I have several problems in showing this point of a problem: we consider $X$ Banach space and $T: D(T) \to X$ a closed operator with domain $D(T) \subseteq X$. Let be $T$ bounded, invertible and suppose the embedding $(D(T),\|\cdot \|_T) \to (X,\|\cdot\|_X)$ is compact. I have to show that $T^{-1}$ is compact.
Firstly I consider $\|\cdot \|_T$ as the graph norm. Then I started thinking that an unbounded operator $T$ with domain $D(T)$ is bounded, invertible if there is a map $T^{-1}$ with image $D(T)$ and $TT^{-1}x = x$ for every $x \in X$ and $T^{-1}Tu = u$ for every $u \in D(T)$.
But I don't have any idea how to proceed. Could someone help me to show the compactness?
Let $G=\{(x,Tx)\;|\;x\in D(T)\}\le X\times X$ be equipped with the graph norm $\|(x,Tx)\|=\|x\|+\|Tx\|$. By the assumption that $G$ is closed, $G$ becomes a Banach space. Consider the map $$ A:G\ni(x,Tx)\mapsto Tx\in X. $$ Then, $A$ is a bounded linear surjection. It is also an injection since $Tx=Tx'$ implies $x=x'$. Hence $$ A^{-1}:Tx \mapsto (x,Tx)\in G $$ is a bounded linear operator by inverse mapping theorem. We observe that $$ i:G\ni (x,Tx)\mapsto x\in X $$ is compact by the assumption. Thus $$ iA^{-1}:X\ni Tx\mapsto x\in X $$ is also compact since it is a product of a bounded linear operator and a compact operator. Compactness of $T^{-1}$ follows from the fact that $T^{-1}y=iA^{-1}y$ for all $y\in T(X)=X$.