Connection between diamond norm and output purity norm

Setting of the problem: Given a quantum channel $\mathcal{E}: \mathcal{H}_A\rightarrow \mathcal{H}_B$ (where $\mathcal{H}$ refers to a Hilbert space and subscript refers to the quantum register associated to the Hilbert space), we can define the following two types of norms:

1. Output purity norms: $\|\mathcal{E}\|_p := \text{max}_{\rho_A}\|\mathcal{E}(\rho_A)\|_p$,
2. Completely bounded norms (or diamond norm when $p=1$): $\||\mathcal{E}\||_p := \text{max}_{\rho_{AA'}}\|(\mathcal{E}\otimes I_{A'})(\rho_{AA'})\|_p$, where $A'$ is a register of same dimension as $A$.

Above, $\|X\|_p$ for a hermitian matrix $X$, is the standard norm: $\text{Tr}(|X|^p)^{\frac{1}{p}}$. All the maximization above is over quantum states. More details on output purity norms can be found at https://arxiv.org/pdf/quant-ph/0203003.pdf and on completely bounded norms can be found at https://cs.uwaterloo.ca/~watrous/CS766/LectureNotes/20.pdf

Question: It is easy to see that $\|\mathcal{E}\|_p \leq \||\mathcal{E}\||_p$, by restricting the maximization in the case of completely bounded norms to product quantum states $\rho_A\otimes \sigma_{A'}$ (with $\sigma_{A'}$ being pure) and then using the fact that $\|\sigma_{A'}\|_p = 1$. My question is how tight is above inequality. That is, if $\||\mathcal{E}\||_p \leq C\times\|\mathcal{E}\|_p$, then how small $C$ can be, in general?