# Entry-wise interpolation of operators

Suppose $A$ and $B$ are two bounded linear operators on $\ell^2(\mathbb{N})$ such that with respect to the standard orthonormal basis, the matrices $A=(a_{mn})_{m,n}$ and $B=(b_{mn})_{m,n}$ consist of non-negative entries. It is not difficult to show that for any $\lambda\in [0,1]$, the matrix $(a_{mn}^{\lambda}b_{mn}^{1-\lambda})_{m,n}$ gives rise to a bounded operator $C_{\lambda}$ on $\ell^2(\mathbb{N})$ and $$\|C_{\lambda}\| \leq \|A\|^{\lambda}\cdot\|B\|^{1-\lambda},$$ where $\|\cdot\|$ denotes the operator norm.

Such an estimate seems to be closely related with interpolation of operators. However the usual interpolation theorems that I know are for ''the same'' operator acting on different spaces. Is there a version of interpolations that could directly prove the above statement (and perhaps in a more general setting)?

Another related question: which functions $\varphi(x,y)$ on two variables that give rise to the norm inequality $$\Big\|\Big(\varphi(a_{mn}, b_{mn})\Big)_{m,n}\Big\| \leq \varphi(\|A\|,\|B\|),$$ where $A=(a_{mn})$ and $B=(b_{mn})$ are two bounded operators on $\ell^2(\mathbb{N})$ whose entries are non-negative? We see above that $\varphi(x,y)=x^{\lambda}y^{1-\lambda}$ for $0\leq\lambda\leq 1$ works. In addition, $\varphi(x,y)=x^{s}y^{t}$ for non-negative integers $s,t$ also works and this follows from properties of Hadamard product.

## 1 Answer

In this setting, this is just the maximum principle (or Hadamard's three-line lemma) in complex analysis. Namely you can define the operator $C_\lambda$ for every complex number $\lambda$ with real part in $[0,1]$. This operator depends holomorphically on $\lambda$. When $\lambda$ has real part $0$, the inequality $\|C_\lambda\| \leq \|C_0\|=\|B\|$ is clear, and so is the inequality $\|C_\lambda\| \leq \|C_1\|=\|A\|$ when $\lambda$ has real part $1$. So by the three line lemma, we obtain $\|C_\lambda\| \leq \|A\|^{Re \lambda} \|B\|^{1-Re \lambda}$.

This is indeed a very particular case of a general interpolation theorem, namely Stein's interpolation method.

• Yes, the proof you gave is what I have (in addition to a direct computational proof). Thank for the reference on Stein's interpolation method. Do you have any insights on what $\varphi$ works? The case $\varphi(x,y)=x^sy^t$ (where $s,t$ are non-negative integers) doesn't seem to follow from Stein's interpolation, does it? – T. Le Feb 7 '18 at 12:03
• I don't know what other $\varphi$ work, but the case $\varphi(x,y)=x^s y^t$ (for $s,t$ non-negative real numbers) follows from Stein interpolation. If $(s,t)$ can be written as $(n\theta,m(1-\theta))$ for $n,m$ integers ans $\theta \in (0,1)$ (for example if $s,t$ are integers, $\theta= \frac 1 2$ and $n=2s,m=2t$), consider the map $z \mapsto A^{nz} B^{m(1-z)}$. – Mikael de la Salle Feb 7 '18 at 12:17
• Interesting point! I believe in the map $z\mapsto A^{nz}B^{m(1-z)}$, you meant entrywise power and multiplication (which is what I'm interested in). But I think you still need Hadamard product to conclude that $\|A^{\circ n}\|\leq \|A\|^{n}$ (I put the circle in the power to indicate that the power is entry wise). – T. Le Feb 7 '18 at 12:51
• Yes, you are correct, I was clumsy in my notation. – Mikael de la Salle Feb 7 '18 at 14:29