# Operator norms and adjoint map: How to show $\lVert\Phi\rVert_{\infty} = \lVert \Phi^*\rVert_1?$

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

$$\lVert\Phi\rVert_{\infty} = \lVert \Phi^*\rVert_1?$$

• Are you familiar with the duality between the matrix norms $\|\cdot\|_1$ and $\|\cdot\|_\infty$? Commented Dec 28, 2020 at 10:24
• @MaoWao Thank you - sorry I missed your comment from many months ago! Commented Apr 5, 2021 at 3:59

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}