Let $\Omega$ be a bounded domain in $\mathbb{R}^3$ and let $u\in H_0^1(\Omega)$. I need to prove $$\|u\|_4 \leq 2^{1/2} \|u\|_2^{1/4}\|\nabla u\|_2^{3/4}. \tag{$*$}\label{aim}$$
First, since $$\frac14 = \frac{3/4}{6} + \frac{1/4}{2},$$ we have by interpolation inequality in $L^p$-spaces that $$\|u\|_4 \leq \|u\|_6^{3/4} \|u\|_2^{1/4}. \tag{1}\label{1}$$ Then by Sobolev-Gagliardo-Nirenberg inequality $$\|u\|_{p^*} \leq \frac{p(n-1)}{n-p} \|\nabla u\|_p$$ (cf. Evans' book, Chapter 5) and the fact that 6 is the Sobolev conjugate of 2 in $\mathbb{R}^3$, we have that $$\|u\|_6 \leq 4\|\nabla u\|_2, \tag{2}\label{2}$$ where the constant 4 is the value of $\frac{p(n-1)}{n-p}$.
By \eqref{1} and \eqref{2} we get $$\|u\|_4 \leq 2^{3/2} \|u\|_2^{1/4} \|\nabla u\|_2^{3/4}.$$
Using this method, one can not obtain the better constant $2^{1/2}$ in \eqref{aim}.
My question is:
- How to obtain the better constant $2^{1/2}$ in \eqref{aim}?
- Is $2^{1/2}$ optimal?