# Does $(q,\nabla u)\lesssim C|u|_1$ implies $\Vert q\Vert_0\lesssim C$?

I am deriving the error estimates of one nonconforming finite element method, and for the error of the gradient of true solution, $\mathbf{q}_h$, which is a piecewise constant vector, I have the following results:

$\Omega$ is an open bounded domain, and for any function $v\in H^1(\Omega)$ and $v=0$ on a nontrivial portion of boundary $\Gamma_1$, we have \begin{equation} \sum_{T\in\mathcal{T}}(\mathbf{q}_h,\nabla v)_{0,T}\lesssim C\Vert\nabla v\Vert_0 \end{equation} where $(\cdot,\cdot)_{0,T}$ denotes the $L^2$ inner product on each element $T$ in triangulation $\mathcal{T}$ of $\Omega$.

From this inequality, can we have any estimate about the norm of $\mathbf{q}_h$. For example, can we say that \begin{equation} \Vert \mathbf{q}_h\Vert_0\lesssim C? \end{equation}

Define $C:=\min\{\tilde C>0:(q,\nabla v)\leq \tilde C\,|u|_1\}$.

We know that, for all $|u|_1\neq 0$,

$$\dfrac{(q\,,\nabla u)}{|u|_1}\leq C$$

and, thanks Cauchy--Schwarz inequality,

$$\dfrac{(q\,,\nabla u)}{|u|_1}\leq \|q\|_0$$

for all $q$.

What happen if we suppose that exists $\tilde q$ such that $\|\tilde q\|_0>C$?

• I don't quite understand your answer, since that $C$ and $\m{q}_h$ is settled and we can only choose different $u$ to test this inequality. I have worked out other ways to estimate my specific $\m{q}_h$ using some additional condition. Thanks for your answer. – Y. Liang Feb 6 '18 at 5:38