{n-1} ~d\theta $$


You also have $|\nabla u|^2 = |\partial\_r u|^2 + \frac1{r^2} |\partial\_\theta u|^2$. Since you are working on the annulus, you have that $r$ is uniformly bounded between $[1,2]$.


On the sector, you further have that $\sin(\theta) \approx \theta$ and $\cos(\theta) \approx $\pi/2 - \theta$. So rescaling once your desired Sobolev inequality is equivalent to


$$ \|u \|\_{L^p\_w} \lesssim \|\nabla u\|\_{L^2\_w} $$


on the domain $(z,\theta) \in [0,1]^2$
where the weighted measure is


$$ \theta^{n-1}(1-\theta)^{m-1} ~dz~d\theta $$.


Now fix a cut-off function $\chi(\theta)$ that is identically 0 for $\theta < 1/3$ and identically $1$ for $\theta > 2/3$.


Now,
$$ \|\partial\_\theta (\chi u)\|\_{L^2\_w} \leq \| \chi \partial\_\theta u\|\_{L^2\_w} + \|\chi' u\|\_{L^2\_w} $$
For the final term you have that since $\chi'$ is supported away from $\theta = 0,1$, you have can apply Poincare's inequality to $u$ which vanishes at the end points, so that
$$ \|\chi' u\|\_{L^2\_w} \lesssim \| \partial\_\theta u\|\_{L^2\_w}$$


And so with the cut-off being non-harmful, it suffices to prove that


$$ \|\chi u\|\_{L^p\_w} \leq \|\nabla (\chi u)\|\_{L^2\_w} $$
and
$$ \|(1-\chi)u\|\_{L^p\_w} \leq \|\nabla (1-\chi)u\|\_{L^2\_w} $$


For each individual one, now you can think of $\chi u$ as a function defined on $[0,1] \times B(0,1)$ where the ball is considered as a subset of $\mathbb{R}^n$. And your claim should now follow from the standard Sobolev inequality in dimension $n+1$.


