# Norm of adjoint operator on normed space

Let $$X$$ and $$Y$$ be normed spaces and $$T:X\to Y$$ a bounded linear operator. The adjoint operator $$T^\times:Y'\to X'$$ of $$T$$ is defined by $$(T^\times g)(x)=g(Tx),\,g\in Y'$$ where $$X'$$ and $$Y'$$ are the dual spaces of $$X$$ and $$Y$$, respectively. I'm trying to prove that $$\|T^\times\|=\|T\|$$. I already got that $$\|T^\times\|\leq\|T\|$$, but I'm struggling to show that $$\|T^\times\|\geq\|T\|$$. In Kreyszig's Functional Analysis book, the author uses the following argument:

However, Theorem 4.3-3 (which resulted from the Hahn-Banach Theorem) states that:

If $$X$$ is a normed space and $$x_0\neq 0$$ is any element of $$X$$, then there exists a bounded linear functional $$\tilde{f}$$ on $$X$$ such that $$\|\tilde{f}\|=1\quad \text{and} \quad \tilde{f}(x_0)=\|x_0\|.$$

We know $$x_0\neq 0$$ does not imply that $$T(x_0)\neq 0$$, so how is this theorem applicable in this case?

You have then $$||T^*||=\sup_{||g||=1} ||Tg||=\sup_{||g||=1} \sup_{||x||=1}|(Tg)(x)|=\sup_{||g||=1} \sup_{||x||=1}|g(Tx)|\geq \sup_{||g||=1} \sup_{||x||=1}||Tx||=\sup_{||x||=1}||Tx||=||T||.$$ When $$Tx=0$$ then of course for any functional $$g$$ we have $$g(Tx) =0 =||Tx||,$$ so the above consideration is true.