Let $E,F$ be normed spaces, $T\in L(E,F)$. Why does $$\overline{\operatorname{ran}(T)}=(\ker(T^*))_\perp$$ holds? Here $T^*$ is the dual operator of $T$ and for $L\subset E^*$ we define $L_\perp:=\{x\in E : \quad l(x)=0\,\forall l\in L\}$. I see why $(\ker(T^*))_\perp\subset \overline{\operatorname{ran}(T)}$ holds, but what is with the other inclusion?
Inclusion for range and kernel of dual operator
functional-analysislinear algebra
Best Answer
Let $x\in \text{ran}(T)$ and $f\in \text{ker}(T^*)$. Then, $f \circ T=0$ implies that $f(T(x))=0$. This implies that $x\in \text{ker}(T^*)_{\perp}$. We deduce that $\text{ran}(T)\subset \text{ker}(T^*)_{\perp}$. Since $\text{ker}(T^*)_{\perp}$ is closed, we deduce that $\overline{\text{ran}(T)}\subset \text{ker}(T^*)_{\perp}$.