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

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 $$l(x)=$$ then $$T'(l)(x)=l(Tx)==0$$ 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 ?

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 $$T'(l)=0$$.
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 $$g$$, $$T^*(g)=0$$.