Prove that $T^*$ is injective iff $ImT$ Is dense

Question

Let X,Y be two normed spaces, and $T:X\rightarrow Y$ a bounded linear operator. prove that the adjoint operator $T^*$ ($T^*f(x)=f(Tx)$ is injective iff $ImT$ is dense

any help would be great guys. I did try a bit to solve it myself, using the deffinition of injective and going straightforward. It didn't work. I suppose that I have to use some theorem in order to solve it, and i hoped you could at least tell me which.

Thanks!

Could anyone give any kind of hint please?

Answer 1

Suppose $\operatorname{Im}T$ is dense and let $f\in Y^*$ with $T^*(f)=0$. This means that $$ f(Tx)=0 $$ for all $x\in X$. Now, let $y\in Y$; by assumption, there is a sequence $(x_n)$ in $X$ such that $y=\lim_{n\to\infty}T(x_n)$. Apply continuity of $f$ and $T$ to conclude.

$$f(y)=\lim_{n\to\infty}f(Tx_n)=0.$$

The converse is (a special case of) the Hahn-Banach theorem: if $y\in Y$ doesn't belong to the closure of a subspace $U$, then there exists a linear functional $f\in Y^*$ such that $f(y)\ne0$, while $f$ vanishes on $U$.

Answer 2

Hint: Show that the kernel of $T^*$ is the annihilator of $\operatorname{Im} T$. Show that the only closed subspace with $0$ annihilator is the whole space.

Here, for a set $S \subset Y$, the annihilator of $Y$ is defined to be $\{\alpha \in Y^*\vert \alpha(y) = 0 \text{ for } y \in S\}$.