If the range of $T:X \to Y$ is not dense , then $T':Y^* \to X^*$ is not injective

Question

Let $X$ be a Banach space . $T:X\to X$ continuous with $T(X)$ not dense , then we can find an element $l \in X^*$ such that $T'(l)=0$ . $T'$ denote de Banach adjoint of $T$.

My attempt :
If $X$ is a Hilbert space , $A:=$ the closure of $T(X)$ , then $A^\perp$ is nonempty . Taking nonzero $a \in A^\perp$ ,define $l(x)=<a,x>$ then $$T'(l)(x)=l(Tx)=<a,Tx>=0$$ So we have $T'(l)=0$ .
However , if we only assume $X$ is a Banach space . Since I can not define $A^\perp$ , I don't know how to do this .
Furthermore , If we assume $T:X \to Y$ with $T(X)$ not dense , can we prove that $T'$ is not injective ?

Answer 1

There is no need to bring in inner products. If $T(X)$ is not dense then its closure is a proper closed subspace . By Hahn Banach Theorem there exists $l$ such that $ l\neq 0$ but $l=0$ on $T(X)$. By definition of adjoint $T'$ this gives $T'(l)=0$.

Answer 2

You can apply a similar idea, if $T(X)$ is not dense, there exists $y$ which is not in the closure of $T(X)$, consider $f:\overline{T(X)}\oplus Vect(y)$ such that $f(y)=1, f(\overline{T(X)})=0$, by Hahn Banach, $f$ can be extended to $Y$ denote this extension by $g$, $T^*(g)=0$.