Lipschitz estimates in quasi-Banach Schatten ideals

Abstract

We study the class of functions f on ${\mathbb {R}}$ satisfying a Lipschitz estimate in the Schatten ideal ${\mathcal {L}}_p$ for $0 < p \le 1$. The corresponding problem with $p\ge 1$ has been extensively studied, but the quasi-Banach range $0< p < 1$ is by comparison poorly understood. Using techniques from wavelet analysis, we prove that Lipschitz functions belonging to the homogeneous Besov class ${\dot{B}}^{\frac{1}{p}}_{\frac{p}{1-p},p}({\mathbb {R}})$ obey the estimate

$$\begin{aligned} \Vert f(A)-f(B)\Vert _{p} \le C_{p}(\Vert f'\Vert _{L_{\infty }({\mathbb {R}})}+\Vert f\Vert _{{\dot{B}}^{\frac{1}{p}}_{\frac{p}{1-p},p}({\mathbb {R}})})\Vert A-B\Vert _{p} \end{aligned}$$

for all bounded self-adjoint operators A and B with $A-B\in {\mathcal {L}}_p$. In the case $p=1$, our methods recover and provide a new perspective on a result of Peller that $f \in {\dot{B}}^1_{\infty ,1}$ is sufficient for a function to be Lipschitz in ${\mathcal {L}}_1$. We also provide related Hölder-type estimates, extending results of Aleksandrov and Peller. In addition, we prove the surprising fact that non-constant periodic functions on ${\mathbb {R}}$ are not Lipschitz in ${\mathcal {L}}_p$ for any $0< p < 1$. This gives counterexamples to a 1991 conjecture of Peller that $f \in {\dot{B}}^{1/p}_{\infty ,p}({\mathbb {R}})$ is sufficient for f to be Lipschitz in ${\mathcal {L}}_p$.

Appendix A: Automatic complete boundedness of \({\mathcal {L}}_p\)-bounded Schur multipliers

The following is a recently published result of Aleksandrov and Peller [5, Theorem 3.1].

Theorem A.0.1

Let $A \in M_n({\mathbb {C}})$ be a matrix, where $1\le n\le \infty$. Let $N\ge 1$ and denote by $\mathrm {id}_{M_N({\mathbb {C}})}$ the $N\times N$ matrix with all entries equal to 1. Then for all $0< p < 1$ we have

$$\begin{aligned} \Vert A\Vert _{\mathrm {m}_p} = \Vert A\otimes \mathrm {id}_{M_{N}({\mathbb {C}})}\Vert _{\mathrm {m}_p}. \end{aligned}$$

In other words, bounded Schur multipliers of ${\mathcal {L}}_p$ are automatically completely bounded. The analogous statement for $p=1$ is well-known, see [39, Theorem 5.1]. For the sake of completeness we include a proof of Theorem A.0.1. The proof is different from that of [5], and is instead closely modelled on a proof for the $p=1$ case due to Smith [48, Theorem 2.1].

Recall that we denote by $\ell _2^n$ the n-dimensional Hilbert space.

Proof of Theorem A.0.1

Let $1\le n\le \infty$ and $N\ge 1$. Let $u \in \ell _2^n\otimes \ell _2^N$ be a unit vector. Write the components of u as $u =\sum _{j,l} u_{j,l}e_j\otimes e_l$. Consider the mapping:

$$\begin{aligned} Q_u:\ell _2^n\rightarrow \ell _2^n\otimes \ell _2^N \end{aligned}$$

given by

$$\begin{aligned} Q_{u}e_j = {\left\{ \begin{array}{ll} (\sum _{l=1}^N |u_{j,l}|^2)^{-1/2}\sum _{l=1}^N u_{j,l}e_j\otimes e_l,&{}\text { if } \sum _{l=1}^N |u_{j,l}|^2 \ne 0\\ 0,&{}\text { otherwise.} \end{array}\right. } \end{aligned}$$

The adjoint of $Q_u$ is easily computed. We have

$$\begin{aligned} Q_u^*(e_{j}\otimes e_l) = e_j{\overline{u}}_{j,l}(\sum _{r=1}^N |u_{j,r}|^2)^{-1/2},\quad 1\le j\le n,\, 1\le l\le N \end{aligned}$$

or $Q_u^*(e_j\otimes e_l) = 0$ if $\sum _{r=1}^N |u_{j,r}|^2 = 0$.

Then we compute $Q_u^*Q_ue_j$. If $\sum _{l=1}^N |u_{j,l}|^2 = 0$ then $Q_u^*Q_ue_j = 0$ and so assuming otherwise we have for every $1\le j\le n$,

$$\begin{aligned} Q_u^*Q_ue_j&= \left( \sum _{l=1}^N |u_{j,l}|^2\right) ^{-1/2}\sum _{r=1}^N u_{j,r}Q_u^*(e_{j}\otimes e_r)\\&= \left( \sum _{l=1}^N |u_{j,l}|^2\right) ^{-1/2}\sum _{r=1}^N u_{j,r}{\overline{u}}_{j,r}(\sum _{l=1}^N |u_{j,l}|^2)^{-1/2}e_j\\&= e_j \cdot \frac{\sum _{r=1}^N |u_{j,r}|^2}{\sum _{r=1}^N |u_{j,r}|^2}\\&= e_j. \end{aligned}$$

So $Q_u$ is indeed a contraction.

Given $u \in \ell _2^n\otimes \ell _2^N$, define ${\widetilde{u}} \in \ell _{2}^n$ as,

$$\begin{aligned} {\widetilde{u}} = \sum _{j=1}^n (\sum _{l=1}^N |u_{j,l}|^2)^{1/2}e_j. \end{aligned}$$

We have that $\Vert {\widetilde{u}}\Vert _{\ell _2^n} = \Vert u\Vert _{\ell _2^n\otimes \ell _2^N}$.

We now assert that

$$\begin{aligned} (\mathrm {id}_{M_N({\mathbb {C}})}\otimes A)\circ (u\otimes v) = Q_u(A\circ {\widetilde{u}}\otimes {\widetilde{v}})Q_v^*. \end{aligned}$$

(A.1)

It suffices to check (A.1) entrywise. On the left hand side, we have

$$\begin{aligned} \langle e_{j,l_1},(\mathrm {id}_{M_N({\mathbb {C}})}\otimes A)\circ (u\otimes v)e_{k,l_2}\rangle = A_{j,k}u_{j,l_1}\overline{v_{k,l_2}}, \quad 1\le j\le n, 1\le k\le n, 1\le l_1,l_2\le N \end{aligned}$$

and on the right

$$\begin{aligned}&\langle e_{j,l_1},Q_u(A\circ ({\widetilde{u}}\otimes {\widetilde{v}}))Q_v^*e_{k,l_2}\rangle = \langle Q_u^*e_{j,l_1},(A\circ {\widetilde{u}}\otimes {\widetilde{v}})Q_v^*e_{k,l_2}\rangle \\&= u_{j,l_1}\overline{v_{k,l_2}}(\sum _{r=1}^N |u_{j,r}|^2)^{-1/2}(\sum _{r=1}^N |u_{k,r}|^2)^{-1/2}\langle e_j,(A\circ {\widetilde{u}}\otimes {\widetilde{v}})e_k\rangle \\&= A_{j,k}u_{j,l_1}\overline{v_{k,l_2}}. \end{aligned}$$

This verifies (A.1).

Since $Q_u$ and $Q_v$ are contractions, (A.1) implies that

$$\begin{aligned} \Vert (\mathrm {id}_{M_N({\mathbb {C}})}\otimes A)\circ (u\otimes v)\Vert _{p} \le \Vert A\Vert _{\mathrm {m}_p}\Vert u\Vert _{\ell _2^n\otimes \ell _2^N}\Vert v\Vert _{\ell _2^n\otimes \ell _2^N}. \end{aligned}$$

Taking the supremum over $u,v \in \ell _2^n\otimes \ell _2^N$ with norm at most 1 and using Lemma 2.2.1 yields the conclusion

$$\begin{aligned} \Vert \mathrm {id}_{M_N({\mathbb {C}})}\otimes A\Vert _{\mathrm {m}_p} \le \Vert A\Vert _{\mathrm {m}_p}. \end{aligned}$$

The reverse inequality is clear. $\square$