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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:

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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?

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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.

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