# Accumulating Knowledge for Lifelong Online Learning

Changjian Shui, Ihsen Hedhli, Christian Gagné
{changjian.shui.1,ihsen.hedhli.1 }<EMAIL_ADDRESS>
Electrical and Computer Engineering Department, Université Laval, Québec
(Québec), Canada

###### Abstract

Lifelong learning can be viewed as a continuous transfer learning procedure
over consecutive tasks, where learning a given task depends on accumulated
knowledge — the so-called _knowledge base_. Most published work on lifelong
learning makes a batch processing of each task, implying that a data
collection step is required beforehand. We are proposing a new framework,
lifelong online learning, in which the learning procedure for each task is
interactive. This is done through a computationally efficient algorithm where
the predicted result for a given task is made by combining two intermediate
predictions: by using only the information from the current task and by
relying on the accumulated knowledge. In this work, two challenges are
tackled: making no assumption on the task generation distribution, and
processing with a possibly unknown number of instances for each task. We are
providing a theoretical analysis of this algorithm, with a cumulative error
upper bound for each task. We find that under some mild conditions, the
algorithm can still benefit from a small cumulative error even when facing few
interactions. Moreover, we provide experimental results on both synthetic and
real datasets that validate the correct behaviour and practical usefulness of
the proposed algorithm.

## Introduction

In the general framework of machine learning, the learning procedure can be
viewed as a system (referred to as _agent_ in this paper) that runs an
algorithm on a given dataset in order to return a hypothesis for predicting
the unseen data. Typically, these algorithms require a large amount of data in
order to make predictions with an acceptable level of performance. Chen and
Liu (2016) termed such a paradigm as _isolated learning_ , since it does not
consider any other related information or previously learned knowledge.
Instead, humans have the ability to continually learn over time by
accommodating new knowledge while retaining previously learned experiences.
Such a continuous learning procedure has represented a challenging problem for
machine learning and for the development of artificial intelligence.

This continual learning idea has inspired various machine learning strategies.
In _domain adaptation_ (Pan and Yang, 2010), the goal is to transfer the
knowledge from a given task (i.e., source domain) to another task with few
labeled or unlabeled observations (i.e., target domain). In the case of
_multi-task learning_ (Zhang and Yang, 2017), the goal is to set a good
performance on different but related tasks simultaneously. Concerning
_lifelong learning_ (Mitchell and Thrun, 1993; Chen and Liu, 2016), the
learning procedure can be viewed as a continuous transfer learning procedure
over incrementally available tasks from the underlying distribution.

Figure 1: Two levels of sequences in lifelong online learning: 1) task
sequence (blue arrows) and 2) instance sequence in each task (green arrows).
For task $j$ at time $t$, the instances $\mathbf{x}_{t}$ also arrive
sequentially, while in the general lifelong learning, each task is processed
in a batch manner.

Lifelong learning has recently received increasing attention due to its
implications in autonomous learning agents and robots. However, in most
published work of lifelong learning, e.g., (Mitchell and Thrun, 1993; Ruvolo
and Eaton, 2013b; Pentina and Lampert, 2014), the learning procedure for each
task is _offline_. That means the algorithm is generally based on the
Empirical Risk Minimization (ERM) rule to obtain a good generalization
performance for the unseen tasks from the same distribution. Nonetheless,
there exist considerable real scenarios in which the learning procedure for
each individual task is _interactive_. For example, in the personalized
product recommendation system, we suppose that a _knowledge base_ from current
clients has been constructed, which can predict the preferences of existing
clients. For an unknown new client, current lifelong learning algorithms must
first collect a batch of data, then train such a batch with the help of the
knowledge base, which cannot effectively handle the interaction demands. From
this point of view, we should consider a new paradigm – _lifelong online
learning_ in which the agent interacts with users in each task, as showed in
Figure 1.

As opposed to the usual theoretical settings of lifelong learning, in lifelong
online learning we do not assume the task generation distribution. Then the
tasks are not necessarily generated in an i.i.d. fashion but are rather
arbitrarily or even adversary generated, which makes this problem more
challenging. Moreover, in the batch lifelong learning framework, only the
total number of tasks may be unknown to us such as in (Ruvolo and Eaton,
2013b). While in the lifelong online learning, both the total number of tasks
and the number of instances in each task can be unknown for us because both of
them arrive sequentially.

Motivated by the practical and theoretical considerations, we develop a new
algorithm in which the agent can interactively learn with the observation in
each task and also benefit from the contextual information learned by the
accumulated knowledge. At each task, the algorithm will predict the arriving
data via a combination of two aspects: the predictions provided by the
accumulating knowledge and the current classifier constructed by the agent
from scratch in the current task. As the interaction times increase, the
prediction by the current classifier will gradually play a more dominant role,
because the agent has continuously learned from the instances and will be more
confident about the current task. This mechanism will be useful when facing a
long term interactions and overcoming the possible _negative transfer_ from
the knowledge base (Pan and Yang, 2010).

The contribution of this paper can be summarized as follows:

  * •

We propose a computationally efficient algorithm in the lifelong online
classification, which effectively leverages the information from the
accumulated knowledge and the classifier which we have constructed for the
current task.

  * •

We theoretically provide an upper bound of the cumulative error of the
proposed algorithm when facing a new task. We find that under some mild
conditions, our approach can still benefit a small cumulative error even in
the non i.i.d. task generation distribution.

  * •

Our empirical results on both synthetic and various real datasets show good
performances compared with some baseline approaches.

The rest of the paper will be organized as follows. We first introduced the
related work in this field. Secondly we set up the problem and discuss the
proposed algorithm. In the third part, we derive the theoretical bound for the
algorithm. Finally the experimental results validate the proposed algorithm.

## Related works

#### Batch Lifelong learning

From the practical point of view, the lifelong learning approaches can be
categorized as two families: _parameter transfer_ and _representation
transfer_ (Chen and Liu, 2016). In parameter transfer, the agent uses
previously learned model parameters to help learn model parameter in the
current task (Evgeniou and Pontil, 2004; Fei, Wang, and Liu, 2016). As for the
representation transfer, each task shares a common representation in a lower
dimensional subspace and generally the _sparse dictionary learning_ approach
is developed, e.g., (Ruvolo and Eaton, 2013b; Maurer, Pontil, and Romera-
Paredes, 2013; Sun, Cong, and Xu, 2018). While in the theoretical
understanding of lifelong learning, Baxter (2000) proposed the generalization
bound on the VC dimension, while Pentina and Lampert (2014) analyzed the
lifelong learning through PAC-Bayesian theory. As for the the non i.i.d. task
generation assumption, Pentina and Lampert (2015) analyzed lifelong learning
with the task environment changing over time but limited in a consistent
manner. Pentina and Urner (2016) proposed a weighted majority vote algorithm
to theoretically prove the sample complexity reduction phenomena in the
_lifelong learning_.

#### Online transfer/multi-task/lifelong learning

Online transfer learning and online multi-task can also be viewed as a
sequential learning procedure from some _prior_ knowledge. In online transfer
learning, the knowledge is transferred only once from the source domain to the
target domain (Zhao et al., 2014; Wang and Pineau, 2015). In online multi-task
learning (Saha et al., 2011; Ruvolo and Eaton, 2013a; Murugesan and Carbonell,
2017), the goal is rather to learn the task relationship and try to minimize
the cumulative errors on all tasks.

There are only few published works in the lifelong online learning. For
instance, Alquier and Pontil (2017) proposed a similar concept of _online-
within-online_ lifelong learning based on the representation transfer, where
the data arrives sequentially in each task. However this approach is not
exactly _online_ since the high computational burden and storage requirements
(i.e., all observed data) make it not scalable for real dataset. Denevi et al.
(2018) proposed an improved algorithm but applied on the linear regression and
adapted to the i.i.d. generation task. Moreover, the representation transfer
based algorithms are still vulnerable when treating the non i.i.d. task
problem. Indeed, a simple adversarial strategy is to generate instances
according to two distributions which hold the same sub-feature space but
different labeling distribution, for which a representation-based approach can
suffer from an important cumulative error.

## Problem setup

Let us define $\\{j\\}_{j=1}^{\infty}$ as an ordered set of tasks. For each
task $j$, let us denote $N_{j}$ the total number of instances
$\\{(\mathbf{x}_{t}^{(j)},y_{t}^{(j)})\\}_{t=1}^{N_{j}}$ where
$\mathbf{x}_{t}^{(j)}\in\mathbb{R}^{d}$, $y_{t}^{(j)}\in\\{-1,+1\\}$, and
$h_{j}$ as the corresponding classifiers.

Suppose that we are at task $T+1$ and the $N_{T+1}$ examples in the task $T+1$
will arrive sequentially ($N_{T+1}$ might be unknown to us), we denote
$\mathrm{K}_{T}$ the current _knowledge base_ that contains all the historical
classifiers up to the last completed task $T$ (i.e.,
$K_{T}=\\{h_{j}\\}_{j=1}^{T}$). In the beginning of the interactive learning
procedure of the current task $T+1$, the classifier $h_{T+1}$ cannot make an
exact decision since it has not observed a sufficient number of instances.
However, $\mathrm{K}_{T}$ can provide some contextual information which can
help the learner to perform better. Therefore, the final prediction rule
$\mathcal{P}_{T+1}$ of the current task $T+1$ will involve two parts, as shown
in the prediction model given in Figure 1.

## Proposed algorithm

The proposed lifelong online algorithm proceeds by using the contextual
information available in the accumulated knowledge to perform better in the
current learning procedure. For the current task $T+1$, two prediction stages
are performed.

Prediction from the knowledge base ($O_{1:T}$):

constructed classifiers $\\{h_{j}\\}_{j=1}^{T}$ are evaluated on the new task
$T+1$ using:

$h_{j}(\mathbf{x}_{t}^{(T+1)})=\langle\mathbf{w}^{(j)},\mathbf{x}_{t}^{(T+1)}\rangle,$
(1)

where $\mathbf{w}^{(j)}$ is the parameter corresponding task $j\in[1,T]$. This
accumulated knowledge is used through the application of an _expert model_ to
make predictions over the arriving observations through the pool of the
previous models, as shown in (Cesa-Bianchi and Lugosi, 2006).

Prediction from the current classifier ($O_{T+1}$):

current model $h_{T+1}$ is being interactively updated by the arriving
observations $\\{\mathbf{x}_{t}\\}_{t=1}^{N_{T+1}}$ and the prediction rule is
the same as Equation 1.

Updates to the current model $h_{T+1}$ are being interactively done via
_online gradient descent_ of the regularized convex loss (Hazan, 2016):

$\mathcal{L}(\mathbf{w})=\ell_{t}(\mathbf{w})+\frac{\lambda}{2}\|\mathbf{w}\|^{2}.$

In the present paper we fix $\ell_{t}$ to be the _hinge loss_ , i.e.:
$\ell_{t}(\mathbf{w}^{(T+1)})~{}=~{}\max\\{0,1-y_{t}\langle\mathbf{w}^{(T+1)},\mathbf{x}_{t}\rangle\\},$
where $\mathbf{w}^{(T+1)}$ is the vector of parameters corresponding task
$(T+1)$.

We propose to balance between the prediction $O_{T+1}$ using only the current
task and the prediction $O_{1:T}$ using the accumulated knowledge base through
a non-increasing series $\\{\alpha_{t}\\}_{t=1}^{N_{T+1}}$ with $\forall
t,~{}0\leq\alpha_{t+1}\leq\alpha_{t}\leq 1$ as the following:

$\mathcal{P}_{T+1}(\mathbf{x})=\alpha_{t}O_{1:T}(\mathbf{x})+(1-\alpha_{t})O_{T+1}(\mathbf{x}).$

Intuitively, when gradually receiving more and more instances, the impact of
the knowledge base on the final prediction will also decrease gradually.
Therefore, the decreasing $\alpha_{t}$ aims at balancing predictions between
the knowledge base and the current classifier $h_{T+1}$. And as for the first
task, we only learn from the current classifier $h_{1}$ since
$K_{0}=\emptyset$, i.e., $\mathcal{P}_{1}(\mathbf{x})=O_{1}(\mathbf{x}).$ In
this case one could see the problem as a classical online learning algorithm,
with the predictions being performed by a classifier interactively updated by
the arriving observations.

For all $T$, to evaluate $\mathcal{P}_{T+1}$, we benefit from the fact that
the proposed updating rule for the knowledge base is additive. Algorithms 1
and 2 present the approach more formally. Moreover, two distinct approaches
(options) are proposed in order to make predictions from the knowledge base
for Accumulated Knowledge Lifelong Online (AKLO) learning: _AKLO Sum_ and
_AKLO Sample_.

AKLO Sum

Predictions are made by a weighted vote from the models in the knowledge base:

$O_{1:T}(\mathbf{x}_{t}^{(T+1)})=\mathcal{T}_{[-1,1]}\big{(}\sum_{i=1}^{T}p_{t}(i)\langle\mathbf{w}^{(i)},\mathbf{x}_{t}^{(T+1)}\rangle\big{)},$
(2)

where $\mathcal{T}_{[a,b]}$ is a piece-wise function defined as:

$\mathcal{T}_{[a,b]}(x)=\left\\{\begin{array}[]{ll}a&\quad x\leq a\\\ x&\quad
a<x<b\\\ b&\quad x\geq b\end{array}.\right.$

AKLO Sample

Predictions are made by sampling a model $i$ from the knowledge base according
to the Categorical distribution $i\sim\mathrm{Cat}(p_{t})$:

$O_{1:T}(\mathbf{x}_{t}^{(T+1)})=\mathcal{T}_{[-1,1]}\big{(}\langle\mathbf{w}^{(i)},\mathbf{x}_{t}^{(T+1)}\rangle\big{)},$
(3)

where the weight $p_{t}$ is defined as the $T$ simplex with
$\sum_{i}^{T}p_{t}(i)~{}=~{}1,~{}\forall i,~{}p_{t}(i)\geq 0$, estimated by
their historical behaviors $p_{t}(i)\propto\exp\\{-\epsilon_{t}L_{t}(i)\\}$.
$L_{t}(i)=\sum_{k=1}^{t}e_{k}(i)$ represents the cumulative error at task
$T+1$ for model $h_{i}$ ($1\leq i\leq T$) in the interactive learning until
time $t$. It is also worth mentioning that since the tasks are not necessarily
i.i.d. generated, the performance of the algorithm is measured by a small
cumulative error in the task $T+1$.

Algorithm 1 Prediction $\mathcal{P}_{T+1}$ at task $T+1$

0: $\eta_{t}>0$, $\forall t\in\\{1,\dots,N_{T+1}\\}$

0: $\mathbf{w}^{(T+1)}_{1}=\mathbf{0}$, $p_{1}=\frac{1}{T}\mathbf{1}$

1: for $t=1$ to $N_{T+1}$ do

2: Observe $\mathbf{x}_{t}^{(T+1)}$

3: Predict using the current classifier $h_{T}$ by computing the
confidence:$O_{T+1}(\mathbf{x}_{t}^{(T+1)})=\mathcal{T}_{[-1,1]}\big{(}\langle\mathbf{w}^{(T+1)}_{t},\mathbf{x}_{t}^{(T+1)}\rangle\big{)}$

4: Predict using cumulative knowledge $K_{T}$ by applying Equation 2 or 3, get
confidence $O_{1:T}(\mathbf{x}_{t}^{(T+1)})$

5: Final Prediction:
$\hat{y}_{t}^{(T+1)}=\mathrm{sign}[(1-\alpha_{t})O_{T+1}(\mathbf{x}_{t}^{(T+1)})+\alpha_{t}O_{1:T}(\mathbf{x}_{t}^{(T+1)})]$

6: Receive the real label $y_{t}^{(T+1)}$

7: Update weight $p_{t}$ by $y_{t}^{(T+1)}$ with Algorithm 2

8: Update $\mathbf{w}^{(T+1)}_{t}$:

9: if
$y_{t}^{(T+1)}\langle\mathbf{w}^{(T+1)}_{t},\mathbf{x}_{t}^{(T+1)}\rangle\geq
1$ then

10: $\mathbf{w}^{(T+1)}_{t+1}=(1-\eta_{t}\lambda)\mathbf{w}^{(T+1)}_{t}$

11: else

12:
$\mathbf{w}^{(T+1)}_{t+1}=(1-\eta_{t}\lambda)\mathbf{w}^{(T+1)}_{t}+\eta_{t}y_{t}^{(T+1)}\mathbf{x}_{t}^{(T+1)}$

13: end if

14: end for

Algorithm 2 Updating the knowledge base $\mathrm{K}_{T}$

0: Knowledge $\mathrm{K}_{T}$, $\epsilon_{t}>0$, $\forall
t\in\\{1,\dots,N_{T+1}\\}$

0: $L_{1}=\mathbf{0}$

1: for $t=1$ to $N_{T+1}$ do

2: Receive label $y_{t}^{(T+1)}$

3: Compute error $e_{t}$ for each model $j\in\\{1,\dots,T\\}$ in the knowledge
base $K_{T}$:
$e_{t}(h_{j})=\Big{(}\mathcal{T}_{[-1,1]}\big{(}\langle\mathbf{w}^{(j)},\mathbf{x}^{(T+1)}_{t}\rangle\big{)}-y_{t}^{(T+1)}\Big{)}^{2}$

4: Update $L_{t+1}=L_{t}+e_{t}$

5: Update
$p_{t}(i)=\frac{\exp\\{-\epsilon_{t}L_{t}(i)\\}}{\sum_{j=1}^{T}\exp\\{-\epsilon_{t}L_{t}(j)\\}}$

6: end for

#### Computational complexity

The computational complexity of updating rule in Algorithm 2 is
$\mathcal{O}(Td+T)$. Then plugging in Algorithm 1 we derive a global
complexity of $\mathcal{O}(N_{T+1}(Td+T+d))$, which corresponds to a _linear
time_ algorithm that is able to efficiently process high dimensionality
datasets.

## Theoretical analysis

In the following, we analyze the error bound of the proposed algorithm at
current task $T+1$. This is coherent with Pentina and Urner (2016), where a
batch non i.i.d. lifelong learning problem is analyzed. Two scenarios are
discussed: a known horizon $N_{T+1}$ and a fixed but unknown horizon
$N_{T+1}$.

### Known horizon $N_{T+1}$

#### Theorem 1

Supposing $\|\mathbf{x}\|\leq X$ and $\|\mathbf{w}\|\leq R$, for any
$\\{\alpha_{t}\\}_{t=1}^{N_{T+1}}$ such that $\alpha_{1}=1$ and $\forall
t,~{}0\leq\alpha_{t+1}\leq\alpha_{t}\leq 1$, we set
$\lambda=\frac{X+R}{R}\sqrt{\frac{\log(N_{T+1})+1}{N_{T+1}}}$,
$\eta_{t}=1/(\lambda t)$,
$\epsilon_{t}=\sqrt{\frac{\log(T)}{8\sum_{t=1}^{N_{T+1}}\alpha_{t}}}$ and we
use Equation 2 (AKLO Sum) as the prediction rule. If the algorithm predicts
$\hat{y}_{t}$, then the cumulative error
$E_{T+1}=\sum_{t=1}^{N_{T+1}}\mathbf{1}\\{\hat{y}_{t}\neq y_{t}\\}$ at task
$T+1$ can be bounded by:

$\begin{split}E_{T+1}\leq&\sum_{t=1}^{N_{T+1}}\alpha_{t}e_{t}(\mathbf{w}^{\star\star})+\sum_{t=1}^{N_{T+1}}(1-\alpha_{t})\ell_{t}(\mathbf{w}^{\star})\\\
&+4\sqrt{2\log(T)\sum_{t=1}^{N_{T+1}}\alpha_{t}}\\\
&+\sum_{t=1}^{N_{T+1}}(1-\alpha_{t})R(X+R)\sqrt{\frac{\log(N_{T+1})+1}{N_{T+1}}},\end{split}$

where:

$\displaystyle\mathbf{w}^{\star}$ $\displaystyle=\underset{\|\mathbf{w}\|\leq
R}{\min}\sum_{t=1}^{N_{T+1}}\ell_{t}(\mathbf{w}),$
$\displaystyle\mathbf{w}^{\star\star}$
$\displaystyle=\underset{\mathbf{w}\in\\{\mathbf{w}^{(1)},\dots,\mathbf{w}^{(T)}\\}}{\min}\sum_{t=1}^{N_{T+1}}e_{t}(\mathbf{w}).$

We provide the complete demonstration in the supplementary material.

If we set $\alpha_{t}\equiv 1$, then the error bound can be simplified as
$\sum_{t=1}^{N_{T+1}}e_{t}(\mathbf{w}^{\star\star})+4\sqrt{2N_{T+1}\log(T)}$,
which exactly recovers to the error bound of the expert problem (Cesa-Bianchi
and Lugosi, 2006). Furthermore, if we set $\alpha_{t}\equiv 0$, the error
bound can be simplified as
$\sum_{t=1}^{N_{T+1}}\ell_{t}(\mathbf{w}^{\star})+R(X+R)\sqrt{N_{T+1}(\log(N_{T+1})+1)}$,
which recovers to the error bound of the Follow The Adaptive Regularized
Leader (FTARL) problem (McMahan, 2017). We should also point out that the
proposed algorithm is not sensible to the size of the knowledge base because
of the $\log(T)$ term in the bound. Besides the direct conclusion from Theorem
1, we can also derive the Corollary 1 if the number of instances $N_{T+1}$ is
small.

#### Corollary 1

For $\gamma\in(0,1)$, if $N_{T+1}$ satisfies $N_{T+1}\leq t_{0}$, with
$t_{0}=\max\\{t|\alpha_{t}\geq\frac{K}{1+K}\\}$, if we assume that
$\zeta=\underset{\mathbf{w}\in\\{\mathbf{w}^{(1)},\dots,\mathbf{w}^{(T)}\\}}{\min}\sum_{t=1}^{N_{T+1}}e_{t}(\mathbf{w})$
is non zero ($\zeta>0$) and
$K\geq\max\\{\frac{1+XR}{\gamma\zeta},\frac{R(X+R)}{4\gamma\sqrt{2\log(T)}}\\}$,
the cumulative error bound $E_{T+1}$ in Theorem 1 can be simplified as:

$E_{T+1}\leq(1+\gamma)\Big{(}\sum_{t=1}^{N_{T+1}}\alpha_{t}e_{t}(\mathbf{w}^{\star\star})+4\sqrt{2\log(T)\sum_{t=1}^{N_{T+1}}\alpha_{t}}\Big{)}.$

The proof of Corollary 1 is also provided in the supplementary material. This
corollary reveals an interesting fact: if $N_{T+1}$ is smaller than a
predefined threshold, then the best model in the knowledge base
$\mathrm{K}_{T}$ will play a dominant role in the error bound. Therefore, in
the current task $T+1$ and even with small interaction times $N_{T+1}$, it is
still possible to obtain a good performance although the current classifier
$h_{T+1}$ is not well trained.

#### Corollary 2

Supposing all the conditions of Theorem 1 hold and now we use Equation 3 (AKLO
Sampling) as the prediction rule. If the algorithm predicts $\hat{y}_{t}$,
then with the probability higher than $1-\delta$, with
$\forall\delta\in(0,1)$, the accumulative error $E_{T+1}$ at task $T+1$ can be
bounded by:

$\begin{split}E_{T+1}\leq&\sum_{t=1}^{N_{T+1}}\alpha_{t}e_{t}(\mathbf{w}^{\star\star})+\sum_{t=1}^{N_{T+1}}(1-\alpha_{t})\ell_{t}(\mathbf{w}^{\star})\\\
&+4\sqrt{2\log(T)\sum_{t=1}^{N_{T+1}}\alpha_{t}}+\sqrt{8\sum_{t=1}^{N_{T+1}}\alpha_{t}^{2}\log(\frac{1}{\delta})}\\\
&+\sum_{t=1}^{N_{T+1}}(1-\alpha_{t})R(X+R)\sqrt{\frac{\log(N_{T+1})+1}{N_{T+1}}}.\end{split}$

The proof of Corollary 2 is based on Theorem 1 and Hoeffding-Azuma inequality.
We also provide the demonstration in the supplementary material.

### Unknown horizon $N_{T+1}$

As we described in the introduction, in the lifelong online learning, both the
number of tasks and the number of examples in each task can all be unknown to
us. However, parameters $\eta_{t}$ and $\epsilon_{t}$ directly depend on
$N_{T+1}$. In the following, we develop some strategies for setting these
hyper-parameters without the knowledge of $N_{T+1}$.

For the current learner $h_{T+1}$, if we directly adjust
$\lambda=\frac{X+R}{R}$, we can also derive a bound in this setting, where the
whole proof procedure is similar to Theorem 1. As for the prediction from
Algorithm 2 in the knowledge base, we apply the _double trick_ for online
learning (Cesa-Bianchi and Lugosi, 2006), which divides the time interval into
periods $I_{m}=[2^{m},2^{m+1}-1]$ of length $2^{m}$, for $m=0,1,\dots$, until
the task completes. The modified algorithm is almost the same as Algorithm 2,
with the difference of a reset $L_{t}=0$ at the beginning of each new interval
$I_{m}$ and the use of
$\epsilon_{t}=\sqrt{\frac{\log(T)}{8\sum_{j=1}^{2^{m}}\alpha_{j}}}$ for $t\in
I_{m}$. Based on such a modified algorithm, we can derive the new error bound
for the unknown $N_{T+1}$ at task $T+1$.

#### Theorem 2

Supposing $\|\mathbf{x}\|\leq X$ and $\|\mathbf{w}\|\leq R$, for any
$\\{\alpha_{t}\\}_{t=1}^{N_{T+1}}$ with $\alpha_{1}=1$ and $\forall
t,~{}0\leq\alpha_{t+1}\leq\alpha_{t}\leq 1$, we set $\lambda=\frac{X+R}{R}$,
$\eta_{t}=1/(\lambda t)$, $\epsilon_{t}$ decided by the double trick and we
choose Equation 2 (AKLO Sum) as the prediction rule. If the algorithm predicts
$\hat{y}_{t}$, then the cumulative error $E_{T+1}$ at task $T+1$ can be
bounded by:

$\begin{split}E_{T+1}\leq&\sum_{t=1}^{N_{T+1}}\alpha_{t}e_{t}(\mathbf{w}^{\star\star})+\sum_{t=1}^{N_{T+1}}(1-\alpha_{t})\ell_{t}(\mathbf{w}^{\star})\\\
&+4\log(N_{T+1})\sqrt{2\log(T)\sum_{t=1}^{N_{T+1}}\alpha_{t}}\\\
&+R(X+R)\sum_{t=1}^{N_{T+1}}(1-\alpha_{t}).\end{split}$

The proof of Theorem 2 is provided in the supplementary material. We should
point out that this error bound is worse than the original bound proposed in
Theorem 1 since we do not know the $N_{T+1}$ in advance. However, in the real
lifelong learning problem, the value $N_{T+1}$ is generally not too large for
each task, such that $\log(N_{T+1})$ and $\sum_{t=1}^{N_{T+1}}(1-\alpha_{t})$
are both small, making the learning procedure effective.

## Empirical evaluations

We evaluate the empirical performance of the proposed algorithm in the online
setting. We will test the proposed algorithm by two synthetic and three real
datasets. Concerning real datasets, the data are already separated in
different categories (i.e., tasks) and for each task a uniform sampling
without replacement is performed on the original data to keep only a portion
of examples. The reason behind this is twofold: 1) in the lifelong learning
context the number of examples for each task is generally not large; and 2)
keeping a relative smaller size theoretically emphasizes the effectiveness of
the proposed algorithm, as shown in Corollary 1. In the following a further
description of the dataset used is given.

### Dataset description

#### Synthetic data 1 (syn1)

For testing the behavior of the proposed algorithm particularity in the non
i.i.d. assumption, we create a set of tasks $\\{j\\}_{1}^{50}$ generated by
two different distributions, namely $\mathcal{D}_{1}$ and $\mathcal{D}_{2}$
(25 tasks for each distribution). Each task $j$ is composed by $N_{j}=100$
instances $\\{(\mathbf{x}_{t}^{j},y_{t}^{j})\\}_{t=1}^{100},$ where
$\mathbf{x}_{t}^{j}\in\mathbb{R}^{2}$ and $y_{t}^{j}\in\\{-1,+1\\}$. For a
given task $T$ from the first distribution $\mathcal{D}_{1}$,
$\mathbf{x}_{t}^{T}$ is generated from the normal multivariate distribution
$\mathbf{x}_{t}^{T}\sim\mathcal{N}(\boldsymbol{\mu}_{1},\sigma_{1}\mathbf{I})$
with $\boldsymbol{\mu}_{1}=[10,10]$ and $\sigma_{1}=1$ and the labeling
function $y_{t}^{T}$ is expressed as
$y_{t}^{T}=\mathrm{sign}(\mathbf{a}_{T}^{\top}\mathbf{x}_{t})$, where
$\mathbf{a}_{T}=[-1+\epsilon_{T},1+\epsilon_{T}]^{\top}$ is the decision
function and $\epsilon_{T}\sim\mathcal{N}(0,10^{-3})$ is the variant part for
task $T$. Similarly, for a given task $T^{\prime}$ from the second
distribution $\mathcal{D}_{2}$ the same strategy is applied using different
parameter settings. Thus, for $\mathcal{D}_{2}$ we set
$\boldsymbol{\mu}_{2}=[20,5]$, $\sigma_{2}=1$,
$y_{t}^{T^{\prime}}=\mathrm{sign}(\mathbf{a}_{T^{\prime}}^{\top}\mathbf{x}_{t})$
and the decision boundary
$\mathbf{a}_{T^{\prime}}=[-0.25+\epsilon_{T^{\prime}},1+\epsilon_{T^{\prime}}]^{\top}$
with $\epsilon_{T^{\prime}}\sim\mathcal{N}(0,10^{-3})$. Furthermore, the agent
has no access to the data structure and generation information and the tasks
from these two distributions will be arbitrarily provided to the agent.

#### Synthetic data 2 (syn2)

With this dataset, we will test the performance of the proposed algorithm in a
non-obvious situation (e.g., adversarial setting). For this, we adopt similar
settings as in the syn1 with the same number of tasks and the same number of
examples for each task. We also keep the same generation technique from
$\mathcal{D}_{1}$ and $\mathcal{D}_{2}$ . However, for all the tasks the
observations have the same feature generation distribution (i.e., the same
marginal distribution)
$\mathbf{x}_{t}\sim\mathcal{N}(\boldsymbol{\mu},\sigma\mathbf{I})$ and a
totally different labeling function. As for the first distribution
$\mathcal{D}_{1}$, the labeling function is
$y_{t}=\mathrm{sign}(\mathbf{a}^{\top}\mathbf{x}_{t})$ and for the second the
distribution $\mathcal{D}_{2}$, the labeling function is adversarial given by
$y_{t}=\mathrm{sign}(-\mathbf{a}^{\top}\mathbf{x}_{t})$, where
$\mathbf{a}=[-1+\epsilon_{T},1+\epsilon_{T}]^{\top}$ and $\epsilon_{T}$ is
kept the same as in syn1. The non-obvious generation is common in the non
i.i.d. settings. For example, in the personalized product recommendation for
two groups of clients, the product might have the same features but these two
groups may have totally different preferences on the same product. In the
experiment, the agent still knows nothing except $(\mathbf{x}_{t},y_{t})$.

#### Landmine detection111 http://www.ee.duke.edu/~lcarin/LandmineData.zip
(Landmine)

This dataset contains 29 binary classification tasks corresponding to 29
geographical regions. For each task, the goal is to detect landmines ($+1$) or
clutters ($-1$). Each example contains $9$ features, we also add a bias term,
resulting in $10$ features during the experiment. We randomly sample $150$
examples for each task.

#### Spam detection222http://ecmlpkdd2006.org/challenge.html (Spam)

We adopt the dataset from ECML PKDD 2006 Discovery challenge for the spam
detection task of $14$ different users. Each user can be viewed as a
individual task and the goal is to build a personalized mail filtering system.
This task is a binary classification problem with label spam ($+1$) and non-
spam ($-1$). Each example has an extremely high number of features ($\approx
1.5\times 10^{5}$) representing the word occurrence frequency (bag of word
model). We randomly select $200$ examples for each task.

#### Shoes data333http://vision.cs.utexas.edu/whittlesearch/ (Shoes)

We used the shoes dataset with attributes from Kovashka, Parikh, and Grauman
(2012) and the same setting as in (Pentina, Sharmanska, and Lampert, 2015). In
this experiment, we study the scenario of learning visual attributes that
characterize shoes across different shoe models. We have $10$ attributes
representing the different tasks in the proposed lifelong online setting
(_pointy at the front, open, bright in color, covered with ornaments, shiny,
high at the heel, long on the leg, formal, sporty, feminine_) which describes
the shoes models. Each attribute (i.e., task) is a binary classification
problem with $100$ examples: $(+1)$ means holding such a property and $(-1)$
means not. Each example has $990$ dimensional features from the original
image. In this dataset, it is worth mentioning that some tasks are clearly
related such as _high heel_ and _shiny_ , and some tasks are not, such as
_high heel_ and _sporty_.

### Methods and measurements

In our experiments, we compare different baseline approaches to verify that
the proposed algorithms can effectively learn from the accumulating knowledge
during the lifelong online learning process:

ITOL:

Independent Task Online Learning — for each task, we train the classifier in
an online way without taking into account the accumulated experience from the
previous tasks;

TOL:

Tasks Online Learning — only one online classifier is performed by
concatenating the data of all the tasks;

AKLO Sum:

Accumulating Knowledge for Lifelong Online learning using the sum updating
rule described in Equation 2 — the prediction is the weighted sum of the
learned classifiers in the _knowledge base_ ;

AKLO Sample:

Accumulating Knowledge for Lifelong Online learning using the sampling
updating rule described in Equation 3 — the prediction comes from a random
sampling w.r.t. the estimated normalized weight;

Unif Sum:

Uniform sum from the _knowledge base_ — this approach is similar to the
proposed algorithm with Equation 2, but the prediction is directly the average
of the advice from the _knowledge base_ , i.e., $p_{t}(i)=\frac{1}{T}$;

Unif Sample:

Uniform sampling from the _knowledge base_ — this approach is similar to the
proposed algorithm with Equation 3, with the uniform sampling strategy.

To fix the hyper-parameters $\alpha_{t}$, $\epsilon_{t}$ and $\lambda$ of the
proposed model, we adopt a simple linear time function for $\alpha_{t}$:
$\alpha_{t}=1-\frac{(t-1)}{N_{T+1}}$ and
$\epsilon_{t}=\sqrt{\frac{\log(T)}{8\sum_{t=1}^{N_{T+1}}\alpha_{t}}}$ as
described in Theorem 1. As for selecting the $\lambda$, we choose it as the
best $\lambda$ from $[10^{-3},10^{-2},\dots,10^{3}]$ obtained from averaging
the performance over ITOL. For some datasets which already have a training and
testing set (such as Spam and Shoes datasets), we validate the best $\lambda$
from the training set. For the other datasets, for each task, we randomly
sample a small portion and we shuffle the sampled examples. Lastly, the
learning rate in Algorithm 1 is set accordingly as $\eta_{t}=1/(\lambda t)$.

### Results and analysis

To measure the performance, we report the average cumulative errors (ACE) over
the tasks in the lifelong learning:
$\mathrm{ACE}=\frac{1}{T}\sum_{j=1}^{T}\frac{1}{N_{j}}\sum_{t=1}^{N_{j}}\mathbf{1}\\{\hat{y}_{t}\neq
y_{t}\\}$.

Table 1 presents 10 repetitions by randomly shuffling the task order and the
example order in each task. The proposed AKLO Sum and AKLO Sample methods
demonstrate their effectiveness by using the accumulating knowledge compared
to the baseline approaches. Particularly the proposed approaches are more
efficient when using datasets where the tasks are not i.i.d. generated (i.e.,
syn1, syn2).

It is worth mentioning that the Unif Sample is even worse than the independent
training (ITOL) as _negative transfers_ are likely to occur with non i.i.d.
datasets. By comparing results obtained with Unif Sum to AKLO Sum, it reveals
that the proposed AKLO Sum is not simply leveraging the information from the
previous experiences, but is able to make an online selection of related tasks
when effectively combining them to make predictions, such as the _shoes_
dataset. AKLO Sum has about $10\%$ lower error rate than the Unif Sum since
some tasks are not related and Unif Sum approach treats all the tasks
equivalently.

Moreover, in the _landmine_ dataset, we find that the performance of the
proposed methods are slightly higher than the results given by for instance
Unif Sum technique compared to other dataset. A possible reason for this is
that _landmine_ is generally regarded as an i.i.d. realization by the task
distribution Pentina and Lampert (2014) and Unif Sum behaves similarly to a
_parameter transfer_ approach in this case.

Fig. 2(a)-(e) show the the evolution of the average cumulative error rate with
the tasks, following the same trends reported in Table 1. Fig. 2(f) reports
the cumulative errors of one realization of tested algorithms for the last
task of the _shoes_ dataset — comparisons of other datasets are provided in
the supplementary material. Such results support the claims that even in the
context of non i.i.d. tasks and a relatively small scale interactions, the
algorithms proposed are still able to provide a good performance for each
task.

Table 2 shows the average running time for each algorithm, where proposed AKLO
Sample and AKLO Sum algorithms demonstrates their time efficiency. We should
also point out that for the _Spam_ dataset, the execution time are longer than
the others datasets given the extremely high dimensionality of the
observations. Moreover, Unif Sample and AKLO Sample approaches take a little
more running time, due to the sampling operation.

Dataset | ITOL | TOL | Unif Sample | Unif Sum | AKLO Sample | AKLO Sum
---|---|---|---|---|---|---
Syn1 |  $41.10$$\pm 1.80$ |  $22.08$$\pm 2.11$ |  $40.77$$\pm 2.38$ |  $35.31$$\pm 1.87$ |  $13.87$$\pm 1.57$ |  $\mathbf{11.00}$$\mathbf{\pm 1.52}$
Syn2 |  $41.85$$\pm 1.35$ |  $43.35$$\pm 2.95$ |  $49.75$$\pm 0.92$ |  $41.84$$\pm 1.29$ |  $16.06$$\pm 2.22$ |  $\mathbf{12.91}$$\mathbf{\pm 2.42}$
Landmine |  $18.81$$\pm 0.33$ |  $14.49$$\pm 0.37$ |  $21.94$$\pm 2.10$ |  $11.14$$\pm 0.91$ |  $13.52$$\pm 0.76$ |  $\mathbf{10.39}$$\mathbf{\pm 0.38}$
Spam |  $24.53$$\pm 1.23$ |  $17.47$$\pm 0.65$ |  $30.02$$\pm 2.92$ |  $16.68$$\pm 1.66$ |  $18.20$$\pm 1.18$ |  $\mathbf{14.30}$$\mathbf{\pm 1.32}$
Shoes |  $31.16$$\pm 1.97$ |  $29.81$$\pm 1.87$ |  $43.27$$\pm 5.10$ |  $31.64$$\pm 3.99$ |  $26.22$$\pm 2.96$ |  $\mathbf{20.74}$$\mathbf{\pm 3.24}$
Table 1: Average cumulative error rate ($\%$) $\pm$ standard deviation ($\%$) over $10$ repetitions. Dataset | ITOL | TOL | Unif Sample | Unif Sum | AKLO Sample | AKLO Sum
---|---|---|---|---|---|---
Syn1 | $0.029$ | $0.023$ | $0.426$ | $0.211$ | $0.425$ | $0.211$
$\pm 1.57\times 10^{-4}$ | $\pm 4.39\times 10^{-4}$ | $\pm 3.63\times 10^{-4}$ | $\pm 7.77\times 10^{-4}$ | $\pm 6.89\times 10^{-4}$ | $\pm 5.47\times 10^{-4}$
Syn2 | $0.028$ | $0.028$ | $0.418$ | $0.205$ | $0.416$ | $0.205$
$\pm 1.25\times 10^{-4}$ | $\pm 5.42\times 10^{-4}$ | $\pm 4.93\times 10^{-4}$ | $\pm 4.00\times 10^{-4}$ | $\pm 3.70\times 10^{-4}$ | $\pm 3.47\times 10^{-4}$
Landmine | $0.022$ | $0.019$ | $0.361$ | $0.179$ | $0.360$ | $0.179$
$\pm 7.82\times 10^{-5}$ | $\pm 7.92\times 10^{-5}$ | $\pm 2.36\times 10^{-3}$ | $\pm 3.30\times 10^{-4}$ | $\pm 1.18\times 10^{-3}$ | $\pm 6.17\times 10^{-4}$
Spam | $0.380$ | $0.398$ | $2.260$ | $2.249$ | $2.467$ | $2.238$
$\pm 0.012$ | $\pm 0.011$ | $\pm 0.015$ | $\pm 0.019$ | $\pm 0.011$ | $\pm 0.011$
Shoes | $0.006$ | $0.006$ | $0.080$ | $0.043$ | $0.080$ | $0.043$
$\pm 5.73\times 10^{-5}$ | $\pm 6.67\times 10^{-5}$ | $\pm 1.44\times 10^{-4}$ | $\pm 1.23\times 10^{-4}$ | $\pm 1.63\times 10^{-4}$ | $\pm 1.02\times 10^{-4}$
Table 2: Average running time (seconds) $\pm$ standard deviation (seconds)
over $10$ repetitions.

(a) syn1 (b) syn2 (c) Landmine

(d) Spam (e) Shoes (f) Shoes’ last task

Figure 2: Evolution of performances of tested algorithms on different
datasets: (a)-(e) evolution of Average Cumulative Error rate (ACE in $\%$)
over the tasks; (f) cumulative errors for the last task in shoes dataset, for
one experiment.

## Conclusion

We are proposing a novel lifelong online learning framework to deal with
consecutive online learning tasks relying on knowledge accumulated during past
experiences. Specific methods are given to effectively leverage the
predictions from the current task classifier and the models built for previous
tasks. A theoretical analysis shows the effectiveness of the proposed method
even without assumptions on the distributions. Several experiments on both
synthetic and real datasets with different contexts are also providing an
empirical validation of the proper behaviour of the proposed algorithms.

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# Weak type commutator and Lipschitz estimates: resolution of the Nazarov-
Peller conjecture

M. Caspers, D. Potapov, F. Sukochev, D. Zanin  M. Caspers, Fachbereich
Mathematik und Informatik der Universität Münster, Einsteinstrasse 62, 48149
Münster, Germany<EMAIL_ADDRESS>D. Potapov, F. Sukochev, D.
Zanin, School of Mathematics and Statistics, UNSW, Kensington 2052, NSW,
Australia<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
(Date: , MSC2000: 47B10, 47L20, 47A30,
Acknowledgement: The first author is supported by the grant SFB 878. The work
of other co-authors is supported by the ARC)

###### Abstract.

Let $\mathcal{M}$ be a semi-finite von Neumann algebra and let
$f:\mathbb{R}\rightarrow\mathbb{C}$ be a Lipschitz function. If
$A,B\in\mathcal{M}$ are self-adjoint operators such that $[A,B]\in
L_{1}(\mathcal{M}),$ then

$\|[f(A),B]\|_{1,\infty}\leq c_{abs}\|f^{\prime}\|_{\infty}\|[A,B]\|_{1},$

where $c_{abs}$ is an absolute constant independent of $f$, $\mathcal{M}$ and
$A,B$ and $\|\cdot\|_{1,\infty}$ denotes the weak $L_{1}$-norm. If
$X,Y\in\mathcal{M}$ are self-adjoint operators such that $X-Y\in
L_{1}(\mathcal{M}),$ then

$\|f(X)-f(Y)\|_{1,\infty}\leq c_{abs}\|f^{\prime}\|_{\infty}\|X-Y\|_{1}.$

This result resolves a conjecture raised by F. Nazarov and V. Peller implying
a couple of existing results in perturbation theory.

##  1\. Introduction

Let $L_{p}(H)$ be the Schatten-von Neumann ideal of $B(H)$. It consists of all
compact operators for which its sequence of singular values lies in
$\ell_{p}.$ Let $F_{p}$ be the class of functions
$f:\mathbb{R}\rightarrow\mathbb{C}$ such that

$f(B)-f(C)\in L_{p}(H),$

for all self-adjoint $B,C$ such that $B-C\in L_{p}(H)$ and set

$\|f\|_{F_{p}}=\sup_{B\neq C}\frac{\|f(B)-f(C)\|_{p}}{\|B-C\|_{p}}.$

It was conjectured by M.G. Krein [17] that whenever the derivative
$f^{\prime}\in L_{\infty}(\mathbb{R})$ we have $f\in F_{1}$. This conjecture
does not hold as was shown by Y.B. Farforovskaya in [11]. Also it was shown
that the analogue of Krein’s problem fails in the case $p=\infty$ (see [9,
10]). In fact already for the absolute value function it was found by T. Kato
that Krein’s problem has a negative answer [15]; and similarly in the case
$p=1$ by E.B. Davies [5].

A positive result in this direction was first obtained by M. Birman and M.
Solomyak [1, Theorem 10] who proved that $C^{1+\epsilon}\subseteq F_{1}$ for
every $\epsilon>0$, and later improved by V. Peller [22] who showed that
$B_{\infty 1}^{1}\subseteq F_{1}$. Here $B_{pq}^{s}$ is the class of Besov
spaces for which we refer to [12]. The Krein problem for the case
$1<p<\infty,p\not=2$ remained open until [24]. In [24] it was shown by the
second and third named author that $F_{p}$ consists exactly of all Lipschitz
functions. Moreover in [3] a quantitative estimate for $\|f\|_{F_{p}}$ was
found, namely $\|f\|_{F_{p}}\simeq p^{2}/(p-1)$. Earlier the same problem had
been considered by M. de la Salle (unpublished, see [27]) who was able to show
already that $\|f\|_{F_{p}}\leq c_{abs}\cdot p^{4}/(p-1)^{2}$ with an absolute
constant $c_{abs},$ which is more optimal than [24].

Other results concerning this problem have been obtained in [8] and [16] and
in the context of this paper we also mention [26] in which weak estimates for
martingale inequalities were obtained.

Using interpolation, the above results would follow from a weak type Lipschitz
estimate between $L_{1}$ and the weak-$L_{1}$ space $L_{1,\infty}$. The
estimate was conjectured in a paper of F. Nazarov and V. Peller [19], and has
remained the major open question in the study of Lipschitz properties of
operator valued functions. Denote $L_{1,\infty}(H)$ for the weak $L_{1}$-space
consisting of all compact operators $A$ whose sequence $\\{\mu(k,A)\\}_{k\geq
0}$ of singular values satisfies $\mu(k,A)=O(\frac{1}{k+1}).$

######  Conjecture 1.1.

Let $f:\mathbb{R}\rightarrow\mathbb{C}$ be Lipschitz. Whenever $A,B\in B(H)$
are self-adjoint operators such that $A-B\in L_{1}(H)$, we have that
$f(A)-f(B)\in L_{1,\infty}(H)$ and

$\|f(A)-f(B)\|_{1,\infty}\leq c_{abs}\|f^{\prime}\|_{\infty}\|A-B\|_{1},$

for some absolute constant $c_{abs}$.

Nazarov and Peller [19] gave an affirmative answer under the assumption that
the rank of $A-B$ equals 1. Since $\|\>\cdot\>\|_{1,\infty}$ is a quasi-norm
and not a norm for $L_{1,\infty}(H)$ it is impossible to extend their result
for when $A-B$ is a general trace class operator.

Another positive result to the conjecture was found by the current authors in
[4] in the special case when $f$ is the absolute value map. The proof relies
on the observation that the Schur multiplier of divided differences

$\left(\frac{f(\lambda)-f(\mu)}{\lambda-\mu}\right)_{\lambda\not=\mu}$

can be written as a finite sum of compositions of a positive definite Schur
multiplier and a triangular truncation operator. For general Lipschitz
functions there is no reason that the latter fact should be true which renders
the technique of [4] inapplicable.

The main result of this paper is a proof of Conjecture 1.1. The importance of
this result lies in the fact that this gives the sharpest possible estimate
for perturbations and commutators. In particular it retrieves
$\|f\|_{F_{p}}\simeq p^{2}/(p-1)$ [3] and the Nazarov–Peller result [19]. A
key ingredient in our proof is the connection with non-commutative Calderón-
Zygmund theory and in particular J. Parcet’s extension of the classical
Calderón–Zygmund theorem (see Theorem 2.1 and [21]).

In the text we prove a somewhat stronger result in the terms of double
operator integrals (see next section for definition), of which Conjecture 1.1
is a corollary.

######  Theorem 1.2.

If $A$ is a self-adjoint operator affiliated with a semifinite von Neumann
algebra $\mathcal{M},$ and if $f:\mathbb{R}\to\mathbb{C}$ is Lipschitz then

$\|T_{f^{[1]}}^{A,A}(V)\|_{1,\infty}\leq
c_{abs}\|f^{\prime}\|_{\infty}\|V\|_{1},\quad V\in(L_{1}\cap
L_{2})(\mathcal{M}).$

Commutator estimate follows from the observation that the double operator
integral $T_{f^{[1]}}^{A,A}([A,B])$ equals $[f(A),B].$ As explained in the
proof of Theorem 5.3, Lipschitz estimates follow from commutator ones.

### Acknowledgements

Authors thank Javier Parcet for a detailed explanation of [21].

##  2\. Preliminaries

###  2.1. General notation

Let $\mathcal{M}$ be a semifinite von Neumann algebra equipped with a faithful
normal semifinite trace $\tau.$ In this paper, we always presume that
$\mathcal{M}$ is represented on a separable Hilbert space.

A (closed and densely defined) operator $x$ affiliated to $\mathcal{M}$ is
called $\tau-$measurable if $\tau(E_{|x|}(s,\infty))<\infty$ for sufficiently
large $s.$ We denote the set of all $\tau-$measurable operators by
$S(\mathcal{M},\tau).$ For every $x\in S(\mathcal{M},\tau),$ we define its
singular value function $\mu(A)$ by setting

$\mu(t,x)=\inf\\{\|x(1-p)\|_{\infty}:\quad\tau(p)\leq t\\}.$

Equivalently, for positive operator $x\in S(\mathcal{M},\tau),$ we have

$n_{x}(s)=\tau(E_{x}(s,\infty)),\quad\mu(t,x)=\inf\\{s:n_{x}(s)<t\\}.$

We have (see e.g. [18, Corollary 2.3.16])

(2.1) $\mu(t+s,x+y)\leq\mu(t,x)+\mu(s,y),\quad t,s>0.$

###  2.2. Non-commutative spaces

For $1\leq p<\infty$ we set,

$L_{p}(\mathcal{M})=\\{x\in S(\mathcal{M},\tau):\
\tau(|x|^{p})<\infty\\},\quad\|x\|_{p}=(\tau(|x|^{p}))^{\frac{1}{p}}.$

The Banach spaces $(L_{p}(\mathcal{M}),\|\cdot\|_{p})$, $1\leq p<\infty$ are
separable.

Define the space $L_{1,\infty}(\mathcal{M})$ by setting

$L_{1,\infty}(\mathcal{M})=\\{x\in S(\mathcal{M},\tau):\
\sup_{t>0}t\mu(t,x)<\infty\\}.$

We equip $L_{1,\infty}(\mathcal{M})$ with the functional
$\|\cdot\|_{1,\infty}$ defined by the formula

$\|x\|_{1,\infty}=\sup_{t>0}t\mu(t,x),\quad x\in L_{1,\infty}(\mathcal{M}).$

It follows from (2.1) that

$\|x+y\|_{1,\infty}=\sup_{t>0}t\mu(t,x+y)\leq\sup_{t>0}t(\mu(\frac{t}{2},x)+\mu(\frac{t}{2},y))\leq$
$\leq\sup_{t>0}t\mu(\frac{t}{2},x)+\sup_{t>0}t\mu(\frac{t}{2},y)=2\|x\|_{1,\infty}+2\|y\|_{1,\infty}.$

In particular, $\|\cdot\|_{1,\infty}$ is a quasi-norm. The quasi-normed space
$(L_{1,\infty}(\mathcal{M}),\|\cdot\|_{1,\infty})$ is, in fact, quasi-Banach
(see e.g. [14, Section 7] or [30]). In view of our main result it is important
to emphasize that the quasi-norm $\|\cdot\|_{1,\infty}$ is not equivalent to
any norm on $L_{1,\infty}(\mathcal{M})$ (see e.g [14, Theorem 7.6]).

###  2.3. Weak type inequalities for Calderón-Zygmund operators

Parcet [21] proved a noncommutative extension of Calderón-Zygmund theory.

Let $K$ be a tempered distribution which we refer to as the convolution
kernel. We let $W_{K}$ be the associated Calderón-Zygmund operator, formally
given by $f\mapsto K\ast f.$ In what follows, we only consider tempered
distributions having local values (that is, which can be identified with
measurable functions $K:\mathbb{R}^{d}\rightarrow\mathbb{C}$).

Let $\mathcal{M}$ be a semi-finite von Neumann algebra with normal, semi-
finite, faithful trace $\tau.$ The operator $1\otimes W_{K}$ can, under
suitable conditions, be defined as noncommutative Calderón-Zygmund operators
by letting them act on the second tensor leg of
$L_{1}(\mathcal{M})\widehat{\otimes}L_{1}(\mathbb{R}^{d}).$ The following
theorem in particular gives a sufficient condition for such an operator to act
from $L_{1}$ to $L_{1,\infty}.$

######  Theorem 2.1 ([21]).

Let $K:\mathbb{R}^{d}\backslash\\{0\\}\rightarrow\mathbb{C}$ be a kernel
satisfying the conditions111Here, $\nabla$ denotes the gradient
$(\frac{1}{i}\frac{\partial}{\partial
x_{1}},\cdots,\frac{1}{i}\frac{\partial}{\partial x_{d}}),$ which is
understood as unbounded self-adjoint operator on $L_{2}(\mathbb{R}^{d}).$

(2.2) $|K|(t)\leq\frac{{\rm const}}{|t|^{d}},\quad|\nabla K|(t)\leq\frac{{\rm
const}}{|t|^{d+1}}.$

Let $\mathcal{M}$ be a semi-finite von Neumann algebra. If $W_{K}\in
B(L_{2}(\mathbb{R}^{d})),$ then the operator $1\otimes W_{K}$ defines a
bounded map from $L_{1}(\mathcal{M}\otimes L_{\infty}(\mathbb{R}^{d}))$ to
$L_{1,\infty}(\mathcal{M}\otimes L_{\infty}(\mathbb{R}^{d})).$

###  2.4. Double operator integrals

Let $A=A^{*}$ be an operator affiliated with $\mathcal{M}.$ Symbolically, a
double operator integral is defined by the formula

(2.3)
$T_{\xi}^{A,A}(V)=\int_{\mathbb{R}^{2}}\xi(\lambda,\mu)dE_{A}(\lambda)VE_{A}(\mu),\quad
V\in L_{2}(\mathcal{M}).$

In the subsequent paragraph, we provide a rigorous definition of the double
operator integral.

Consider projection valued measures on $\mathbb{R}$ acting on the Hilbert
space $L_{2}(\mathcal{M})$ by the formulae $x\to E_{A}(\mathcal{B})x$ and
$x\to xE_{A}(\mathcal{B}).$ These spectral measures commute and, hence (see
Theorem V.2.6 in [2]), there exists a countably additive (in the strong
operator topology) projection-valued measure $\nu$ on $\mathbb{R}^{2}$ acting
on the Hilbert space $L_{2}(\mathcal{M})$ by the formula

(2.4) $\nu(\mathcal{B}_{1}\otimes\mathcal{B}_{2}):x\to
E_{A}(\mathcal{B}_{1})xE_{A}(\mathcal{B}_{2}),\quad x\in L_{2}(\mathcal{M}).$

Integrating a bounded Borel function $\xi$ on $\mathbb{R}^{2}$ with respect to
the measure $\nu$ produces a bounded operator acting on the Hilbert space
$L_{2}(\mathcal{M}).$ In what follows, we denote the latter operator by
$T_{\xi}^{A,A}$ (see also [20, Remark 3.1]).

In the special case when $A$ is bounded and ${\rm spec}(A)\subset\mathbb{Z},$
we have

(2.5)
$T_{\xi}^{A,A}(V)=\sum_{k,l\in\mathbb{Z}}\xi(k,l)E_{A}(\\{k\\})VE_{A}(\\{l\\}).$

We are mostly interested in the case $\xi=f^{[1]}$ for a Lipschitz function
$f.$ Here,

$f^{[1]}(\lambda,\mu)=\begin{cases}\frac{f(\lambda)-f(\mu)}{\lambda-\mu},\quad\lambda\neq\mu\\\
0,\quad\lambda=\mu.\end{cases}$

##  3\. Approximate intertwining properties of Fourier multipliers

We prove intertwining properties of Fourier multipliers partly inspired by K.
de Leeuw’s proof of his restriction theorem for $L^{p}$-multipliers [6,
Section 2].

In what follows,

$G_{l}(s)=\frac{1}{l\sqrt{\pi}}e^{-(\frac{s}{l})^{2}},\quad
s\in\mathbb{R},\quad l>0.$

That is, $G_{l}$ is a probability density function for certain Gaussian random
variable. The notation $G_{l}^{\otimes d}$ stands for the function from
$L_{1}(\mathbb{R}^{d})$ given by the tensor product of $G_{l}$ with itself
repeated $d$ times.

######  Lemma 3.1.

For every $f\in L_{1}(\mathbb{R})$ with $\int_{-\infty}^{\infty}f(s)ds=0,$ we
have $f\ast G_{l}\to 0$ in $L_{1}(\mathbb{R})$ as $l\to\infty.$

###### Proof.

Suppose first that $f$ is a step function of the form
$f=\sum_{k=1}^{m}\alpha_{k}\chi_{I_{k}},$ $m\geq 1$ where
$I_{k}=[a_{k},b_{k}],\ 1\leq k\leq m$ are disjoint intervals and
$\sum_{k=1}^{m}\alpha_{k}m(I_{k})=0.$

We have

$(f\ast
G_{l})(t)=\sum_{k=1}^{m}\alpha_{k}\int_{a_{k}}^{b_{k}}G_{l}(t-s)ds=\sum_{k=1}^{m}\alpha_{k}\int_{t-b_{k}}^{t-a_{k}}G_{l}(u)du=$
$=\sum_{k=1}^{m}\alpha_{k}\int_{\frac{t-b_{k}}{l}}^{\frac{t-a_{k}}{l}}G_{1}(s)ds=\sum_{k=1}^{m}\alpha_{k}(F(\frac{t-a_{k}}{l})-F(\frac{t-b_{k}}{l})),$

where $F(t)=\int_{-\infty}^{t}G_{1}(s)ds.$ To prove the assertion for our $f,$
it suffices to show that

$l\int_{-\infty}^{\infty}\left|\sum_{k=1}^{m}\alpha_{k}(F(t-\frac{a_{k}}{l})-F(t-\frac{b_{k}}{l}))\right|dt\to
0$

Clearly,

$\left|F(t-\frac{a_{k}}{l})-F(t)+\frac{a_{k}}{l}F^{\prime}(t)\right|\leq\frac{a_{k}^{2}}{2l^{2}}\max\limits_{s\in[t-\frac{a_{k}}{l},t]}|F^{\prime\prime}(s)|.$

If $l>\max_{1\leq k\leq m}|a_{k}|$ and $l>\max_{1\leq k\leq m}|b_{k}|,$ then

$\left|\sum_{k=1}^{m}\alpha_{k}(F(t-\frac{a_{k}}{l})-F(t-\frac{b_{k}}{l}))\right|\leq\frac{1}{2l^{2}}(\sum_{k=1}^{m}|\alpha_{k}|(a_{k}^{2}+b_{k}^{2}))\max\limits_{s\in[t-1,t+1]}|F^{\prime\prime}(s)|.$

This proves the assertion for $f$ as above.

To prove the assertion in general, fix $f_{m}$ as above (i.e., mean zero step
functions) such that $f_{m}\to f$ in $L_{1}(\mathbb{R}).$ Since
$\|G_{l}\|_{1}=1,$ it follows from Young’s inequality that

$\|f\ast G_{l}\|_{1}\leq\|(f-f_{m})\ast G_{l}\|_{1}+\|f_{m}\ast
G_{l}\|_{1}\leq\|f-f_{m}\|_{1}+\|f_{m}\ast G_{l}\|_{1}.$

Therefore,

$\limsup_{l\to\infty}\|f\ast G_{l}\|_{1}\leq\|f-f_{m}\|_{1}.$

Passing $m\to\infty,$ we conclude the proof. ∎

By Fubini Theorem, linear span of elementary tensors

$(f_{1}\otimes\cdots\otimes f_{d}):(t_{1},\cdots,t_{d})\to f_{1}(t_{1})\cdots
f_{d}(t_{d}),\quad f_{1},\cdots,f_{d}\in L_{1}(\mathbb{R})$

is dense in $L_{1}(\mathbb{R}^{d}).$

######  Lemma 3.2.

For every $f\in L_{1}(\mathbb{R}^{d})$ with $\int_{\mathbb{R}^{d}}f(s)ds=0,$
we have $f\ast G_{l}^{\otimes d}\to 0$ in $L_{1}(\mathbb{R}^{d})$ as
$l\to\infty.$

###### Proof.

Suppose first that $f$ is a linear combination of elementary tensors. That is,

(3.1) $f=\sum_{k=1}^{m}\bigotimes_{j=1}^{d}f_{jk},\quad f_{jk}\in
L_{1}(\mathbb{R}).$

Firstly, we consider the case when for every $k$, $1\leq k\leq m$ there exists
$j$, $1\leq j\leq d$ such that $\int_{\mathbb{R}}f_{jk}(s)=0.$ In this case,
by Lemma 3.1 we have that

$\|f\ast G_{l}^{\otimes d}\|_{1}\leq\sum_{k=1}^{m}\prod_{j=1}^{d}\|f_{jk}\ast
G_{l}\|_{1}\to 0.$

Now, we show that the case of $f$ given by (3.1) satisfying

(3.2) $\sum_{k=1}^{m}\prod_{j=1}^{d}\int_{\mathbb{R}}f_{jk}(s)ds=0.$

can be reduced to the just considered case when for every $k$, $1\leq k\leq m$
there exists $j$, $1\leq j\leq d$ such that $\int_{\mathbb{R}}f_{jk}(s)=0.$ To
this end, for every subset $\mathscr{A}\subset\\{1,\cdots,d\\},$ we set

$f_{j,k,\mathscr{A}}=\begin{cases}f_{jk}-(\int_{\mathbb{R}}f_{jk}(s)ds)\chi_{(0,1)},\quad
j\in\mathscr{A}\\\ (\int_{\mathbb{R}}f_{jk}(s)ds)\chi_{(0,1)},\quad
j\notin\mathscr{A}.\end{cases}$

By the linearity, we can rewrite (3.1) as

$f=\sum_{k=1}^{m}\sum_{\mathscr{A}\subset\\{1,\cdots,d\\}}\bigotimes_{j=1}^{d}f_{j,k,\mathscr{A}}$

Observing now that

$\sum_{k=1}^{m}\bigotimes_{j=1}^{d}f_{j,k,\varnothing}=(\sum_{k=1}^{m}\prod_{j=1}^{d}\int_{\mathbb{R}}f_{jk}(s)ds)\chi_{(0,1)}^{\otimes
d}$

and appealing to (3.2), we arrive at

(3.3)
$f=\sum_{k=1}^{m}\sum_{\varnothing\neq\mathscr{A}\subset\\{1,\cdots,d\\}}\bigotimes_{j=1}^{d}f_{j,k,\mathscr{A}}.$

Note that $f_{j,k,\mathscr{A}}$ is mean zero for $j\in\mathscr{A}.$ Using
representation (3.3) instead of (3.1) for $f$, we may assume without loss of
generality that for every $k$, $1\leq k\leq m$ there exists $j$, $1\leq j\leq
d$ such that $\int_{\mathbb{R}}f_{jk}(s)=0.$ This completes the proof of the
lemma in the special case when $f$ is given by (3.1) and satisfies (3.2).

To prove the general case, fix $f\in L_{1}(\mathbb{R}^{d})$ with
$\int_{\mathbb{R}^{d}}f(s)ds=0,$ and select a sequence
$\\{f_{m}\\}_{m=1}^{\infty}$ of mean zero sums of elementary tensors such that
$f_{m}\to f$ in $L_{1}(\mathbb{R}^{d})$ as $m\to\infty$. Since
$\|G_{l}^{\otimes d}\|_{1}=1,\ l\geq 1$ it follows from Young inequality that

$\|f\ast G_{l}^{\otimes d}\|_{1}\leq\|(f-f_{m})\ast G_{l}^{\otimes
d}\|_{1}+\|f_{m}\ast G_{l}^{\otimes d}\|_{1}\leq\|f-f_{m}\|_{1}+\|f_{m}\ast
G_{l}^{\otimes d}\|_{1}.$

Therefore,

$\limsup_{l\to\infty}\|f\ast G_{l}^{\otimes d}\|_{1}\leq\|f-f_{m}\|_{1}.$

Passing $m\to\infty,$ we conclude the proof. ∎

In what follows,

(3.4) $e_{k}(t):=e^{i\langle k,t\rangle},\quad k,t\in\mathbb{R}^{d}$

and $\mathcal{F}$ stands for the Fourier transform.

######  Lemma 3.3.

If $g\in L_{\infty}(\mathbb{R}^{d})$ is such that $\mathcal{F}(g)\in
L_{1}(\mathbb{R}^{d}),$ then for every $k\in\mathbb{R}^{d}$ we have

$(g(\nabla))(G_{l}^{\otimes d}e_{k})-g(k)G_{l}^{\otimes d}e_{k}\to 0$

in $L_{1}(\mathbb{R}^{d})$ as $l\to\infty.$

###### Proof.

Fix $k\in\mathbb{R}^{d}.$ Set $h_{1}(t):=g(k)e^{-|t-k|^{2}}$ and
$h_{0}(t):=g(t)-h_{1}(t)$, $t\in\mathbb{R}^{d}.$ Observe that, for every
$t\in\mathbb{R}^{d},$ we have

$\mathcal{F}(G_{l}^{\otimes d})(t)=\pi^{-d/2}e^{-l^{2}|t|^{2}}.$

Since $h_{1}(\nabla)$ on the Fourier side is a multiplier on $h_{1},$ it
follows that, for every $t\in\mathbb{R}^{d},$

$\mathcal{F}(G_{l}^{\otimes
d}e_{k})(t)=\pi^{-d/2}e^{-l^{2}|t-k|^{2}},\quad(\mathcal{F}((h_{1}(\nabla))(G_{l}^{\otimes
d}e_{k})))(t)=g(k)\pi^{-d/2}e^{-(l^{2}+1)|t-k|^{2}}.$

Applying the inverse Fourier transform to the second equality, we arrive at

$(h_{1}(\nabla))(G_{l}^{\otimes d}e_{k})=g(k)G_{(l^{2}+1)^{1/2}}^{\otimes
d}e_{k}.$

A direct computation yields $G_{(l^{2}+1)^{1/2}}^{\otimes d}-G_{l}^{\otimes
d}\to 0$ in $L_{1}(\mathbb{R}^{d})$ as $l\to\infty.$ We conclude that

$(h_{1}(\nabla))(G_{l}^{\otimes d}e_{k})-g(k)G_{l}^{\otimes d}e_{k}\to 0$

in $L_{1}(\mathbb{R}^{d})$ as $l\to\infty.$ It, therefore, suffices to show
that

$(h_{0}(\nabla))(G_{l}^{\otimes d}e_{k})\to 0$

in $L_{1}(\mathbb{R}^{d})$ as $l\to\infty.$ Define the function $f\in
L_{1}(\mathbb{R}^{d})$ by setting $f(t)=e^{i\langle
k,t\rangle}(\mathcal{F}h_{0})(t),$ $t\in\mathbb{R}^{d}.$ We rewrite the latter
equation as $f\ast G_{l}^{\otimes d}\to 0$ as $l\to\infty.$ Note that

$\int_{\mathbb{R}^{d}}f(s)ds=\int_{\mathbb{R}^{d}}e^{i\langle
k,s\rangle}(\mathcal{F}h_{0})(s)ds=h_{0}(k)=0.$

The assertion follows now from the Lemma 3.2. ∎

##  4\. Proof of Theorem 1.2 in the special case

For $s>0,$ the dilation operator $\sigma_{s}$ acts on the space of Lebesgue
measurable functions on $\mathbb{R},$ by the formula
$(\sigma_{s}x)(t)=x(t/s).$

######  Lemma 4.1.

Let $x,y$ be measurable and $\theta$ be integrable functions on $\mathbb{R}.$
Let $z(t):=t^{-1},$ $t>0,$ $z(t)=0,$ $t<0,$ and let $u>0$. For Lebesgue
measurable functions $x\otimes y$ and $\theta\otimes z$ on $\mathbb{R}^{2},$
we have

$\mu(\sigma_{u}(x)\otimes y)=\sigma_{u}\mu(x\otimes
y),\quad\mu(t,\theta\otimes z)=\|\theta\|_{1}t^{-1},\quad t>0.$

###### Proof.

Denoting Lebesgue measure on $\mathbb{R}^{2}$ by $m$, we have for every $t>0$

$\displaystyle m(\\{\sigma_{u}(x)\otimes y>t\\})$
$\displaystyle=m(\\{(s_{1},s_{2}):\ x(\frac{s_{1}}{u})y(s_{2})>t\\})$
$\displaystyle=um(\\{(s_{1},s_{2}):\ x(s_{1})y(s_{2})>t\\})$
$\displaystyle=um(\\{x\otimes y>t\\}).$

This proves the first assertion.

Firstly, we prove the second assertion for simple function $x\in
L_{1}(\mathbb{R}).$ If $x=\sum_{k}a_{k}\chi_{B_{k}}$ with $B_{k}$ being
pairwise disjoint sets, then222The notation $\bigoplus_{k}x_{k}$ stands for
disjoint sum of the functions $x_{k},$ that is $\sum_{k}z_{k},$ where
functions $z_{k}$ have pairwise disjoint support and $\mu(z_{k})=\mu(x_{k}).$
We refer the reader to the Definition 2.4.3 in [18] and subsequent comments.

$\mu(x\otimes z)=\mu(\bigoplus_{k}(a_{k}\chi_{B_{k}}\otimes
z))=\mu(\bigoplus_{k}\mu((a_{k}\chi_{B_{k}}\otimes z)).$

If $B$ is a set of finite measure, then there exists a measure preserving
bijection from $B$ to $(0,m(B))$ (see [13]). Therefore, we have

$\mu(\chi_{B}\otimes z)=\mu(\chi_{(0,m(B))}\otimes z)=m(B)z.$

Thus,

$\mu(x\otimes z)=\mu(\bigoplus_{k}|a_{k}|m(B_{k})z)=(\sum_{k}a_{k}m(B_{k}))z.$

The second assertion follows now by approximation. ∎

######  Lemma 4.2.

For every $X\in L_{1,\infty}(\mathcal{M})$ and every $l>0$, we have

$e^{-d}\pi^{-\frac{d}{2}}\|X\|_{1,\infty}\leq\|X\otimes G_{l}^{\otimes
d}\|_{1,\infty}\leq\|X\|_{1,\infty}.$

###### Proof.

For every operator $A\in S(\mathcal{M},\tau)$ and for every function $g\in
L_{\infty}(0,\infty),$ we have333Without loss of generality, $\mathcal{M}$ is
atomless. Suppose first that $x\in\mathcal{M}$ is $\tau-$compact. By Theorem
2.3.11 in [18], there exists a trace preserving $*-$isomorphism
$i:L_{\infty}(0,\infty)\to\mathcal{M}_{1}$ such that $i_{1}(\mu(A))=|A|.$
Consider trace preserving isomorphism $i\otimes 1:L_{\infty}(0,\infty)\otimes
L_{\infty}(0,\infty)\to\mathcal{M}\otimes L_{\infty}(0,\infty).$ We have
$i(\mu(A)\otimes g)=|A|\otimes g.$ Since every trace preserving
$*-$isomorphism preserves singular value function, the claim follows for
$\tau-$compact operators. The general case follows by approximation.

(4.1) $\mu(A\otimes g)=\mu(|A|\otimes g)=\mu(\mu(A)\otimes
g)=\mu(\mu(A)\otimes\mu(g)).$

Let $z$ be as in Lemma 4.1. It follows from the definition
$\|\cdot\|_{1,\infty}$ that $\mu(X)\leq\|X\|_{1,\infty}z$ and, therefore,

$\mu(X\otimes G_{l}^{\otimes d})\stackrel{{\scriptstyle\eqref{mu
tensor}}}{{=}}\mu(\mu(X)\otimes G_{l}^{\otimes
d})\leq\|X\|_{1,\infty}\mu(z\otimes G_{l}^{\otimes d})\stackrel{{\scriptstyle
L.\ref{tensor mu}}}{{=}}\|X\|_{1,\infty}\mu(z).$

This proves the right hand side inequality.

On the other hand, $\mu(G_{l})=l^{-1}\sigma_{l}\mu(G).$ By Lemma 4.1, we have
$\mu(G_{l}^{\otimes d})=l^{-d}\sigma_{l^{d}}\mu(G_{1}^{\otimes d}).$ Thus,

$\mu(X\otimes G_{l}^{\otimes d})\stackrel{{\scriptstyle\eqref{mu
tensor}}}{{=}}\mu(X\otimes l^{-d}\sigma_{l^{d}}\mu(G_{1}^{\otimes
d}))\stackrel{{\scriptstyle L.\ref{tensor
mu}}}{{=}}l^{-d}\sigma_{l^{d}}\mu(X\otimes G_{1}^{\otimes d}).$

Therefore, we have

$\|X\otimes G_{l}^{\otimes
d}\|_{1,\infty}=\sup_{t>0}\frac{t}{l^{d}}\mu(\frac{t}{l^{d}},X\otimes
G_{1}^{\otimes d})=\sup_{s>0}s\mu(s,X\otimes G_{1}^{\otimes d})=\|X\otimes
G_{1}^{\otimes d}\|_{1,\infty}.$

Clearly, $\mu(G_{1})\geq\frac{1}{e\sqrt{\pi}}\chi_{(0,1)}.$ It follows that

$\|X\otimes G_{1}^{\otimes d}\|_{1,\infty}\stackrel{{\scriptstyle\eqref{tensor
mu}}}{{=}}\|X\otimes\mu(G_{1})^{\otimes
d}\|_{1,\infty}\geq\|X\otimes(\frac{1}{e\sqrt{\pi}}\chi_{(0,1)})^{\otimes
d}\|_{1,\infty}=e^{-d}\pi^{-\frac{d}{2}}\|X\|_{1,\infty}.$

This proves the left hand side inequality. ∎

The following lemma is ideologically similar to Theorem II.4.3 in [28].

######  Lemma 4.3.

If $g$ is a smooth homogeneous function on $\mathbb{R}^{2}\backslash\\{0\\},$
then $\mathcal{F}g$ satisfies (possibly, after some $\delta$ distribution is
subtracted) the conditions (2.2).

###### Proof.

By assumption, $g$ is a smooth function on the circle $\\{|z|=1\\}.$ Thus,

$g(e^{i\theta})=\sum_{k\in\mathbb{Z}}\alpha_{k}e^{ik\theta},$

where Fourier coefficients decrease faster than every power. Therefore,

$g=\sum_{k\in\mathbb{Z}}\alpha_{k}g_{k},\quad
g_{k}(z)=\frac{z^{k}}{|z|^{k}},\quad 0\neq z\in\mathbb{C}.$

For every $k\neq 0,$ we have444This can be checked e.g. by substituting
$m=\Omega=g_{k}$ into the formula (26) in Theorem II.4.3 in [28].

$(\mathcal{F}g_{k})(z)=\frac{|k|}{2\pi
i^{k}}\cdot\frac{g_{k}(z)}{|z|^{2}},\quad 0\neq z\in\mathbb{C}.$

Hence,

$(\mathcal{F}g)(z)=\alpha_{0}\delta+\frac{1}{|z|^{2}}h(e^{i{\rm Arg}(z)}),$

where the smooth function $h$ on the circle is defined by the formula

$h(e^{i\theta})=\sum_{0\neq k\in\mathbb{Z}}\frac{|k|}{2\pi
i^{k}}\alpha_{k}e^{ik\theta}.$

So, $(\mathcal{F}g-\alpha_{0}\delta)(z)=O(|z|^{-2}).$ Furthermore, have

$\nabla(\frac{h(e^{i{\rm Arg(z)}})}{|z|^{2}})=h(e^{i{\rm
Arg(z)}})\cdot\nabla(\frac{1}{|z|^{2}})+\frac{1}{|z|^{2}}\cdot\frac{dh(e^{i\theta})}{d\theta}|_{\theta={\rm
Arg}(z)}\cdot\nabla({\rm Arg}(z))=O(\frac{1}{|z|^{3}}).$

This completes the verification that $\mathcal{F}g-\alpha_{0}\delta$ satisfies
condition (2.2). ∎

######  Theorem 4.4.

For every $A=A^{*}\in\mathcal{M}$ with ${\rm spec}(A)\subset\mathbb{Z}$ and
for every Lipschitz function $f,$ we have

$\|T_{f^{[1]}}^{A,A}(V)\|_{1,\infty}\leq
c_{abs}\|f^{\prime}\|_{\infty}\|V\|_{1},\quad V\in L_{1}(\mathcal{M}).$

###### Proof.

Fix a smooth homogeneous function $g$ on $\mathbb{R}^{2}$ such that
$g(e^{i\theta})=\tan(\theta)$ for $\theta\in(-\frac{\pi}{4},\frac{\pi}{4})$
and for $\theta\in(\frac{3\pi}{4},\frac{5\pi}{4}).$ Without loss of
generality, $g$ is mean zero on the circle $\\{|z|=1\\}.$ By Lemma 4.3,
$\mathcal{F}g$ satisfies the conditions (2.2). The operator $g(\nabla)\in
B(L_{2}(\mathbb{R}^{2}))$ since $g$ is bounded. Recall that
$(g(\nabla))(x)=(\mathcal{F}g)\ast x.$ By Theorem 2.1, we have

$1\otimes g(\nabla):L_{1}(\mathcal{M}\otimes L_{\infty}(\mathbb{R}^{2}))\to
L_{1,\infty}(\mathcal{M}\otimes L_{\infty}(\mathbb{R}^{2})).$

Consider Schwartz functions555Let $\psi$ be a Schwartz function on
$\mathbb{R}$ which is $1$ on $(-1,1)$ and which is supported on $(-2,2).$ Set
$\psi_{m}=\sigma_{m}\psi\cdot(1-\sigma_{\frac{1}{m}}\psi).$ It follows that
$\mathcal{F}(\psi_{m})=\mathcal{F}(\sigma_{m}\psi)\ast\mathcal{F}(1-\sigma_{\frac{1}{m}}\psi)=m\sigma_{\frac{1}{m}}(\mathcal{F}(\psi))-m\sigma_{\frac{1}{m}}(\mathcal{F}(\psi))\ast\frac{1}{m}\sigma_{m}(\mathcal{F}(\psi)).$
Applying Young’s inequality, we conclude that
$\|\mathcal{F}(\psi_{m})\|_{1}\leq\|m\sigma_{\frac{1}{m}}(\mathcal{F}(\psi))\|_{1}+\|m\sigma_{\frac{1}{m}}(\mathcal{F}(\psi))\|_{1}\|\frac{1}{m}\sigma_{m}(\mathcal{F}(\psi))\|_{1}=\|\mathcal{F}(\psi)\|_{1}+\|\mathcal{F}(\psi)\|_{1}^{2}.$
Consider the functions $\phi_{m}=\psi_{3m}^{\otimes 2}.$ By Fubini Theorem,
$\sup_{m\geq 1}\|\mathcal{F}(\phi_{m})\|_{1}<\infty.$ Clearly, $\psi_{m}=1$ on
the set $[-m,m]\backslash[-\frac{2}{m},\frac{2}{m}].$ Thus, $\phi_{m}(t)=1$ if
$t\in 3mK$ and $3mt\notin 2K,$ where $K=[-1,1]\times[-1,1].$ Thus,
$\phi_{m}(t)=1$ whenever $|t|\in(\frac{1}{m},m).$ $\phi_{m}$ on
$\mathbb{R}^{2}$ which vanish near $0,$ such that $\phi_{m}(t)=1$ for
$|t|\in(\frac{1}{m},m)$ and such that $\|\mathcal{F}\phi_{m}\|_{1}\leq
c_{abs}$ for all $m\geq 1.$ It follows that

$\|1\otimes(g\phi_{m})(\nabla)\|_{L_{1}\to L_{1,\infty}}\leq\|1\otimes
g(\nabla)\|_{L_{1}\to L_{1,\infty}}\|1\otimes\phi_{m}(\nabla)\|_{L_{1}\to
L_{1}}\leq$ $\leq\|1\otimes g(\nabla)\|_{L_{1}\to
L_{1,\infty}}\|\mathcal{F}\phi_{m}\|_{1}\leq c_{abs}\|1\otimes
g(\nabla)\|_{L_{1}\to L_{1,\infty}}=c_{abs},\quad m\geq 1.$

The last equality holds because $g$ is fixed.

By assumption, $A=\sum_{j\in\mathbb{Z}}jp_{j},$ where
$\\{p_{j}\\}_{j\in\mathbb{Z}}$ are pairwise orthogonal projections such that
$\sum_{j\in\mathbb{Z}}p_{j}=1.$ Since $A$ is bounded, it follows that
$p_{j}=0$ for all but finitely many $j\in\mathbb{Z}.$ Hence, sums are, in
fact, finite. Consider a unitary operator

$u=\sum_{j\in\mathbb{Z}}p_{j}\otimes e_{(j,f(j))},$

where $e_{(j,f(j))}$ is given in (3.4). Without loss of generality,
$\|f^{\prime}\|_{\infty}\leq 1.$ For every $m\geq\|A\|_{\infty},$ we have
$|i-j|,|f(i)-f(j)|\leq 2m$ for every $i,j\in{\rm spec}(A).$ Hence,

$(g\phi_{m})(i-j,f(i)-f(j))=g(i-j,f(i)-f(j))=\frac{f(i)-f(j)}{i-j},\quad
i,j\in{\rm spec}(A),\quad i\neq j.$

It follows from the preceding paragraph and from the equality
$\|G_{l}^{\otimes 2}\|_{1}=1$ that

(4.2) $\|(1\otimes(g\phi_{m})(\nabla))(u(V\otimes G_{l}^{\otimes
2})u^{*})\|_{1,\infty}\leq c_{abs}\|u(V\otimes G_{l}^{\otimes
2})u^{*}\|_{1}=c_{abs}\|V\|_{1}.$

It is clear that

$(1\otimes(g\phi_{m})(\nabla))(u(V\otimes G_{l}^{\otimes
2})u^{*})=\sum_{i,j}p_{i}Vp_{j}\otimes(g\phi_{m}(\nabla))(G_{l}^{\otimes
2}e_{(i-j,f(i)-f(j))}).$

Since there are only finitely many summands, it follows from Lemma 3.3 (as
applied to the Schwartz function $g\phi_{m}$) that

$(1\otimes(g\phi_{m})(\nabla))(u(V\otimes G_{l}^{\otimes 2})u^{*})-\sum_{i\neq
j}p_{i}Vp_{j}\otimes\frac{f(i)-f(j)}{i-j}G_{l}^{\otimes
2}e_{(i-j,f(i)-f(j))}\to 0$

in $L_{1}(\mathcal{M}\otimes L_{\infty}(\mathbb{R}^{2}))$ as $l\to\infty.$ It
is immediate that

$\sum_{i\neq j}p_{i}Vp_{j}\otimes\frac{f(i)-f(j)}{i-j}G_{l}^{\otimes
2}e_{(i-j,f(i)-f(j))}=$ $=\Big{(}\sum_{k\in\mathbb{Z}}p_{k}\otimes
e_{(k,f(k))}\Big{)}\cdot\Big{(}\sum_{i\neq
j}p_{i}Vp_{j}\otimes\frac{f(i)-f(j)}{i-j}G_{l}^{\otimes
2}\Big{)}\cdot\Big{(}\sum_{l\in\mathbb{Z}}p_{l}\otimes e_{(-l,-f(l))}\Big{)}=$
$\stackrel{{\scriptstyle\eqref{doi
special}}}{{=}}u(T_{f^{[1]}}^{A,A}(V)\otimes G_{l}^{\otimes 2})u^{*}.$

Therefore,

(4.3) $(1\otimes(g\phi_{m})(\nabla))(u(V\otimes G_{l}^{\otimes
2})u^{*})-u(T_{f^{[1]}}^{A,A}(V)\otimes G_{l}^{\otimes 2})u^{*}\to 0$

in $L_{1}(\mathcal{M}\otimes L_{\infty}(\mathbb{R}^{2}))$ (and, hence, in
$L_{1,\infty}(\mathcal{M}\otimes L_{\infty}(\mathbb{R}^{2}))$) as
$l\to\infty.$

Combining (4.2) and (4.3), we arrive at

$\limsup_{l\to\infty}\|u(T_{f^{[1]}}^{A,A}(V)\otimes G_{l}^{\otimes
2})u^{*}\|_{1,\infty}\leq c_{abs}\|V\|_{1}.$

Since $u$ is unitary, it follows that

$\limsup_{l\to\infty}\|T_{f^{[1]}}^{A,A}(V)\otimes G_{l}^{\otimes
2}\|_{1,\infty}\leq c_{abs}\|V\|_{1}.$

The assertion follows now from Lemma 4.2. ∎

##  5\. Proof of the main results

In this section we collect the results announced in the abstract and its
corollaries. Throughout this section fix a semi-finite von Neumann algebra
$\mathcal{M}$ with normal, semi-finite, faithful trace $\tau.$

######  Lemma 5.1.

Let $A=A^{*}\in\mathcal{M}.$ If $\\{\xi_{n}\\}_{n\geq 0}$ is a uniformly
bounded sequence of Borel functions on $\mathbb{R}^{2}$ such that
$\xi_{n}\to\xi$ everywhere, then

(5.1) $T_{\xi_{n}}^{A,A}(V)\to T_{\xi}^{A,A}(V),\quad V\in L_{2}(\mathcal{M})$

in $L_{2}(\mathcal{M})$ as $n\to\infty.$

###### Proof.

Let $\nu$ be a projection valued measure on $\mathbb{R}^{2}$ considered in
Subsection 2.4 (see (2.4)). Let $\gamma:\mathbb{R}\to\mathbb{R}^{2}$ be a
Borel measurable bijection. Clearly, $\nu\circ\gamma$ is a projection valued
measure on $\mathbb{R}.$ Hence, there exists a self-adjoint operator $B$
acting on the Hilbert space $L_{2}(\mathcal{M})$ such that
$E_{B}=\nu\circ\gamma.$

Set $\eta_{n}=\xi_{n}\circ\gamma$ and $\eta=\xi\circ\gamma.$ We have
$\eta_{n}\to\eta$ everywhere on $\mathbb{R}.$ Thus,

$T_{\xi_{n}}^{A,A}=\int_{\mathbb{R}^{2}}\xi_{n}d\nu=\int_{\mathbb{R}}\eta_{n}(\lambda)dE_{B}(\lambda)=\eta_{n}(B)\to\eta(B)=$
$=\int_{\mathbb{R}}\eta(\lambda)dE_{B}(\lambda)=\int_{\mathbb{R}^{2}}\xi
d\nu=T_{\xi}^{A,A}.$

Here, the convergence is understood with respect to the strong operator
topology on the space $B(L_{2}(\mathcal{M})).$ In particular, (5.1) follows. ∎

###### Proof of Theorem 1.2.

Step 1. Let $A$ is bounded. For every $n\geq 1,$ set

$A_{n}\stackrel{{\scriptstyle
def}}{{=}}\sum_{k\in\mathbb{Z}}\frac{k}{n}E_{A}([\frac{k}{n},\frac{k+1}{n})),$
$\xi_{n}(t,s)=f^{[1]}(\frac{k}{n},\frac{l}{n}),\quad
t\in[\frac{k}{n},\frac{k+1}{n}),s\in[\frac{l}{n},\frac{l+1}{n}).$

It is immediate that (see e.g. Lemma 8 in [25] for much stronger assertion)

$T_{\xi_{n}}^{A,A}(V)=T_{f^{[1]}}^{A_{n},A_{n}}(V)=T_{(n\sigma_{n}f)^{[1]}}^{nA_{n},nA_{n}}(V).$

It follows from Theorem 4.4 that

$\|T_{\xi_{n}}^{A,A}(V)\|_{1,\infty}\leq
c_{abs}\|(n\sigma_{n}f)^{\prime}\|_{\infty}\|V\|_{1}=c_{abs}\|f^{\prime}\|_{\infty}\|V\|_{1}.$

Note that $\xi_{n}\to f^{[1]}$ everywhere. It follows from Lemma 5.1 that

$T_{\xi_{n}}^{A,A}(V)\to T_{f^{[1]}}(V),\quad V\in L_{2}(\mathcal{M})$

in $L_{2}(\mathcal{M})$ (and, hence, in measure — see e.g [20]) as
$n\to\infty.$ Since the quasi-norm in $L_{1,\infty}(\mathcal{M})$ is a Fatou
quasi-norm, it follows that

$\|T_{f^{[1]}}^{A,A}(V)\|_{1,\infty}\leq
c_{abs}\|f^{\prime}\|_{\infty}\|V\|_{1},\quad V\in(L_{1}\cap
L_{2})(\mathcal{M}).$

Step 2. Let now $A$ be an arbitrary operator affiliated with $\mathcal{M}.$
Set $A_{n}=AE_{A}([-n,n]).$ By Step 1, we have

$\|T_{f^{[1]}}^{A_{n},A_{n}}(V)\|_{1,\infty}\leq
c_{abs}\|f^{\prime}\|_{\infty}\|V\|_{1}.$

It follows immediately from the definition of the double operator integral
that

$T_{f^{[1]}}^{A_{n},A_{n}}(V)=E_{A}([-n,n])\cdot T_{f^{[1]}}^{A,A}(V)\cdot
E_{A}([-n,n])\to T_{f^{[1]}}^{A,A}(V)$

in $L_{2}(\mathcal{M})$ (and, hence, in measure) as $n\to\infty.$ Since the
quasi-norm in $L_{1,\infty}(\mathcal{M})$ is a Fatou quasi-norm, the assertion
follows. ∎

The following lemma is ideologically similar to Theorem 7.4 in [20].

######  Lemma 5.2.

If $A,B\in\mathcal{M}$ are such that $[A,B]\in L_{2}(\mathcal{M}),$ then, for
every Lipschitz function $f,$ we have

$T_{f^{[1]}}^{A,A}([A,B])=[f(A),B].$

###### Proof.

By definition of double operator integral given in Subsection 2.4, we have

(5.2) $T_{\xi_{1}}^{A,A}T_{\xi_{2}}^{A,A}=T_{\xi_{1}\xi_{2}}^{A,A}.$

Let $\xi_{1}=f^{[1]}$ and let $\xi_{2}(\lambda,\mu)=\lambda-\mu$ when
$|\lambda|,|\mu|\leq\|A\|_{\infty}.$ $\xi_{2}(\lambda,\mu)=0$ when
$|\lambda|>\|A\|_{\infty}$ or $|\mu|\leq\|A\|_{\infty}.$

If $p$ is a $\tau-$finite projection, then $pB\in L_{2}(\mathcal{M})$ and

$T_{\xi_{1}\xi_{2}}^{A,A}(pB)=f(A)pB-pBf(A),\quad T_{\xi_{2}}^{A,A}(pB)=ApB-
pBA,$

Applying (5.2) to the operator $pB\in L_{2}(\mathcal{M}),$ we obtain

(5.3) $T_{f^{[1]}}^{A,A}(ApB-pBA)=f(A)pB-pBf(A).$

Applying Proposition 6.6 in [20] to the operator $nA,$ we construct a sequence
$\\{p_{n,k}\\}_{k\geq 0}$ of $\tau-$finite projections such that
$p_{n,k}\uparrow 1$ as $k\to\infty$ and such that $\|[nA,p_{n,k}]\|_{2}\leq
1.$ Let $\\{\eta_{m}\\}_{m\geq 0}$ be an orthonormal basis in
$L_{2}(\mathcal{M}).$ Fix $k_{n}$ so large that

(5.4) $\|(1-p_{n,k_{n}})\eta_{m}\|_{2}\leq\frac{1}{n},\quad 0\leq m<n.$

Set $q_{n}=p_{n,k_{n}}.$ It follows from (5.4) that $q_{n}\to 1$ in the strong
operator topology (in the left regular representation of $\mathcal{M}$).
Clearly, $[A,q_{n}]\to 0$ in $L_{2}(\mathcal{M}).$

By construction,

$Aq_{n}B-q_{n}BA=[A,q_{n}]B+q_{n}[A,B]\to[A,B],\quad n\to\infty,$

in $L_{2}(\mathcal{M}).$ Since $T_{f^{[1]}}^{A,A}$ is bounded, it follows that

$f(A)q_{n}B-q_{n}Bf(A)\stackrel{{\scriptstyle\eqref{finite
p}}}{{=}}T_{f^{[1]}}^{A,A}(Aq_{n}B-q_{n}BA)\to T_{f^{[1]}}^{A,A}(AB-BA),\quad
n\to\infty,$

in $L_{2}(\mathcal{M}).$ On the other hand,

$f(A)q_{n}B-q_{n}Bf(A)\to[f(A),B],\quad n\to\infty,$

in the strong operator topology (in the left regular representation of
$\mathcal{M}$). This concludes the proof. ∎

######  Theorem 5.3.

For all self-adjoint operators $A,B\in\mathcal{M}$ such that $[A,B]\in
L_{1}(\mathcal{M})$ and for every Lipschitz function $f,$ we have

$\|[f(A),B]\|_{1,\infty}\leq
c_{abs}\|f^{\prime}\|_{\infty}\cdot\|[A,B]\|_{1}.$

For all self-adjoint operators $X,Y\in\mathcal{M}$ such that $X-Y\in
L_{1}(\mathcal{M})$ and for every Lipschitz function $f,$ we have

$\|f(X)-f(Y)\|_{1,\infty}\leq c_{abs}\|f^{\prime}\|_{\infty}\|X-Y\|_{1}.$

###### Proof.

By assumption, $[A,B]\in(L_{1}\cap L_{2})(\mathcal{M}).$ The first assertion
follows by combining Lemma 5.2 and Theorem 1.2. Applying the first assertion
to the operators

$A=\begin{pmatrix}X&0\\\ 0&Y\end{pmatrix},\quad B=\begin{pmatrix}0&1\\\
1&0\end{pmatrix},$

we obtain the second assertion. ∎

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