Prove that the linear isomorphism $T_0$ form $X/\ker(T)$ to $T(X)$ such that $T_0(x+\ker(T))$ = $T(x)$, $\|T_0\|$ = $\|T\|$.

Question

Here is the question:

Suppose $T$ is a bounded linear operator from Banach space $X$ to Banach space $Y$. Prove that if $T_0$ is the natural linear isomorphism form $X/\ker(T)$ to $T(X)$ such that $T_0(x+\ker(T))$ = $T(x)$ $\forall$ $x \in X$, then $\|T_0\|$ = $\|T\|$.

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I would like to know how to prove $\|T_0\| \le \|T\|$. Thanks.

Answer 1

For any $y \in \ker (T)$ we have $$\|T_0(x+\ker (T)\|=\|Tx\|=\|T(x+y)\|\leq \|T\| \|x+y\|.$$ Taking infimum over all $y \in \ker(T)$ finishes the proof.