# Showing compact operator

Let $$X,Y,Z$$ be Banach spaces and let $$T:X\to Y$$ be a compact linear operator, $$S:Z\to Y$$ be a bounded linear operator such that $$S(Z)\subset T(X)$$. I have to show that $$S$$ is a compact linear operator.

I proceed like this. Let $$(z_n)$$ be a bounded sequence in $$Z$$. Thus for all $$n\in \mathbb{N}$$, there exists $$x_n\in X$$ such that $$S(z_n)=T(x_n)$$. But now how to show that $$(x_n)$$ is bounded? This can be shown if I can prove that $$T$$ is injective. But I failed to do that. I tried to use open mapping theorem, but could not do so as I don't know whether $$T(X)$$ is a Banach space. Any hint will be appreciated.

Case 1: $$T$$ and $$S$$ both one-to-one. In this case define $$W:Z\rightarrow X$$ by $$Wz=x$$ if $$Sz=Tx$$. If $$z\in Z$$ then $$Sz\in S(Z)\subseteq T(X)$$ so there exists $$x\in X$$ such that $$Sz=Tx$$. Further $$x$$ is uniquely determined by $$z.$$ Thus $$W$$ is a well defined linear map. Note that $$T\circ W=S$$ so it is enough to show that $$W$$ is bounded. Suppose $$z_{n}\rightarrow z$$ and $$% Wz_{n}\rightarrow x$$. Let $$x_{n}=Wz_{n}$$. Then $$z_{n}$$ $$\rightarrow z,x_{n}\rightarrow x$$ and $$Sz_{n}=Tx_{n}$$. Since $$T$$ and $$S$$ are bounded, $$% Tx_{n}=Sz_{n}\rightarrow Sz$$ and $$Tx_{n}\rightarrow Tx$$. Hence $$Sz=Tx$$ which im plies $$Wz=x$$. Closed Graph Theorem shows that $$W$$ is bounded.
Case 2 (general case): let $$M$$ and $$N$$ be the null spaces of $$T$$ and $$S$$ respectively. Define $$T_{1}:X|M\rightarrow Y$$ and $$S_{1}:Z|N\rightarrow Y$$ by $$T_{1}(x+M)=Tx$$ and $$S_{1}(z+N)=Sz$$. Then $$T_{1}$$ and $$S_{1}$$ are bounded operators. Further they are one-to-one. If $$\{x_{n}+M\}$$ is a norm bounded sequence in $$X|M$$ then there exists $$\{m_{n}\}\subseteq M$$ such that $$% \left\Vert x_{n}+m_{n}\right\Vert$$ is bounded. Since $$T$$ is compact $$% \{T_{1}(x_{n}+M)\}=\{Tx_{n}\}=\{T(x_{n}+m_{n})\}$$ has a convergent subsequence. We have proved that $$T_{1}$$ is compact. Also $$% S(Z|N)=S(Z)\subseteq T(X)=T(X|M)$$. Case 1 now shows that $$S_{1}$$ is compact. However $$S$$ is the composition of the canonical map $$z\rightarrow z+N$$ from $$% z\rightarrow Z|N$$ and $$S_{1}$$. Hence $$S$$ is compact.
A less technicl variant of the other answer: $$T(X)$$ endowed with the quotient topology from $$X$$ is a Banach space and compactness of $$T:X\to Y$$ is equivalent to the compactness of the inclusion $$J: T(X) \hookrightarrow Y$$. By assumption, $$S$$ has values in $$T(X)$$ and the closed graph theorem gives the continuity of $$S:X\to T(X)$$ (because $$S$$ is continuous w.r.t. the coarser topology from $$Y$$). Hence, as an operator from $$X$$ to $$Y$$, $$S$$ factorizes over the compact $$J$$ and is thus compact by the ideal property of compact operators.