# $\tilde{T} : X/X_0 \to Y$ is also compact

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!

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 $$T(B)=\bar T(B')$$.
• So you are saying that $\tilde{T} (B^{\})=T(B)$ readily follows from definitions? – reflexive Feb 15 at 14:56
• 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. – Tsemo Aristide Feb 15 at 15:01