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Let $\Omega$ be a regular domain of $\mathbb{R}^d$, $d=2,3$. Let $\mathcal{T}_h$ be a triangulation of $\Omega$ of size $h>0$. Assume we can prove \begin{equation*} \begin{aligned} \|v\|_{L^2(\Omega)}&\le C h\|v\|_{H^1(\Omega)}\\ \|v\|_{L^2( \Omega)}&\le C \|v\|_{L^2(\Omega)}, \end{aligned} \end{equation*} then by interpolation we get \begin{equation} \|v\|_{L^2(\Omega)}\le C h^s\|v\|_{H^s_i(\Omega)}\qquad(1), \end{equation} where $H^s_i(\Omega)=H^s(\Omega)$ if $0\le s<\frac{1}{2}$, $H^{\frac{1}{2}}_i(\Omega)=H^{\frac{1}{2}}_{00}(\Omega)$ and $H^s_i(\Omega)=H^s_{0}(\Omega)$ if $\frac{1}{2}< s\le 1$.

How to prove the inequality $(1)$?

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If you have two couple of Banach spaces $(X_{0},X_{1})$ and $(Y_{0},Y_{1})$ and $T:X_{0}+X_{1}\rightarrow Y_{0}+Y_{1}$ a linear operator such that $T$ is bounded from $X_{i}$ to $Y_{i}$, $i=0,1$, then if you consider the interpolation spaces $X_{\theta}:=(X_{0},X_{1})_{q,\theta}$ and $Y_{\theta }:=(Y_{0},Y_{1})_{q,\theta}$ you have the interpolation inequality$$ \Vert T\Vert_{L(X_{\theta};Y_{\theta})}\leq C\Vert T\Vert_{L(X_{0};Y_{0}% )}^{1-\theta}\Vert T\Vert_{L(X_{1};Y_{1})}^{\theta}. $$ See for example wikipedia-interpolation From this inequality you get \begin{align*} \Vert T(x)\Vert_{Y_{\theta}} & \leq\Vert T\Vert_{L(X_{\theta};Y_{\theta}% )}\Vert x\Vert_{X_{\theta}}\\ & \leq C\Vert T\Vert_{L(X_{0};Y_{0})}^{1-\theta}\Vert T\Vert_{L(X_{1};Y_{1}% )}^{\theta}\Vert x\Vert_{X_{\theta}}% \end{align*} for all $x\in X_{\theta}$. Now you have to use the facts that (see the book of Lions and Magenes Non-homogeneous boundary problems and applications, volume I) \begin{align*} L^{2}(\Omega) & =(L^{2}(\Omega),L^{2}(\Omega))_{2,s},\\ H^{s}(\Omega) & =(L^{2}(\Omega),H_{0}^{1}(\Omega))_{2,s},\quad0\leq s<\frac{1}{2},\\ H_{00}^{1/2}(\Omega) & =(L^{2}(\Omega),H_{0}^{1}(\Omega))_{2,1/2},\\ H_{0}^{s}(\Omega) & =(L^{2}(\Omega),H_{0}^{1}(\Omega))_{2,s},\quad s>\frac {1}{2}. \end{align*} So you will take $X_{0}=L^{2}(\Omega)$ and $X_{1}=H_{0}^{1}(\Omega)$ and $Y_{0}=L^{2}(\Omega)$ and $Y_{1}=L^{2}(\Omega)$ and consider the operator $T:L^{2}(\Omega)+H_{0}^{1}(\Omega)\rightarrow L^{2}(\Omega)+L^{2}(\Omega)$ to be the identity $T(v):=v$. Then by your hypothesis $T:H_{0}^{1}(\Omega )\rightarrow L^{2}(\Omega)$ is bounded with$$\Vert T(v)\Vert_{L^{2}}\leq Ch\Vert v\Vert_{H^{1}}% $$ for all $v\in H_{0}^{1}(\Omega)$, which gives that $\Vert T\Vert_{L(H_{0}% ^{1}(\Omega);L^{2}(\Omega))}\leq Ch$. Similarly $\Vert T\Vert_{L(L^{2}% (\Omega);L^{2}(\Omega))}=1$. Since $H_{s}=H_{i}^{s}(\Omega)$ and $Y_{s}% =L^{2}(\Omega)$ it follows from the inequality% \begin{align*} \Vert v\Vert_{L^{2}} & =\Vert T(v)\Vert_{L^{2}(\Omega)}\leq C\Vert T\Vert_{L(L^{2}(\Omega);L^{2}(\Omega))}^{1-s}\Vert T\Vert_{L(H_{0}^{1}% (\Omega);L^{2}(\Omega))}^{s}\Vert v\Vert_{H_{i}^{s}(\Omega)}\\ & \leq C1^{1-\theta}(Ch)^{s}\Vert v\Vert_{H_{i}^{s}(\Omega)}=C(Ch)^{s}\Vert v\Vert_{H_{i}^{s}(\Omega)}% \end{align*} for all $v\in H_{i}^{s}(\Omega)$.

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