A sequence $T:X$ $\rightarrow Y$ and $S:Y$ $\rightarrow Z$ is called exact of $RanT=kerS$
Let X,Y and Z be Banach spaces. T is bounded operator from $X$ $\rightarrow$Y
$T$ $\in B(X,Y)$ and $S$ $\in B(Y,Z)$ And suppose S has a closed range. prove that if
$X-$ $^T$ $-Y$$-$ $^S$ $-Z$ is exact then $Z^*$ $-$ $^S*$ $-Y^*$$-$ $^T*$ $-X^*$ is exact.
I have to show that $RanS^*=kerT^*$. Here * represents the dual space.
I thought of showing two inclusions. That is $RanS^*\subseteq kerT^*$ and $kerT^*\subseteq RanS^*$.
first let $\varphi\in RanS^*$ be a functional then show that $\varphi\in kerT^*$. I need to show that $\varphi(y^*)=0$ where $y^*\in Y^*$.
Is my approach correct. How I'm going to use the fact that S has a closed range?
I suppose you want to show that if we have an exact sequence, then the dual one is exact.Let's look at general case, because it's not difficult, but i will give you a scratch. Let be $$ ... \longrightarrow L_{k-1} \underset{T_k}{\longrightarrow} L_k \underset{T_{k+1}}{\longrightarrow} L_{k+1} \longrightarrow ...$$ be an exact sequence, so the dual one $$ ... \longleftarrow L^*_{k-1} \underset{T*_k}{\longleftarrow} L^*_k \underset{T^*_{k+1}}{\longrightarrow} L^*_{k+1} \longleftarrow ... $$ The first part $ker T_k^* \subset Im T_{k+1}^*$ is obvious, because $0 = (T_{k+1} T_k)^* = T_k^*T^*_{k+1}$.Let's show other inclusion, let $f$ be an arbitary element of $kerT^*_k$, then $ker T_{k+1} = ImT_k \subset kerf$, then $f$ defines a linear functional F on $ImT_{k+1}$ by formula $$F(T_{k+1}(x)) = f(x)$$ This functional is continuous, so by Hahn-Banach theorem it can be continued by $F_0 \in L^*_{k+1}$ and $T^*_{k+1}F_0 = f$. Proof that $F$ is continuous uses the fact that $T_{k+1}$ is bounded