# Regular $p$-norm of a matrix

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 to this question 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 $\mathbb{Z}/n\mathbb{Z}$). 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 quoted above.

Note: A loosely related question concerned with the Schatten $p$-norm of a matrix can be found here.

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.

## 1 Answer

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...

• +1 Thank you very much for your answer. The whole story is indeed much simpler than I (and several colleagues I asked) thought. Concerning your last paragraph: Sometimes, whether a question is trivial or not simply depends on the perspective from which we look at it. As I wrote, one can indeed use the Riesz-Thorin interpolation theorem together with Hölder's inequality to prove the (suboptimal) estimate I stated in the question. Your argument, though, shows that this interpolation approach is not only an overkill; it even misses the target :-). – Jochen Glueck Jun 3 '18 at 20:32