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.
Best Answer
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.