Denote linear operators on a Hilbert space $\mathcal{X}$ by $L(\mathcal{X})$. For any $A\in L(\mathcal{X})$, we have the trace norm given by
$$\lVert A\rVert_{1}=\operatorname{Tr} \sqrt{A^{*} A}$$
and the operator norm given by
$$\lVert A\rVert_{\infty}=\max _{u \in \mathcal{X}, \lVert u\rVert = 1}\lVert A u\rVert,$$
where $\lVert u\rVert$ is the Euclidean 2-norm on vectors.
A superoperator is a linear mapping of the form $\Phi: \mathrm{L}(\mathcal{X}) \rightarrow \mathrm{L}(\mathcal{Y})$. One has the following superoperator norms
$$\lVert\Phi\rVert_{1} = \max \{ \lVert\Phi(X)\rVert_{1}: X \in L(\mathcal{X}),\lVert X\rVert_{1} \leq 1 \}$$
and
$$\lVert\Phi\rVert_{\infty} = \max \{ \lVert\Phi(X)\rVert_{\infty}: X \in L(\mathcal{X}),\lVert X\rVert_{\infty} \leq 1 \} $$
Finally, the adjoint of a superoperator $\Phi$ is defined as the unique superoperator $\Phi^\star$ that satisfies
$$\langle B, \Phi(A)\rangle = \langle \Phi^*(B), A\rangle$$
for any $A\in L(\mathcal{X})$ and $B\in L(\mathcal{Y})$. How does one show that
$$\lVert\Phi\rVert_{\infty} = \lVert \Phi^*\rVert_1?$$
Best Answer
As noted in the comments, there is a duality between $\|\cdot \|_1$ and $\|\cdot\|_{\infty}$. Namely, $$ \|X\|_1 = \max \{ |\langle X, Y \rangle| : \|Y\|_{\infty} \leq 1\} $$ and $$ \|X\|_{\infty} = \max \{ |\langle X, Y \rangle| : \|Y\|_{1} \leq 1\} $$
So then we have $$ \begin{aligned} \lVert\Phi\rVert_{\infty} &= \max_X \{ \lVert\Phi(X)\rVert_{\infty}: \lVert X\rVert_{\infty} \leq 1 \} \\ &= \max_{X,Y} \{|\langle \Phi(X), Y\rangle| : \|X\|_{\infty} \leq 1, \|Y\|_1 \leq 1\} \\ &= \max_{X,Y} \{|\langle X, \Phi^*(Y)\rangle| : \|X\|_{\infty} \leq 1, \|Y\|_1 \leq 1\} \\ &= \max_{Y} \{\|\Phi^*(Y)\|_1 : \|Y\|_1 \leq 1\} \\ &= \|\Phi^*\|_1 \end{aligned} $$