# Compact embedding of the domain and compact inverse

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 $$y\in T(X)=X$$.