I'd like to show the two following inequalities:
For $s=0,1$, for all $v$ in $H^{s+1}(\Omega) \cap H_0^1(\Omega)$: $\|v\|^2_s \leq \sqrt2 \|v\|_{s-1}\|v\|_{s+1}$.
I recall that $H^{-1}(\Omega)=(H^1_0(\Omega))'$ and $\|v\|_{-1}=\sup_{w \in H^1_0(\Omega)} \frac{(v,w)_{L^2}}{\|w\|_1}.$
I don't have much idea yet, so I'd appreciate some help. For $s=1$, I've been thinking of dividing my space $\Omega$ into a triangular mesh $g$, so that I get: $\|v\|^2_{H^1(\Omega)}=\sum_{\tau \in g} \|v\|^2_{H^1(\tau)}=\sum_{\tau \in g} \frac{\lvert \tau \rvert}{\lvert \hat{\tau} \rvert}\|\hat{v}\|^2_{H^1(\tau)}=2\sum_{\tau \in g} \lvert \tau \rvert \|\hat{v}\|^2_{H^1(\hat{\tau})}$,
where $\hat{\tau}$ is the reference triangles of vertices $\{(0,0),(1,0),(0,1)\}$ and $\hat{v}$ is the tranformation of $v$ onto $\hat{\tau}$.
This makes a 2 pop, which is a good sign, but still I don't know how to continue. Thanks.
For $s=0$, we have $$ \|v\|_{0}^2 = (v,v)_{L^2} = \|v\|_1 \frac{(v,v)_{L^2}}{\|v\|_1} \le \|v\|_1 \sup_{0\neq w\in H^1_0} \frac{(v,w)_{L^2}}{\|w\|_1} = \|v\|_1\|v\|_{-1}.$$ For $s=1$, I can get a constant that depends on the dimension $d$ by elementary means. This constant depends on the choice of Sobolev norm; if you choose $\|v\|_k=\sum_{|\alpha|\le k} \|D^\alpha v\|_0$ then the below gives the constant 1. Since $v|_{\partial \Omega}=0,$ $$ \|\nabla v\|_{0}^2 = \int_{\Omega} \nabla v\cdot\nabla v dx = -\int_{\Omega}(\Delta v)v dx \le \|\Delta v\|_0\|v\|_0,$$ Hence $$ \|v\|_1^2 = \|v\|_0^2 + \|\nabla v\|_ 0^2 \le (\|v\|_0 + \|\Delta v\|_0 )\|v\|_0 \le \sqrt{2d} \|v\|_2\|v\|_0$$ where we used
\begin{align} \|\Delta v\|_0^2 &= \int_{\Omega} \Big|\sum_{i=1}^d \partial_i^2v\Big|^2 dx \le d\int_{\Omega}\sum_{i=1}^d|\partial_i^2 v|^2dx, \end{align} so that (with $a+b\le\sqrt 2\sqrt{a^2+b^2}$) $$ \| v\|_0+\|\Delta v\|_0\le \sqrt2\sqrt{\| v\|_0^2+\|\Delta v\|_0^2} \le \sqrt{2d}\sqrt{\sum_{|\alpha|\le 2} \|D^\alpha v\|_0^2}=\sqrt{2d}\|v\|_2.$$