Let $n \in \mathbb{N}$ and $p \in [1,\infty]$ be fixed and endow $\mathbb{C}^n$ with the $p$-norm $\|\cdot\|\_p$. For every matrix $A \in \mathbb{C}^{n \times n}$ we denote the operator norm of $A$ as an operator on $\mathbb{C}^n$ by $\|A\|\_p$, too. Moreover, let $|A|$ denote the matrix whose entries are the absolute values of the entries of $A$. 


The number $\| \,|A|\,\|\_p$ is sometimes called the **regular norm** of $A$ (in particular in Banach lattice theory, where a norm on the space of regular operators is constructed this way).


We clearly have $\|A\|\_p \le \| \,|A|\,\|\_p$, and equality holds for $p = 1$ and $p = \infty$. For general $p$, Mark Meckes explained in [his answer](https://mathoverflow.net/a/111658/102946) to [this question](https://mathoverflow.net/q/110299/102946) that the Riesz-Thorin theorem implies the estimate
\begin{align\*}
 \| \,|A|\,\|\_p \le c\_{n,p} \|A\|\_p, \tag{E}
\end{align\*}
where $c\_{n,p} = n^{\frac{2}{p}(1 - \frac{1}{p})}$. 


**Question:** Is the estimate (E) sharp? If not, what is the optimal constant for the estimate (depending on $n$ and $p$)?


*Remarks:*


* Clearly, the inequality (E) is sharp (and actually an equality) for $p = 1$ and $p = \infty$.
* Let $U \in \mathbb{C}^{n\times n}$ be such that $\frac{1}{\sqrt{n}}U$ is a unitary matrix and such that each entry of $U$ has modulus $1$ (such a matrix exists for every $n$; for instance, we can choose $U$ to be the transformation matrix of the [Fourier transform on $\mathbb{Z}/n\mathbb{Z}$](https://en.wikipedia.org/wiki/Discrete_Fourier_transform)). Then one can prove by employing Hölder's inequality several times that $\|\, |U| \, \|\_p = n$ for each $p \in [1,\infty]$, and that $\|U\|\_p = n^{1/p}$ for $p \in [1,2]$, while $\|U\|\_p = n^{1-1/p}$ for $p \in [2,\infty]$. Thus, we obtain $\|\,|U|\,\|\_p = \tilde c\_{n,p} \|U\|\_p$, where
\begin{align\*}
 \tilde c\_{n,p} = 
 \begin{cases}
 n^{1-1/p} \quad & \text{for } p \in [1,2], \\
 n^{1/p} \quad & \text{for } p \in [2,\infty].
 \end{cases}
\end{align\*} 
We have $\tilde c\_{n,p} \le c\_{n,p}$, and the existence of $U$ shows that the optimal constant for (E) is somewhere in between.
* Note that $\tilde c\_{n,2} = c\_{n,2} = n^{1/2}$, so we know that the estimate (E) is sharp for $p=2$; this is also mentioned in [Mark Meckes' answer](https://mathoverflow.net/a/111658/102946) quoted above.


Note: A loosely related question concerned with the Schatten $p$-norm of a matrix can be found [here](https://mathoverflow.net/q/108155/102946).


**Edits made 2018-06-03:** 


* I've replaced the real scalar field with $\mathbb{C}$ since this seems to be more appropriate for the question.
* I've added the example matrix $U$ to show that the number $\tilde c\_{n,p}$ is a lower bound for the optimal constant.



Regular $p$-norm of a matrix
It is trivially $n^{1/p}$ for $p>2$ (and, therefore, $n^{1-\frac 1p}$ for $1<p<2$ because the norm of the adjoint operator in the dual space is the same as the norm of the operator in the space itself). To see it, choose the vector $x$ of $\ell^p$ norm $1$ so that
$$
\sum\_i\left|\sum\_j |a\_{ij}|x\_j\right|^p=\||A|\|^p
$$
Now choose $i$ so that 
$$
\left|\sum\_j |a\_{ij}|x\_j\right|^p\ge \frac 1n\||A|\|^p
$$
Rotate $x\_j$ to $y\_j$ preserving the absolute values so that for this particular $i$, we have 
$$
\sum\_j |a\_{ij}|x\_j=\sum\_j a\_{ij}y\_j
$$
Then 
$$
\|A\|^p\ge \sum\_i\left|\sum\_j a\_{ij}y\_j\right|^p\ge \frac 1n\||A|\|^p
$$
and we are done.


What "Riesz-Thorin" are we talking about? Much less trivial questions get closed in no time and this one got 8 upvotes and 2 favorites...


