I am interested in the operator $T: v\mapsto u$ where $u$ is the weak solution of the following well-studied Elliptic PDE \begin{align} -\text{div} (A\nabla u) &=v \text{ in } \Omega,\\ u & =0 \text{ on } \partial \Omega. \end{align} Here, for each $x\in \Omega$, $A(x)$ is a symmetric matrix such that, exists $c>0$, $y^TA(x)y>c||y||^2_2$ for all vector $y$.
It is well known that the operator $T$ is upper bounded (continuous). I am wondering if it is lower bounded or not, i.e. does there exist a constant $\gamma>0$ such that \begin{align} ||T v||\geq \gamma ||v|| \text{ for all } v. \end{align}
I know that an operator is lower bounded iff it is injective and has closed range (see Theorem 2.5 here). However, since I am not familiar with PDE, I don't know where to begin. Could any one give me some clues to prove it? Thank you!
No, this operator is not lower bounded in your sense. If it were, we would have $$ \gamma \int_\Omega |\nabla u|^2\leq c\gamma\int_\Omega A\nabla u\cdot\nabla u=c\gamma\int_\Omega vu\leq c\gamma\|u\|_2\|v\|_2\leq c\|u\|_2^2 $$ for $u\in C_c^\infty(\Omega)$. To see that this is not true, take any $u\in L^2(\Omega)\setminus H^1(\Omega)$ and a sequence $(u_n)$ in $C_c^\infty(\Omega)$ such that $u_n\to u$ in $L^2$.