Suppose we consider in $\mathbb R^n$, then how to show $\Vert f \Vert_{L^{p}} \leq C\Vert \nabla^{s}f \Vert_{L^{q}}^{\alpha}$, where $s>0$ is noninteger and $\alpha \in (0,1)$?

$\begingroup$ Can you be more precise: what are $p$ and $q$ and what do you mean by $\nabla^s f$? $\endgroup$ – Piotr Hajlasz Nov 21 '18 at 14:31
Such an inequality cannot be true unless $f=0$. That can be proved by a standard homogeneity argument. Suppose that $\Vert \nabla^sf\Vert_p<\infty$. Replacing $f$ by $tf$, where $t>0$ we have $$ t\Vert f\Vert_p=\Vert tf\Vert_p\leq C\Vert \nabla^s(tf)\Vert_p^\alpha=t^\alpha C\Vert \nabla^sf\Vert_p^\alpha, \quad \Vert f\Vert_p\leq t^{\alpha1} C\Vert \nabla^sf\Vert_p^{\alpha}, $$ Since $\alpha1<0$, letting $t\to\infty$, the right hand side will converge to $0$ and hence $\Vert f\Vert_p=0$, $f=0$.

$\begingroup$ And on the positive side, one can find a dozen of related (and true!) inequalities of similar sort for example in Samko's book Hypersingular Integrals and Their Applications. $\endgroup$ – Mateusz Kwaśnicki Dec 18 '18 at 21:02
I guess that your question is illformulated: you have to respect the homogeneity on both sides of the inequality. Let us say that for $f$ in the Schwartz space you always have $$ \Vert f\Vert_{W^{t,q}(\mathbb R^n)}\lesssim \Vert \vert D\vert^s f \Vert_{L^{p}(\mathbb R^n)}\quad\text{(and also}\quad \Vert f\Vert_{L^{n/(n1)}(\mathbb R^n)}\lesssim \Vert D f\Vert_{L^{1}(\mathbb R^n)}), $$ provided $ \frac{st}{n}=\frac{1}{p}\frac{1}{q},\quad 1<p<q<+\infty. $ Applying the second inequality above to $f=u^2$, you obtain for $p>1$, $$ \Vert u\Vert_{L^{2n/(n1)}(\mathbb R^n)}^2\lesssim \Vert uD u\Vert_{L^{1}(\mathbb R^n)}\lesssim \Vert u\Vert_{L^{p'}(\mathbb R^n)} \Vert D u\Vert_{L^{p}(\mathbb R^n)} \lesssim \Vert u\Vert_{L^{p'}(\mathbb R^n)} \Vert \vert D\vert u\Vert_{L^{p}(\mathbb R^n)}, $$ so that $$ \Vert u\Vert_{L^{2n/(n1)}(\mathbb R^n)} \lesssim \Vert u\Vert_{L^{p'}(\mathbb R^n)}^{1/2} \Vert \vert D\vert u\Vert_{L^{p}(\mathbb R^n)}^{1/2}, $$ an inequality resembling yours. Choosing $f= u^\alpha$ leads to more general interpolation inequalities.