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In a normed linear space N, let $B=\{x \in N : |x| <1\}$ . Then define a linear map $T :N \to N_1$ ,( $N_1$ some normed linear space ) to be compact if closure of $T(\bar{B})$ is compact in $N_1$ . With this definition I was trying to prove :

X,Y Banach space. $T : X \to Y$ compact, $X_0$ closed subspace of $X$ s.t. $X_0 \subset Ker(T)$, Then $\tilde{T} : X/X_0 \to Y $ is also compact.

I couldn't figure out how to proceed. Thanks in advance for help!

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Let $B'=\{x\in X/X_0: |x|<1\}$, $\bar T(B')=T(B)$ since by definition of the norm of a quotient space if $p:X\rightarrow X/X_0$ is the quotient map, $p(B)=B'$. Since $T=\bar T\circ p$ we deduce that $T(B)=\bar T(B')$.

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  • $\begingroup$ Can you please provide some more details. $\endgroup$ – reflexive Feb 15 at 14:53
  • $\begingroup$ math.stackexchange.com/questions/1968606/… $\endgroup$ – Tsemo Aristide Feb 15 at 14:53
  • $\begingroup$ So you are saying that $\tilde{T} (B^{\})=T(B)$ readily follows from definitions? $\endgroup$ – reflexive Feb 15 at 14:56
  • $\begingroup$ yes, it comes from definition. Firstly show that $p(B)=B'$ with the definition of the norm of $X/X_0$ provided in the link. $\endgroup$ – Tsemo Aristide Feb 15 at 15:01

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