Can I use inverse inequality for an infinite space, $H^1$? Let $u \in H^1(\Omega)$ and $Q_0 u$ is a $L^2$ projection of $u$ to a polynomial finite space $P_k(T)$ where $T \in \mathcal{T}_h$ is a finite element and $\mathcal{T}_h$ is a set of all the elements.
Can I combine two terms by using inverse inequality as below:
\begin{eqnarray*}
\sum_{T \in \mathcal{T}_h} \|Q_0 u - u \|_{\partial T} &\le& C(\sum_{T \in \mathcal{T}_h} h ^{-1} \| Q_0 u - u \|_T ^2  + \sum_{T \in \mathcal{T}_h} h \| \nabla ( Q_0 - u)\|_T ^2)^{\frac{1}{2}}\\ 
&\le& C(\sum_{T \in \mathcal{T}_h}h \| \nabla ( Q_0 - u)\|_T ^2)^{\frac{1}{2}} \\
&\le& Ch^k \| u \|_{k+1}
\end{eqnarray*}
I used trace inequality to a boundary term and try to use an inverse inequality to bound my error.  Is this correct?  If it is not, can someone explain to me why not?
Thanks
Anna 
 A: The magnitude of your final estimate does not look correct.
Rule of thumb: the approximation property loses $1/2$ when moving to the boundary because of the regularity decreasing of $1/2$ for the function $u$.
Let's say $\Omega\subset \mathbb{R}^2$:
$$
\|Qu - u\|_{0,\partial T} \lesssim h^{1/2}\|u\|_{\frac12,\partial T} 
\lesssim h^{1/2}|u|_{1, T}. 
$$ 
Then you lose another order of $h$ when you consider the $L^1$-sum of $\|Qu - u\|_{\partial T}$. If we follow what you did in the question, the estimate should be something like
$$
\sum_{T \in \mathcal{T}_h} \|Q u - u \|_{0,\partial T}  \lesssim \sum_{T \in \mathcal{T}_h} \|Q u - u \|_{0, T}^{1/2} \|Q u - u \|_{1, T}^{1/2} 
\\
\lesssim  \sum_{T \in \mathcal{T}_h} \left(h^{-1} \|Q u - u \|_{0, T} + h\|Q u - u \|_{1, T}\right) ,
$$
which will not give you an $h$ in front of the final estimate if finally you wanna use $W^{k,2}$-norm ($H^k$-norm) on the whole domain. 
But if it is $L^2$-sum, the standard argument is:
$$
\sum_{T \in \mathcal{T}_h} \|Q u - u \|_{0,\partial T}^2 \lesssim \sum_{T \in \mathcal{T}_h} \|Q u - u \|_{0, T} \|Q u - u \|_{1, T} 
\\
\lesssim  \left(\sum_{T \in \mathcal{T}_h}  \|Q u - u \|_{0, T}^2\right)^{1/2}  \left(\sum_{T \in \mathcal{T}_h}  \|Q u - u \|_{1, T}^2\right)^{1/2} \lesssim h |u|_{1,\Omega}^2.
$$
Take square root of both sides you will see that missing $h^{1/2}$ in the estimate vs the interior estimate in each element. Above is just the standard result of $L^2$-projection's approximation property, for example this widely cited paper in the finite element community.
Last thing is that what you used is not the inverse inequality either, the inverse inequality applies only on polynomial space, and it is bounding the $H^{k+1}$-norm by a lower order norm, for example
$$
|Qu|_{1,T} \leq h^{-1} \|Qu\|_{0,T},
$$
it is called "inverse" because normally by Poincaré type inequality, the estimate is the other way around.
