Let $$\frac{1}{\tau}=a\left(\frac{1}{p}-\frac{1}{d}\right)+\frac{1-a}{q},$$ $\tau>0, p\geq 1, a\in [0,1]$, $q\geq 1$ and $d\geq 1$. I know that $$\|u\|_{L^\tau(\mathbb R^d)}\leq C\|\nabla u\|_{L^p(\mathbb R^d)}^a\|u\|_{L^q(\mathbb R^d)}^{1-a}.\tag{E}$$
I want to prove that
$$\|u-\bar u\|_{L^\tau(\mathcal D)}\leq C\|\nabla u\|_{L^p(\mathcal D)}^a\|u-\bar u\|_{L^q(\mathcal D)}^{1-a},$$ where $$\mathcal D=\{x\in\mathbb R^d\mid r<|x|<R\},$$ and $$\bar u=\frac{1}{|\mathcal D|}\int_{\mathcal D}u.$$
Question) If $\frac{1}{p}-\frac{1}{d}>0$ it's a consequence of Sobolev inequality. But how can I conclude when $\frac{1}{p}-\frac{1}{d}\leq 0$ ? It's probably using a prolongement extension of $u-\bar u$ on $\mathbb R^d$ and using (E) but I don't see how to make it.