The norm of conjugate operator in Banach space I want to prove that $ \begin {Vmatrix} T \end {Vmatrix} = \begin {Vmatrix} T^{*} \end {Vmatrix}$ where $ T: E_1 \to E_2$, $T^{*}: E^{*}_2 \to E^{*}_1$ are  operators between two Banach spaces and its dual spaces respectively.
Let $f \in E^{*}_2$, $x \in E_1$
From $T^{*}f(x) = f(Tx)$ one gets by composition rule that $\begin {Vmatrix} T^{*}f(x) \end {Vmatrix} \leq \begin {Vmatrix} T \end {Vmatrix} \begin {Vmatrix} f \end {Vmatrix}\begin {Vmatrix} x \end {Vmatrix} $ and taking $\sup$  concludes $\begin {Vmatrix} T^{*} \end {Vmatrix} \leq \begin {Vmatrix} T \end {Vmatrix}$.
Now I need to show converse inequality, but how can I do that? Hint says that Hahn-Banach theorem is needed. Or  is there easier way than I chose? 

Fixing $x\in E_1$ with $\|x\|\leq1$, the Hahn-Banach theorem implies there is some $f\in E_2^*$ such that $\|f\|=1$ and $|T^*f(x)|=|f(Tx)|=\|Tx\|$. Thus
$$\|Tx\|=|T^*f(x)|\leq\|T^*\|\|f\|\|x\|\leq\|T^*\|$$
Now taking the supremum over $x$, we obtain $\|T\|\leq\|T^*\|$.
