Interpolation inequalities 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 $(1)$?
 A: 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 $v\in H_{i}^{s}(\Omega)$.
