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I am reviewing functional analysis and getting stuck in this problem. Let $X,Y$ be two Banach spaces and $A,B\in L(X,Y)$. Prove that if $A$ is a compact operator and $R(B)\subset R(A)$ then $B$ is also a compact operator.

Can anyone give me some hints for this question ? Thank you very much.

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  • $\begingroup$ I think you need to assume that $B$ is bounded $\endgroup$ – Ben Grossmann Jun 26 '16 at 4:11
  • $\begingroup$ Yes. B is a bounded linear operator $\endgroup$ – Omega Jun 26 '16 at 4:13
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Endow $R(A)$ with the quotient norm having the unit ball $A(K_X)$ where $K_X$ is the unit ball of $X$. This is a Banach space and $\tilde B:X\to R(A)$, $x\mapsto B(x)$ is well-defined by assumption, linear, and has closed graph (because the norm of $Y$ gives a coarser norm on $R(A)$ which makes $B$ continuous). The closed graph theorem implies the continuity of $\tilde B$ so that $\tilde B(K_X) =B(K_X) \subseteq c A(K_X)$ for some $c>0$. Since $A(K_X)$ is relatively compact in $Y$ the same holds for $B(K_X)$.

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  • $\begingroup$ What is the quotient norm on R(A) #Jochen $\endgroup$ – Omega Jun 27 '16 at 23:58
  • $\begingroup$ The quotient norm of $z\in R(A)$ is $\inf\lbrace \|x\|_X: A(x)=z\rbrace$. The open unit ball of the quotient norm is the image of the open unit ball of $X$. $\endgroup$ – Jochen Jun 28 '16 at 10:23

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