Let $$\frac{1}{r}+\frac{\gamma }{d}=a\left(\frac{1}{p}+\frac{\alpha -1}{d}\right)+(1-a)\left(\frac{1}{q}+\frac{\beta }{d}\right),$$
where $d\geq 1$, $a\in [0,1]$, $\alpha ,\beta,\gamma \in \mathbb R$.

There exists $C$ independent of $u$ s.t. $$\||x|^\gamma u\|\_{L^r(\mathbb R^d)}\leq C\||x|^\alpha \nabla u\|\_{L^p(\mathbb R^d)}^{a}\||x|^\beta u\|\_{L^q(\mathbb R^d)}^{1-a}$$ for all $u\in\mathcal C\_0^1(\mathbb R^d)$. How can I interpret this ? 

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For example, if $\frac{1}{r}=\frac{a}{p}+\frac{1-a}{q}$, I know that for all $u\in L^p\cap L^q$ we have that $$\|u\|\_{L^r}\leq \|u\|\_{L^p}^a\|u\|\_{L^q}^{1-a}$$
what can be interpreted as if $u\in L^p\cap L^q$, then $u\in L^r$ for all $r\in [p,q]$. Or in other word $L^r\supset L^p\cap L^q$ for all $r\in [p,q]$.

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**Q1)** For my case, we have that $u\in \mathcal C\_0^1 (\mathbb R^d)$ so it seems to be normal that $u\in L^p$ and $u\in W^{1,p}$ for all $p\geq 1$... but may be there is a density argument behind ? 

**Q2)** So may be if $\alpha =\gamma =\beta =0$ I can say that $L^r\supset W^{1,q}\cap L^{p^\*}$ for all $r\in [p^\*,q]$ where $\frac{1}{p^\*}=\frac{1}{p}-\frac{1}{d}$ ? But it doesn't look incredible...

**Q3)** And what are those $|x|^\gamma, |x|^\alpha $ and $|x|^\beta $ ? Which information do they give us ?

**Q4)** Could you also give me an application of such an inequality ?