# Compactness criteria via Laguerre and Hankel transformations

Á. P. Horváth

###### Abstract.

The aim of this paper is to prove Kolmogorov-Riesz type theorems via Bessel
and Laguerre translations, and Pego-type theorems by the corresponding
transformations.

###### Key words and phrases:

Kolmogorov-Riesz theorem, Pego’s theorem, Bessel translation, Laguerre
translation

###### 2020 Mathematics Subject Classification:

46B50, 46E30, 42C20

Supported by the NKFIH-OTKA Grants K128922 and K132097.

##  1\. Introduction

The populous family of compactness theorems is established by Arzelá and
Ascoli. Our starting point, the Kolmogorov-Riesz theorem characterizes
precompact (totally bounded) sets of $L^{p}(\mathbb{R}^{n})$, see e.g. [1].
Besides, such a characterization is interesting in itself, it has several
applications to differential and integral equations. Compactness criteria were
studied in particular non-standard function spaces, e.g. in Sobolev spaces
[17], variable Lebesgue spaces [13], or weighted variable exponent amalgam and
Sobolev spaces [5], and also in more general circumstances, see e.g. [10],
[25] and [14]. Sudakov-type improvements of the classical theorem are derived
in [18] (see [26] as well).
The useful criterion of compactness given by Pego via Fourier transformation
(see [23]) has also great influence, see e.g. [9], [11], [12]. In the
inspirating paper [19], compactness criterion is given by Laplace
transformation.

Below we offer a new aspect of characterization of precompact sets in certain
Banach spaces. Instead of deriving similar theorems by related, for instance
Mellin or cosine transformations, we investigate and extend the notion of
equicontinuity. Motivated by the effect of translation, $\tau_{y}f(x)=f(x-y)$,
on exponential functions, different translation operators were introduced by
orthogonal systems $\\{\varphi_{n}\\}_{n}$ as
$T_{y}\varphi_{n}(x)=\varphi_{n}(x)\varphi_{n}(y)$. Hereinafter we deal with
Laguerre and Bessel translations.

In the next section a Riesz-Kolmogorov type theorem is derived by Laguerre
translation to weighted $L^{p}$ spaces on the half-line. The corresponding
transformation, the discrete Laguerre-Fourier transformation implies a
remarkable simplification, since the structure of the corresponding
$l^{p^{\prime}}$ spaces are simpler than the original $L^{p}$ ones. In this
section we also introduce and study Laguerre translation on sequences.
In the third section the method is presented by Bessel translation. The
corresponding transformation is the Hankel (Bessel-Fourier) transformation
which establishes a Pego-type theorem.

##  2\. Laguerre translation method

Laguerre translation is developed by product formulae, see e.g. [28] and by
investigation of the related Cauchy problem, see e.g. [7]. We mention that it
is a natural idea to handle Bessel and Laguerre translations in parallel,
since the translated functions in both cases are the solutions to very similar
Cauchy problems,
$u_{xx}-u_{tt}+\frac{2\alpha+1}{x}u_{x}-\frac{2\alpha+1}{t}u_{t}-ru=0$, where
$r(x,t)=0$ in Bessel case and $r(x,t)=x^{2}-t^{2}$ in Laguerre case. The
derived convolution structure is examined e.g. in [15]. The norm of the
translation operator ensures a maximum principle to the corresponding
hyperbolic problem, and it implies Nikol’skii-type norm-estimations on the
half-line, see [2]. Convolution is applicable to study of best approximation
in certain spaces, see [16]. In this section compactness is investigated by
Laguerre translation. The main result of the section is the characterization
of precompact sets in $L^{2}_{\alpha}$.

###  2.1. Laguerre translation and precompact sets in $L^{p}_{\alpha}$

We introduce weighted $L^{p}$-spaces on the half-line similar to the ones
given in the previous section.
Let $1\leq p\leq\infty$ and $\alpha>-1$. $f:\mathbb{R}_{+}\to\mathbb{R}\in
L^{p}_{\alpha}$ if

$\|f\|_{p,\alpha}:=\left(\int_{0}^{\infty}\left|f(x)e^{-\frac{x}{2}}\right|^{p}x^{\alpha}dx\right)^{\frac{1}{p}}=\left(\int_{0}^{\infty}|\tilde{f}|^{p}d\mu_{\alpha}\right)^{\frac{1}{p}}<\infty.$
$\|f\|_{\infty,\alpha}=\|\tilde{f}\|_{\infty}$

independently of $\alpha$.

The translation operator acting on the space given above can be defined as
follows, see e.g. [15] .
Let $\alpha>-\frac{1}{2}$

(1) $T_{t}^{\alpha}(f,x)$
$=c_{\alpha}\int_{0}^{\pi}f(x+t+2\sqrt{xt}\cos\vartheta)e^{-\sqrt{xt}\cos\vartheta}j_{\alpha-\frac{1}{2}}(\sqrt{xt}\sin\vartheta)\sin^{2\alpha}\vartheta
d\vartheta,$

where
$c_{\alpha}=\frac{\Gamma(\alpha+1)}{\Gamma\left(\alpha+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}$
and $j_{\alpha-\frac{1}{2}}$ is the entire Bessel function, cf. (30).

By symmetry of the definition in $t,x\geq 0$,

$T_{t}^{\alpha}(f,x)=T_{x}^{\alpha}(f,t),$

and again by definition

$T_{t}^{\alpha}(f,0)=f(t),\hskip 4.0pt\hskip 4.0pt\hskip
4.0ptT_{0}^{\alpha}(f,x)=f(x).$

Denoting by $L_{n}^{\alpha}(x)$ the $n^{th}$ Laguerre polynomial orthogonal on
$(0,\infty)$ with respect to the Laguerre weight, that is

$\int_{0}^{\infty}L_{n}^{\alpha}(x)L_{k}^{\alpha}(x)e^{-x}x^{\alpha}dx=\Gamma(\alpha+1)\binom{n+\alpha}{n}\delta_{n,k},$

and taking into consideration that

$L_{n}^{\alpha}(0)=\binom{n+\alpha}{n}=w(\alpha,n)=:w(n),$

we denote by

$R_{n}^{\alpha}(x)=\frac{L_{n}^{\alpha}(x)}{\binom{n+\alpha}{n}}.$

Thus $R_{n}^{\alpha}(0)=1$ and

(2)
$\left\|\sqrt{\frac{w(n)}{\Gamma(\alpha+1)}}R_{n}^{\alpha}\right\|_{2,\alpha}=1,\hskip
4.0pt\hskip
4.0pt\|\tilde{R}_{n}^{\alpha}\|_{\infty}=\tilde{R}_{n}^{\alpha}(0)=1.$

With this notation we have that

(3) $T_{t}^{\alpha}(R_{n}^{\alpha},x)=R_{n}^{\alpha}(x)R_{n}^{\alpha}(t),$

see e.g. [15].

According to [15, Theorem 1] for all $\alpha\geq 0$ and $1\leq p\leq\infty$,
considering the translated function as a function of the variable $x$,

(4) $\|e^{-\frac{t}{2}}T_{t}^{\alpha}(f,x)\|_{p,\alpha}\leq\|f\|_{p,\alpha}.$

The corresponding convolution is

(5) $(f*g)(t):=\int_{0}^{\infty}T_{t}^{\alpha}(f,x)g(x)e^{-x}x^{\alpha}dx.$

Again by [15, Theorem 1], for all $\alpha\geq 0$ and $1\leq p,q,r\leq\infty$
with $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}-1$; if $f\in L^{p}_{\alpha}$ and
$g\in L^{q}_{\alpha}$, then

(6) $\|f*g\|_{r,\alpha}\leq\|f\|_{p,\alpha}\|g\|_{q,\alpha}.$

The first theorem of this section is a Kolmogorov -Riesz type theorem, where
the standard equicontinuity property is replaced by the one based on Laguerre
translation.

######  Theorem 1.

Let $1\leq p<\infty$ and $\alpha\geq 0$. A bounded set $K\subset
L^{p}_{\alpha}$ is precompact if and only if the properties below are
fulfilled.
${\bf P_{a}}.$ For all $\varepsilon>0$ there is an $R>0$ such that for all
$f\in K$

(7)
$\left(\int_{R}^{\infty}|\tilde{f}|^{p}d\mu_{\alpha}\right)^{\frac{1}{p}}<\varepsilon.$

${\bf P_{b}.}$ For all $\varepsilon$ and $M_{0}$ positive numbers there is a
$\delta>0$ such that for all $0\leq t\leq M_{0}$, $0\leq h\leq\delta$ and
$f\in K$

(8)
$\left(\int_{0}^{\infty}\left|\left(T_{t+h}^{\alpha}f(x)-T_{t}^{\alpha}f(x)\right)e^{-\frac{x}{2}}\right|^{p}d\mu_{\alpha}(x)\right)^{\frac{1}{p}}<\varepsilon.$

###### Proof.

First we assume that $K$ is precompact. Then for an $\varepsilon>0$ there is a
finite $\frac{\varepsilon}{2}$-net,
$H_{\frac{\varepsilon}{2}}=\\{u_{1},\dots,u_{j}\\}$ in $K$
$(j=j(\varepsilon))$. Since $C_{0}$ \- the continuous functions with compact
support - are dense in $L_{p,\alpha}$, there is an $S_{\varepsilon}\subset
C_{0}$, such that $S_{\varepsilon}=\\{\Phi_{1},\dots,\Phi_{j}\\}$ and
$\|u_{l}-\Phi_{l}\|_{p,(\alpha)}<\frac{\varepsilon}{2}$ for all $u_{l}\in
H_{\frac{\varepsilon}{2}}$. That is there is an $R=R_{\varepsilon}>0$ such
that the closed ball $\overline{B(0,R)}$ contains the support of each $\Phi\in
S_{\varepsilon}$ wich ensures ${\bf P_{a}}$. To prove ${\bf P_{b}}$ take an
$f\in L^{p}_{\alpha}$, $t\in[0,M_{0}]$ with some $M_{0}$, and let $\Phi$ be
the element of $S_{\varepsilon}$ closest to $f$ with
$\mathrm{supp}\Phi\in(0,R)$. In view of (1)

$\left(\int_{0}^{\infty}\left|\left(T_{t+h}^{\alpha}\Phi(x)-T_{t}^{\alpha}\Phi(x)\right)e^{-\frac{x}{2}}\right|^{p}d\mu_{\alpha}(x)\right)^{\frac{1}{p}}$
$=\left(\int_{0}^{\infty}\left|c_{\alpha}\int_{0}^{\pi}\left(\Phi(x+t+h+2\sqrt{x(t+h)}\cos\vartheta)e^{-\sqrt{x(t+h)}\cos\vartheta}j_{\alpha-\frac{1}{2}}(\sqrt{x(t+h)}\sin\vartheta)\right.\right.\right.$
$\left.\left.\left.-\Phi(x+t+2\sqrt{xt}\cos\vartheta)e^{-\sqrt{xt}\cos\vartheta}j_{\alpha-\frac{1}{2}}(\sqrt{xt}\sin\vartheta)\right)\sin^{2\alpha}\vartheta
d\vartheta e^{-\frac{x}{2}}\right|^{p}x^{\alpha}dx\right)^{\frac{1}{p}}=(*).$

For sake of simplicity let us denote by

$\Psi(x,t,\vartheta):=\Phi(x+t+2\sqrt{xt}\cos\vartheta)e^{-\sqrt{xt}\cos\vartheta}j_{\alpha-\frac{1}{2}}(\sqrt{xt}\sin\vartheta).$

Recalling that $\mathrm{supp}\Phi\subset(0,R)$ and $t\in[0,M_{0}]$, if
$R<|\sqrt{x}-\sqrt{t}|$ then $\Psi(x,t,\vartheta)=0$, that is if
$\sqrt{x}>\sqrt{M_{0}}+R$ then $\Psi(x,t,\vartheta)=0$. Considering that
difference of the arguments of the three functions above are at most
$2\sqrt{xh}\leq c(R,M_{0})\sqrt{h}$ and $\Psi(x,t,\vartheta)$ is compactly
supported and continuous, there is a $\delta>0$ such that if $0\leq
h\leq\delta$

$\left|\Psi(x,t+h,\vartheta)-\Psi(x,t,\vartheta)\right|\leq\varepsilon$

and so

$(*)\leq\varepsilon\left(\int_{0}^{\infty}e^{-p\frac{x}{2}}x^{\alpha}dx\right)^{\frac{1}{p}}.$

Then, because the set of $\Phi$-s in question is finite we can choose $R$ and
$\delta$ uniformly. Since the norm of translation is bounded on $[0,M_{0}]$,
cf. (4) (and it is a linear operator), we can finish this part with triangle
inequality.

Vice versa assuming ${\bf P_{a}}$ and ${\bf P_{b}}$ let

$V_{a}f(x):=\frac{1}{A}\int_{0}^{a}T_{t}^{\alpha}f(x)e^{-\frac{t}{2}}t^{\alpha}dt$

if $0\leq x\leq R$ with some finite $R$, and we define $V_{a}f(x)=0$ if $x>R$.
where $A=\int_{0}^{a}e^{-\frac{t}{2}}t^{\alpha}dt$. Then, applying Hölder’s
inequality and the symmetry of translation

(9)
$|V_{a}f(x+u)-V_{a}f(x)|\leq\frac{1}{A}\int_{0}^{a}|T_{t}^{\alpha}f(x+u)-T_{t}^{\alpha}f(x)|e^{-\frac{t}{2}}t^{\alpha}dt$
$\leq\frac{1}{A^{\frac{1}{p}}}e^{\frac{a}{2p^{\prime}}}\left(\int_{0}^{\infty}|T_{x+u}^{\alpha}f(t)-T_{x}^{\alpha}f(t)|^{p}e^{-\frac{pt}{2}}t^{\alpha}dt\right)^{\frac{1}{p}},$

and similarly

(10)
$|V_{a}f(x)|\leq\frac{1}{A^{\frac{1}{p}}}e^{\frac{a}{2p^{\prime}}}\left(\int_{0}^{\infty}|T_{x}^{\alpha}f(t)|^{p}e^{-\frac{pt}{2}}t^{\alpha}dt\right)^{\frac{1}{p}}.$

Let

$F_{a,R}:=\left\\{V_{a}f(x):f\in K,\hskip 4.0ptx\leq R\right\\}.$

By (10) and (4)

$|e^{-\frac{x}{2}}V_{a}f(x)|\leq\frac{1}{A^{\frac{1}{p}}}e^{\frac{a}{2p^{\prime}}}\|f\|_{p,\alpha}.$

Thus with fixed $a$ and $R$, $F_{a,R}$ is bounded (by
$M\frac{1}{A^{\frac{1}{p}}}e^{\frac{a}{2p^{\prime}}+\frac{R}{2}}$ if $K$ is
bounded by $M$) and (choosing $M_{0}=R$) by (9) and assumption ${\bf P_{b}}$
it is equicontinuous. Thus for an arbitrary $\varepsilon>0$ there is an
$\varepsilon$-net, $V_{a}f_{1},\dots,V_{a}f_{n}$, in $F_{a,R}$ such that
$f_{i}\in K$, $i=1,\dots,n$.
Let $f\in K$ be arbitrary, and for an $\varepsilon>0$ choose $R$ according to
property ${\bf P_{a}}$. Then again by Hölder and Fubini theorems

$\|V_{a}f-f\|_{p,\alpha}$
$\leq\left(\int_{0}^{R}\left|e^{-\frac{x}{2}}\int_{0}^{a}\frac{1}{A}\left(T_{t}^{\alpha}f(x)-f(x)\right)e^{-\frac{t}{2}}t^{\alpha}dt\right|^{p}x^{\alpha}dx\right)^{\frac{1}{p}}+\left(\int_{R}^{\infty}\left|e^{-\frac{x}{2}}f(x)\right|^{p}x^{\alpha}dx\right)^{\frac{1}{p}}$
$\leq\varepsilon+e^{\frac{a}{2p^{\prime}}}\left(\frac{1}{A}\int_{0}^{\infty}e^{-\frac{px}{2}}\int_{0}^{a}\left|T_{t}^{\alpha}f(x)-f(x)\right|^{p}e^{-\frac{t}{2}}t^{\alpha}dtx^{\alpha}dx\right)^{\frac{1}{p}}$
$\leq\varepsilon+e^{\frac{a}{2p^{\prime}}}\left(\frac{1}{A}\int_{0}^{a}e^{-\frac{t}{2}}t^{\alpha}\int_{0}^{\infty}e^{-\frac{px}{2}}\left|T_{t}^{\alpha}f(x)-f(x)\right|^{p}x^{\alpha}dxdt\right)^{\frac{1}{p}}$
$\leq\varepsilon+e^{\frac{a}{2p^{\prime}}}\sup_{0\leq t\leq
a}\left(\int_{0}^{\infty}e^{-\frac{px}{2}}\left|T_{t}^{\alpha}f(x)-f(x)\right|^{p}x^{\alpha}dx\right)^{\frac{1}{p}}.$

Thus choosing $a$ according to property ${\bf P_{b}}$ small enough (assume
that $a<1$), $\|V_{a}f-f\|_{p,\alpha}<2\varepsilon$.
With these chosen $a$ and $R$ construct $F_{a,R}$ and for the previous
$\varepsilon$ and $f$, from the $\varepsilon$-net in question choose
$V_{a}f_{i}$ such that $|V_{a}f(x)-V_{a}f_{i}(x)|<\varepsilon$ on $[0,R]$.
Then we have

$\|V_{a}f-V_{a}f_{i}\|_{p,\alpha}=\left(\int_{0}^{R}\left|e^{-\frac{x}{2}}(V_{a}f(x)-V_{a}f_{i}(x))\right|^{p}x^{\alpha}dx\right)^{\frac{1}{p}}$
$\leq\varepsilon\left(\int_{0}^{\infty}e^{-\frac{px}{2}}x^{\alpha}dx\right)^{\frac{1}{p}}\leq\varepsilon
c(\alpha,p).$

That is by the triangle inequality $\\{f_{i}\\}_{i=1}^{n}$ is a
$(4+c(\alpha,p))\varepsilon$-net in $K$.

###  2.2. Laguerre translation on sequences

Laguerre translation on sequences, to the best of my knowledge, have not
appeared explicitly just as convolution of sequences, see e.g. [4]. The
corresponding algebras are investigated in [20]. Below we derive the
translation from convolution and investigate its properties.

Let $\alpha>-1$. Recalling the notation above we introduce the discrete
weights

$w(k)=w(\alpha,k)=\binom{k+\alpha}{k},$

and the space of real sequences, $a=\\{a(k)\\}_{k=0}^{\infty}$. For $1\leq
p<\infty$ $a\in l^{p}_{\alpha}$ if

$\|a\|_{p,\alpha}:=\left(\sum_{k=0}^{\infty}|a(k)|^{p}w(k)\right)^{\frac{1}{p}}<\infty.$
$\|a\|_{\infty,\alpha}=\|a\|_{\infty}$

independently of $\alpha$.

If $a\in l^{p}_{\alpha}$ and $b\in l^{p^{\prime}}_{\alpha}$ we denote by

$\langle a,b\rangle=\sum_{k=0}^{\infty}a(k)b(k)w(k).$

Remark.

Certainly, in discrete case the criterion of precompactness is simpler, cf.
[17, Theorem 4]. It is as follows.

For any $1\leq p<\infty$ and $\alpha\geq 0$ a set $K\subset l^{p}_{\alpha}$ is
precompact if and only if it is pointwise bounded (i.e. for all
$n\in\mathbb{N}$ there is an $M(n)$ such that for all $a\in K$ $a(n)\leq
M(n)$), and the next property fulfils.

${\bf P_{as}.}$ For all $\varepsilon>0$ there is an $N\in\mathbb{N}$ such that
for all $a\in K$

(11)
$\left(\sum_{k=N+1}^{\infty}|a(k)|^{p}w(k)\right)^{\frac{1}{p}}<\varepsilon.$

Indeed, suppose, that $K$ is precompact. Then it is obviously bounded in
$l^{p}_{\alpha}$. Since $w(k)\geq 1$, it is pointwise bounded as well. For
$\frac{\varepsilon}{2}$ it has a finite $\frac{\varepsilon}{2}$-net,
$b_{1},\dots,b_{n}$, say. Since the finite sequences are dense in
$l^{p}_{\alpha}$, there are $s_{i}$ finite sequences, such that
$\|p_{i}-s_{i}\|_{p,\alpha}<\frac{\varepsilon}{2}$, $i=1,\dots,n$. Thus the
maximal length of $s_{i}$-s is an appropriate choice of $N$.

Assume now, that $K$ is pointwise bounded and fulfils ${\bf P_{as}}$. Choose
an $N$ for an arbitrary $\varepsilon$ ensured by ${\bf P_{as}}$. Let
$S_{N}:=\\{a^{N}:=a(0),\dots,a(N):a\in K\\}$ be the set of the $N+1$-long
initial parts of the sequences in $K$. Then the distance of $K$ and $S_{N}$ is
at most $\varepsilon$. Because $S_{N}$ is finite dimensional it is bounded in
$l^{p}_{\alpha}$ ($S_{N}$ is bounded by $cM_{N}N^{\alpha+1}$, where
$M_{N}:=\max_{n=1}^{N}M(n)$) and (again by being finite dimensional) $S_{N}$
contains a finite $\varepsilon$-net, and the corresponding sequences form a
$2\varepsilon$-net in $K$.

To define translation on the spaces of sequences, our starting point is the
convolution defined in [4].
Let $\alpha>\alpha_{0}=\frac{-5+\sqrt{17}}{2}$. Then by [4, Theorem 1]

(12)
$\gamma(n,m,k):=\gamma(n,m,k,\alpha)=\int_{0}^{\infty}R_{n}^{\alpha}(x)R_{m}^{\alpha}(x)R_{k}^{\alpha}(x)e^{-2x}x^{\alpha}dx>0,$

for all $k,m,n\in\mathbb{N}$. In view of [4, (4.2)]

(13) $\sum_{k=0}^{\infty}\gamma(n,m,k)w(k)=1$

for all $m,n\in\mathbb{N}$.

Let $a\in l^{p}_{\alpha}$, $b\in l^{q}_{\alpha}$, where
$\frac{1}{p}+\frac{1}{q}\geq 1$. In [4, (4.5)] the next convolution is
defined.

(14)
$(a*b)(k):=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a(m)b(n)\gamma(n,m,k)w(n)w(m).$

Similarly to the $L^{p}$-cases if $1\leq p,q,r\leq\infty$ and
$\frac{1}{r}=\frac{1}{p}+\frac{1}{q}-1$,

(15) $\|a*b\|_{r,\alpha}\leq\|a\|_{p,\alpha}\|b\|_{q,\alpha},$

see [4, (4.6)].

Thus, fixing $\alpha>\alpha_{0}$, we can define the corresponding translation
as

(16) $T_{k}(a)(n):=\sum_{m=0}^{\infty}a(m)w(m)\gamma(n,m,k).$

By symmetry we immediately have that

(17) $T_{k}(a)(n)=T_{n}(a)(k).$

So (14) can be written as

(18) $(a*b)(k)=\sum_{n=0}^{\infty}b(n)T_{k}(a)(n)w(n)=\langle
b,T_{k}(a)\rangle=\langle a,T_{k}(b)\rangle.$

As on functions, Laguerre translation on sequences is also a bounded operator.

######  Proposition 1.

Let $\alpha>\alpha_{0}$ and $1\leq p\leq\infty$ Then for all $k$

$\|T_{k}(a)\|_{p,\alpha}\leq\|a\|_{p,\alpha}.$

###### Proof.

Let $a\in l^{\infty}$. In view of (16), (12) and (13)

$|T_{k}(a)(n)|=|\sum_{m=0}^{\infty}a(m)w(m)\gamma(n,m,k)|\leq\|a\|_{\infty}\sum_{m=0}^{\infty}\gamma(n,m,k)w(m)=\|a\|_{\infty}.$

If $a\in l^{1}_{\alpha}$

$\sum_{n=0}^{\infty}|T_{k}(a)(n)|w(m)=\sup_{b,\|b\|_{\infty}\leq
1}\sum_{n=0}^{\infty}b(n)T_{k}(a)(n)w(n)$ $=\sup_{b,\|b\|_{\infty}\leq
1}\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a(m)b(n)\gamma(n,m,k)w(n)w(m)=\sup_{b,\|b\|_{\infty}\leq
1}(a*b)(k).$

According to (15)

$\|a*b\|_{\infty}\leq\|a\|_{1,\alpha},$

which implies the statement in $l^{1}_{\alpha}$. Finally interpolation ensures
the result.

To prepare the next subsection, we mention the standard connection between the
corresponding spaces of functions and sequences.
For a function $f$ in some $L^{p}_{\alpha}$ let us denote the corresponding
sequence by $\hat{f}=a_{f}$, where

(19)
$\hat{f}(n):=\int_{0}^{\infty}\tilde{f}\tilde{R}_{n}^{\alpha}d\mu_{\alpha},$

for $n\in\mathbb{N}$.
Let $a$ be a sequence. The corresponding function can be defined as
$\check{a}=f_{a}(x)$, where

$\tilde{f}_{a}\sim\sum_{m=0}^{\infty}a(m)w(m)\tilde{R}_{m}^{\alpha}$

if the series is convergent in some sense.
Let $1\leq p\leq 2$. Then, as usual, if $f\in L^{p}_{\alpha}$ then $\hat{f}\in
l^{p^{\prime}}_{\alpha}$, and if $a\in l^{p}_{\alpha}$ then $\check{a}\in
L^{p^{\prime}}_{\alpha}$ and the operators map $L^{p}_{\alpha}$ to
$l^{p^{\prime}}_{\alpha}$ and $l^{p}_{\alpha}$ to $L^{p^{\prime}}_{\alpha}$
are bounded.
Indeed, by interpolation it is a consequence of Parseval’s formula and the
next inequalities:

$|\hat{f}(k)|=\left|\int_{0}^{\infty}\tilde{f}\tilde{R}^{\alpha}_{k}d\mu_{\alpha}\right|\leq\|f\|_{1,\alpha},$

and

$\|\tilde{f}_{a}\|_{\infty}\leq\sum_{m=0}^{\infty}|a(m)|w(m)\|\tilde{R}_{m}^{\alpha}\|_{\infty}=\|a\|_{1,\alpha},$

where (2) is considered.

Let $a$ and $b$ are in $l^{1}_{\alpha}$, say, then the series $\tilde{f}_{a}$
and $\tilde{g}_{b}$ are uniformly convergent. Thus by (14)

(20)
$\tilde{f}_{a}(x)\tilde{g}_{b}(x)e^{-\frac{x}{2}}=\sum_{k=0}^{\infty}(a*b)(k)w(k)\tilde{R}_{k}^{\alpha},$

that is

(21) $\check{a}\check{b}\sim a*b,$

see [4].

Similarly (14) ensures that if $\hat{f}=a_{f}$ then

(22) $\widehat{\tilde{f}\tilde{R}_{n}^{\alpha}}(k)=T_{n}(a_{f})(k).$

Moreover (3) implies that

(23) $\widehat{f*g}(n)=\hat{f}(n)\hat{g}(n),\hskip 4.0pt\hskip
4.0pt\widehat{T_{t}^{\alpha}(f)}(n)=\hat{f}(n)R_{n}^{\alpha}(t),$

provided that $\alpha\geq 0$, $f\in L^{p}_{\alpha}$ and $g\in L^{q}_{\alpha}$
and $\frac{1}{p}+\frac{1}{q}\geq 1$.

###  2.3. Pego-type theorem with Laguerre transformation

At first we introduce the notion of equicontinuity in mean with respect to
sequences.

${\bf P_{bs}}.$ A set $K\subset l^{p}_{\alpha}$ is equicontinuous in mean if
for all $\varepsilon>0$ there is an $N\in\mathbb{N}$ such that for all $j>N$
and $a\in K$

(24)
$\left(\sum_{k=0}^{\infty}|T_{j}(a)(k)-a(k)|^{p}w(k)\right)^{\frac{1}{p}}<\varepsilon.$

Subsequently we need the the next extra property with respect to functions.

${\bf P_{a0}}.$ A set $K\subset L^{p}_{\alpha}$ is equivanishing at zero if
for all $\varepsilon>0$ there is a $\delta>0$ such that for all $f\in K$

(25)
$\left(\int_{0}^{\delta}|\tilde{f}|^{p}d\mu_{\alpha}\right)^{\frac{1}{p}}\leq\varepsilon.$

After this preparation we are in position to state a Pego-type theorem.

######  Theorem 2.

Let $1\leq p\leq 2$ and $\alpha\geq 0$.
$(\mathrm{a})$ If $K\subset l^{p}_{\alpha}$ is bounded and fulfils ${\bf
P_{as}}$, then $\check{K}\subset L^{p^{\prime}}_{\alpha}$ fulfils ${\bf
P_{b}}$.
$(\mathrm{b})$ If $K\subset L^{p}_{\alpha}$ fulfils ${\bf P_{b}}$, then
$\hat{K}\subset l^{p^{\prime}}_{\alpha}$ fulfils ${\bf P_{as}}$.
$(\mathrm{c})$ If $K\subset l^{p}_{\alpha}$ fulfils ${\bf P_{bs}}$, then
$\check{K}\subset L^{p^{\prime}}_{\alpha}$ fulfils ${\bf P_{a}}$.
$(\mathrm{d})$ If in addition $\alpha>\frac{1}{2}$, $K\subset L^{p}_{\alpha}$
is bounded and fulfils ${\bf P_{a}}$ and ${\bf P_{a0}}$, then $\hat{K}\subset
l^{p^{\prime}}_{\alpha}$ fulfils ${\bf P_{bs}}$.

###### Proof.

$(\mathrm{a})$: Let $f=f_{a}\in\check{K}\subset L^{p^{\prime}}_{\alpha}$. Then
by (23) and (2)

$\left(\int_{0}^{\infty}\left|\left(T_{t+h}^{\alpha}f(x)-T_{t}^{\alpha}f(x)\right|e^{-\frac{x}{2}}\right)^{p^{\prime}}d\mu_{\alpha}(x)\right)^{\frac{1}{p^{\prime}}}$
$\leq
c\left(\sum_{k=0}^{\infty}\left|a(k)\left(R_{k}^{\alpha}(t+h)-R_{k}^{\alpha}(t)\right)\right|^{p}w(k)\right)^{\frac{1}{p}}$
$\leq
c\left(\sum_{k=0}^{N}\left|a(k)\left(R_{k}^{\alpha}(t+h)-R_{k}^{\alpha}(t)\right)\right|^{p}w(k)\right)^{\frac{1}{p}}+2ce^{\frac{t}{2}}\left(\sum_{k=N+1}^{\infty}|a(k)|^{p}w(k)\right)^{\frac{1}{p}}$
$=S_{1}+S_{2}.$

Since $\left\|\left(R_{k}^{\alpha}\right)^{\prime}\right\|_{\infty}\leq
ck\left\|R_{k-1}^{(\alpha+1)}\right\|_{\infty}$, see [27, (5.1.14)],

$S_{1}\leq cNMe^{\frac{t}{2}}h,$

where $K$ is bounded by $M$. Thus, recalling that $t\in[0,M_{0}]$ by ${\bf
P_{as}}$ if $N$ is large enough, $S_{2}\leq\frac{\varepsilon}{2}$, and we can
choose $\delta$ so small such that $S_{1}\leq\frac{\varepsilon}{2},$ if $0\leq
h\leq\delta$.
$(\mathrm{b})$: Let $f\in K$. Take a function $g\in C_{0}$ such that
$g\geq 0$, $\mathrm{supp}g\subset[0,\delta]$ and
$\int_{0}^{\infty}g(x)e^{-x}d\mu_{\alpha}(x)=1$. As $\hat{g}(k)\to 0$, we can
choose $N$ so large that $|\hat{g}(k)|<\frac{1}{2}$ if $k\geq N$. Thus
applying (23), (5), Minkowski inequality and the symmetry of translation we
have

$\left(\sum_{k=N}^{\infty}|\hat{f}(k)|^{p^{\prime}}w(k)\right)^{\frac{1}{p^{\prime}}}\leq
2\left(\sum_{k=N}^{\infty}|\hat{f}(k)(1-\hat{g}(k))|^{p^{\prime}}w(k)\right)^{\frac{1}{p^{\prime}}}$
$\leq
c\left(\int_{0}^{\infty}\left|(f(t)-f*g(t))e^{-\frac{t}{2}}\right|^{p}d\mu_{\alpha}(t)\right)^{\frac{1}{p}}$
$=c\left(\int_{0}^{\infty}\left|\left(f(t)-\int_{0}^{\infty}T_{t}^{\alpha}f(x)g(x)e^{-x}d\mu_{\alpha}(x)\right)e^{-\frac{t}{2}}\right|^{p}d\mu_{\alpha}(t)\right)^{\frac{1}{p}}$
$\leq
c\int_{0}^{\infty}\left(\int_{0}^{\infty}\left(|f(t)-T_{x}^{\alpha}f(t)|e^{-\frac{t}{2}}\right)^{p}d\mu_{\alpha}(t)\right)^{\frac{1}{p}}g(x)e^{-x}d\mu_{\alpha}(x)$
$\leq c\sup_{0\leq x\leq\delta}\|f-T_{x}^{\alpha}f\|_{p,\alpha}.$

$(\mathrm{c})$: Let $N$ be the appropriate index for $\varepsilon$ in ${\bf
P_{bs}}$, and let us define a finite sequence $b=(b(N),\dots,b(N+n))$ such
that $b_{k}\geq 0$, $k=N,\dots,N+n$ and $\sum_{N}^{N+n}b(k)w(k)=1$. Let us
denote by $\check{b}=g$ and for an $a\in K$ $\check{a}=f$. For $\varepsilon>0$
we can take an apropriate initial part of $a$,
$a_{L}:=(a(0),a(1),\dots,a(L))$, and $\check{a}_{L}=p_{L}$ such that
$\|f-p_{L}\|_{p^{\prime},\alpha}\leq\varepsilon$.
Since $\lim_{x\to\infty}g(x)e^{-x}=0$, there is an $R$ such that
$|g(x)e^{-x}|\leq\frac{1}{2}$ if $x>R$. Considering that $|g(x)e^{-x}|\leq 1$
on $x\geq 0$, in view of (21), as above

$\left(\int_{R}^{\infty}|\tilde{f}(x)|^{p^{\prime}}d\mu_{\alpha}(x)\right)^{\frac{1}{p^{\prime}}}\leq\varepsilon+\left(\int_{R}^{\infty}|\tilde{p_{L}}(x)|^{p^{\prime}}d\mu_{\alpha}(x)\right)^{\frac{1}{p^{\prime}}}$
$\leq\varepsilon+2\left(\int_{R}^{\infty}|\tilde{p_{L}}(x)(1-g(x)e^{-x})|^{p^{\prime}}d\mu_{\alpha}(x)\right)^{\frac{1}{p^{\prime}}}$
$=\varepsilon+\left(\int_{R}^{\infty}|\tilde{p_{L}}(x)-\tilde{p_{L}}(x)g(x)e^{-x}|^{p^{\prime}}d\mu_{\alpha}(x)\right)^{\frac{1}{p^{\prime}}}\leq\varepsilon+c\|a_{L}-a_{L}*b\|_{p,\alpha}$
$=\varepsilon+c\left(\sum_{k=0}^{L}\left|a(k)-\sum_{j=N}^{N+n}b(j)T_{k}(a)(j)w(j)\right|^{p}w(k)\right)^{\frac{1}{p}}$
$\leq\varepsilon+c\left(\sum_{k=0}^{\infty}\left|\sum_{j=N}^{N+n}\left[a(k)-T_{j}(a)(k)\right]b(j)w(j)\right|^{p}w(k)\right)^{\frac{1}{p}}$
$\varepsilon+\leq c\sup_{j\geq
N}\left(\sum_{k=0}^{\infty}|a(k)-T_{j}(a)(k)|^{p}w(k)\right)^{\frac{1}{p}}\leq(1+c)\varepsilon.$

$(\mathrm{d})$: According to (22) and (2)

$\left(\sum_{k=0}^{\infty}|a(k)-T_{j}(a)(k)|^{p^{\prime}}w(k)\right)^{\frac{1}{p^{\prime}}}=\left(\sum_{k=0}^{\infty}\left|\hat{f}(k)-\widehat{\tilde{R}_{j}^{\alpha}\tilde{f}}(k)\right|^{p^{\prime}}w(k)\right)^{\frac{1}{p^{\prime}}}$
$\leq
c\left(\int_{0}^{\infty}\left|f(x)\left(1-R_{j}^{\alpha}(x)e^{-x}\right)e^{-\frac{x}{2}}\right|^{p}d\mu_{\alpha}(x))\right)^{\frac{1}{p}}\leq
2c\left(\int_{0}^{\delta}\left|f(x)e^{-\frac{x}{2}}\right|^{p}d\mu_{\alpha}(x))\right)^{\frac{1}{p}}$
$+c\left(\int_{\delta}^{R}\left|f(x)\left(1-R_{j}^{\alpha}(x)e^{-x}\right)e^{-\frac{x}{2}}\right|^{p}d\mu_{\alpha}(x))\right)^{\frac{1}{p}}$
$+2c\left(\int_{R}^{\infty}\left|f(x)e^{-\frac{x}{2}}\right|^{p}d\mu_{\alpha}(x))\right)^{\frac{1}{p}}=I+II+III.$

By the assumption $I+III\leq 4c\varepsilon$. To estimate $II$ we consider that

$\left(R_{j}(x)e^{-x}\right)^{\prime}=e^{-x}\frac{-L_{j-1}^{(\alpha+1)}(x)-L_{j}^{\alpha}(x)}{\binom{j+\alpha}{j}}=\frac{-L_{j}^{(\alpha+1)}(x)e^{-x}}{\binom{j+\alpha}{j}},$

see [27, (5.1.13)]. In view of [22, (2.8)]

(26)
$\tilde{R}_{n}^{\alpha}(x)x^{\frac{\alpha}{2}}\leq\frac{C}{n^{\frac{\alpha}{2}}}\left\\{\begin{array}[]{ll}(x\nu){\frac{\alpha}{2}},\hskip
4.0pt\hskip 4.0pt0\leq x\leq\frac{1}{\nu}\\\ (x\nu)^{-\frac{1}{4}},\hskip
4.0pt\hskip 4.0pt\frac{1}{\nu}<x\leq\frac{\nu}{2}\\\
\left(\nu\left(\nu^{\frac{1}{3}}+|x-\nu|\right)\right)^{-\frac{1}{4}},\hskip
4.0pt\hskip 4.0pt\frac{\nu}{2}<x\leq\frac{3\nu}{2}\\\ e^{-\gamma x},\hskip
4.0pt\hskip 4.0pt\frac{3\nu}{2}<x,\end{array}\right.$

where $\nu=4n+2\alpha+2$. Thus

(27) $\frac{\left|L_{j}^{(\alpha+1)}(x)\right|e^{-x}}{\binom{j+\alpha}{j}}\leq
c\frac{j^{\frac{1}{4}-\frac{\alpha}{2}}}{\delta^{\frac{\alpha+1}{2}+\frac{1}{4}}},\hskip
4.0pt\hskip 4.0ptx\in(\delta,R).$

Thus, denoting the bound of $K$ with $M$

$II\leq c(\delta)Rj^{\frac{1}{4}-\frac{\alpha}{2}}\|f\|_{p,\alpha}\leq
c(\delta,R,M)j^{\frac{1}{4}-\frac{\alpha}{2}},$

which is small if $j$ is large enough.

The computation in proof of (d) ensures the regular behavior of translation on
sequences.

######  Proposition 2.

Let $\alpha>\frac{1}{2}$, $1\leq p<\infty$. If $K\subset l^{p}_{\alpha}$ is
precompact, then $K$ fulfils ${\bf P_{bs}}$.

###### Proof.

By property ${\bf P_{as}}$ it is enough to prove the statement for finite sets
of finite sequences. Let $S:=s_{1},\dots,s_{n}$ and
$s_{i}=\\{a_{ik}\\}_{k=0}^{n_{i}}$ and
$p_{i}=\sum_{k=0}^{n_{i}}a_{ik}R_{k}^{\alpha}$. For any $\varepsilon>0$ there
are $\delta_{i}$ and $R_{i}$ such that
$\left(\int_{0}^{\delta_{i}}|\tilde{p}_{i}|^{2}d\mu_{\alpha}\right)^{\frac{1}{2}}\leq\varepsilon$
and
$\left(\int_{R_{i}}^{\infty}|\tilde{p}_{i}|^{2}d\mu_{\alpha}\right)^{\frac{1}{2}}\leq\varepsilon$,
$i=1,\dots,n$. Let $R:=R(\varepsilon_{1},\varepsilon)=\max_{i=1}^{n}R_{i}$,
$N:=N(\varepsilon_{1})=\max_{i=1}^{n}n_{i}$,
$\delta:=\delta(\varepsilon_{1},\varepsilon)=\min_{i=1}^{n}\delta_{i}$,
$M:=M(\varepsilon_{1})=\max_{i=1}^{n}\|p_{i}\|_{2,\alpha}$. With the
abbreviation $a_{ik}=\hat{p}_{i}(k)$, as above we have

$\left|\hat{p}_{i}(k)-T_{j}(\hat{p}_{i})(k)\right|=\left|\int_{0}^{\infty}\left(p_{i}(x)-R_{j}^{\alpha}(x)p_{i}(x)e^{-x}\right)R_{k}(x)e^{-x}x^{\alpha}dx\right|$
$\leq\|R_{k}^{\alpha}\|_{2,\alpha}\left(2\left(\int_{0}^{\delta}|\tilde{p}_{i}|^{2}d\mu_{\alpha}\right)^{\frac{1}{2}}+\|p_{i}\|_{2,\alpha}\sup_{[\delta,R]}\left|1-R_{j}(x)e^{-x}\right|+2\left(\int_{R}^{\infty}|\tilde{p}_{i}|^{2}d\mu_{\alpha}\right)^{\frac{1}{2}}\right).$

By (27) and (2)

$\left|\hat{p}_{i}(k)-T_{j}(\hat{p}_{i})(k)\right|\leq
ck^{-\frac{\alpha}{2}}\left(4\varepsilon+MR\frac{j^{\frac{1}{4}-\frac{\alpha}{2}}}{\delta^{\frac{\alpha+1}{2}+\frac{1}{4}}}\right)\leq
5c\varepsilon k^{-\frac{\alpha}{2}}$

if $j$ is large enough. Thus

$\|\hat{p}_{i}-T_{j}(\hat{p}_{i})\|_{p,\alpha}\leq
5c\varepsilon\left(\sum_{k=0}^{N}k^{-p\frac{\alpha}{2}}w(k)\right)^{\frac{1}{p}}\leq
5c\varepsilon N^{\frac{\alpha+1}{p}-\frac{\alpha}{2}}.$

Since $N$ is independent of $\varepsilon$, it is arbitrary small if $j$ is
large enough. According to Proposition 1 the proof can be finished with a
triangle inequality.

The main theorem of this section os the next corollary.

######  Corollary 1.

Let $\alpha\geq 0$. A set $K\subset L^{2}_{\alpha}$ is precompact if and only
if it satisfies ${\bf P_{b}}$, and the corresponding set $\hat{K}$ is
pointwise bounded.

###### Proof.

If $K$ is precompact in $L^{2}_{\alpha}$, Theorem 1 ensures equicontinuity in
mean. Moreover $K$ is bounded in $L^{2}_{\alpha}$, so $\hat{K}$ is bonded in
$l^{2}_{\alpha}$ and as above, pointwise too.
If $K$ is equicontinuous in mean, by Theorem 2 $\hat{K}$ is equivanishing in
$l^{2}_{\alpha}$, and the pointwise boundedness ensures precompactness of
$\hat{K}$ and of $K$ too.

##  3\. Bessel translation method

Bessel translation and Hankel transformation are widely examined by several
authors. We mention here only a few examples. One of our main sources is an
early paper, [21]. By Bessel translation modulus of smoothness and the related
best approximation can be investigated, see [24], and Nikol’skii inequalities
for entire functions can be proved, see [3]. For Fourier-Bessel
transformation, similarly to the standard Fourier transformation, uncertainty
results are derived, see [8], which are somehow in concordance with Pego-type
theorems. Hereinafter we concentrate to compactness criteria.

Let $L_{p,\alpha}$ be the space of measurable functions on $\mathbb{R}_{+}$
equipped with the norm

$\|f\|_{p,(\alpha)}=\left(\int_{0}^{\infty}|f(x)|^{p}x^{2\alpha+1}dx\right)^{\frac{1}{p}},\hskip
4.0pt\hskip 4.0pt\alpha\geq-\frac{1}{2},\hskip 4.0pt1\leq p<\infty,\hskip
4.0pt\hskip 4.0pt\hskip
4.0pt\|f\|_{\infty,(\alpha)}=\|f\|_{\infty,\mathbb{R}_{+}}.$

The dual space of $L_{p,\alpha}$ is denoted by $L_{p^{\prime},\alpha}$, where

$\frac{1}{p}+\frac{1}{p^{\prime}}=1.$

First we introduce the Bessel translation of an integrable function, see [21,
(5.19)]:

(28)
$T_{t,\alpha}f(s)=\int_{0}^{\pi}f(\sqrt{t^{2}+s^{2}-2st\cos\varphi})d\mu(\varphi),$

where $d\mu(\varphi)=c_{\alpha}\sin^{2\alpha}\varphi d\varphi$, and
$c_{\alpha}=\left(\int_{0}^{\pi}\sin^{2\alpha}\varphi d\varphi\right)^{-1}$.
The symmetry of the definition implies

(29) $T_{t,\alpha}f(s)=T_{s,\alpha}f(t).$

We need the entire Bessel functions which are

(30)
$j_{\alpha}(z)=\Gamma(\alpha+1)\left(\frac{2}{z}\right)^{\alpha}J_{\alpha}(z)=\sum_{k=0}^{\infty}\frac{(-1)^{k}\Gamma(\alpha+1)}{\Gamma(k+1)\Gamma(k+\alpha+1)}\left(\frac{z}{2}\right)^{2k}.$

Subsequently the next properties will be necessary. The norm of an entire
Bessel function on the half-line is attained at zero:

(31) $\|j_{\alpha}\|_{\infty,\mathbb{R}_{+}}=j_{\alpha}(0)=1.$

The derivative of $j_{\alpha}$ can be expressed as

(32) $j_{\alpha}^{\prime}(z)=-\frac{1}{2(\alpha+1)}zj_{\alpha+1}(z),$

see [6].

As it is mentioned in the introduction, similarly to the exponential case, the
Bessel translation fulfils

$T_{t,\alpha}j_{\alpha}(\lambda x)=j_{\alpha}(\lambda t)j_{\alpha}(\lambda
x),$

see [21].

The operator norm of Bessel translation is one, see [24, (2.24)]

(33) $\|T_{t}f\|_{p,(\alpha)}\leq\|f\|_{p,(\alpha)},\hskip 4.0pt\hskip
4.0pt1\leq p\leq\infty.$

The Kolmogorov-Riesz type theorem of this section is as follows.

######  Theorem 3.

$1\leq p<\infty$ $K\subset L_{p,\alpha}$ is bounded. $K$ is precompact if and
only if the two conditions below are fulfiled.
${\bf P_{A}.}$ $K$ is equivanishing, i.e.

(34) $\forall\varepsilon>0\hskip 4.0pt\exists R>0,\hskip 4.0pt\forall f\in
K\hskip
4.0pt\left(\int_{R}^{\infty}|f(x)|^{p}x^{2\alpha+1}dx\right)^{\frac{1}{p}}<\varepsilon.$

${\bf P_{B}.}$ $K$ is $L_{p,\alpha}$-equicontinuous, i.e.
$\forall\varepsilon>0$ and $\forall$ $M_{0}>0$ there is a $\delta>0$ such that
$\forall$ $0\leq h\leq\delta$, $\forall$ $f\in K$, $\forall$ $t\in[0,M_{0}]$

(35)
$\left(\int_{0}^{\infty}|T_{t+h,\alpha}f(s)-T_{t,\alpha}f(s)|^{p}s^{2\alpha+1}ds\right)^{\frac{1}{p}}<\varepsilon.$

###### Proof.

To prove property ${\bf P_{A}}$ we can proceed just as in the proof of Theorem
1.
To prove ${\bf P_{B}}$, with the notation of Theorem 1 we take an $f\in
L_{p,\alpha}$ and a $\Phi$ in $S_{\frac{\varepsilon}{3}}$ closest to $f$.
Denoting by $(s,t,\varphi):=(\sqrt{s^{2}+t^{2}-2st\cos\varphi})$, in view of
(28)

$\|T_{t+h,\alpha}\Phi-T_{t,\alpha}\Phi\|_{p,(\alpha)}$
$=\left(\int_{0}^{\infty}\left|\int_{0}^{\pi}\left(\Phi(s,t+h,\varphi)-\Phi(s,t,\varphi)\right)d\mu(\varphi)\right|^{p}s^{2\alpha+1}ds\right)^{\frac{1}{p}}$
$\leq\int_{0}^{\pi}\left(\int_{0}^{\infty}\left|\Phi(s,t+h,\varphi)-\Phi(s,t,\varphi)\right|^{p}s^{2\alpha+1}ds\right)^{\frac{1}{p}}d\mu(\varphi).$

Since $||s-t|-h|\leq(s,t,\varphi),(s,t+h,\varphi)\leq s+t+h$, recalling that
$\Phi\in S_{\frac{\varepsilon}{3}}$ and $t<M_{0}$

$\|T_{t+h,\alpha}\Phi-
T_{t,\alpha}\Phi\|_{p,(\alpha)}\leq\int_{0}^{\pi}\left(\int_{0}^{R}\left|\Phi(s,t+h,\varphi)-\Phi(s,t,\varphi)\right|^{p}s^{2\alpha+1}ds\right)^{\frac{1}{p}}d\mu(\varphi),$

where $R=R_{\frac{\varepsilon}{3}}+M_{0}+1$.
$|\sqrt{s^{2}+(t+h)^{2}-2s(t+h)\cos\varphi}-\sqrt{s^{2}+t^{2}-2st\cos\varphi}|\leq
h$ (cf. [24, (2.27)]). Recall that there are finitely many uniformly
continuous $\Phi$ in $S_{\frac{\varepsilon}{3}}$. Since they are
equicontinuous (in standard sense) we can choose an appropriate $\delta$ to
$\frac{\varepsilon}{3B}$, where
$B=B(\varepsilon)=\left(\int_{0}^{R}s^{2\alpha+1}ds\right)^{\frac{1}{p}}$ such
that $|\Phi(s,t+h,\varphi)-\Phi(s,t,\varphi)|<\frac{\varepsilon}{3B}$, thus
considering that $\mu$ is a probability measure, $\|T_{t+h,\alpha}\Phi-
T_{t,\alpha}\Phi\|_{p,(\alpha)}\leq\frac{\varepsilon}{3}$. Considering the
norm of the linear operator, (33), by a triangle inequality (35) is proved.

On the other hand supposing ${\bf P_{A}}$ and ${\bf P_{B}}$, we show that $K$
is precompact. To this we define

(36) $M_{a}f(s):=\frac{1}{A}\int_{0}^{a}T_{t,\alpha}f(s)t^{2\alpha+1}dt,$

where $A=\int_{0}^{a}t^{2\alpha+1}dt$. By Hölder’s inequality, then applying
the symmetry of translation, cf. (29)

(37)
$|M_{a}f(s+u)-M_{a}f(s)|\leq\frac{1}{A}\int_{0}^{a}|T_{t,\alpha}f(s+u)-T_{t,\alpha}f(s)|t^{2\alpha+1}dt$
$\leq\frac{1}{A^{\frac{1}{p}}}\left(\int_{0}^{\infty}|T_{t,\alpha}f(s+u)-T_{t,\alpha}f(s)|^{p}t^{2\alpha+1}dt\right)^{\frac{1}{p}}$
$=\frac{1}{A^{\frac{1}{p}}}\left(\int_{0}^{\infty}|T_{s+u,\alpha}f(t)-T_{s,\alpha}f(t)|^{p}t^{2\alpha+1}dt\right)^{\frac{1}{p}}.$

Similarly, and recalling (33)

(38)
$|M_{a}f(s)|\leq\frac{1}{A^{\frac{1}{p}}}\left(\int_{0}^{\infty}|T_{s,\alpha}f(t)|^{p}t^{2\alpha+1}dt\right)^{\frac{1}{p}}\leq\frac{1}{A^{\frac{1}{p}}}\|f\|_{p,(\alpha)}.$

Let $a$ and $R$ be fixed positive numbers, and let $F_{a,R}:=\\{M_{a}f(s):f\in
K,s<R\\}$. Then by (37) and (35) $F_{a,R}$ is equicontinuous (in standard
sense), and by (38) and boundedness of $K$ $F_{a,R}$ is uniformly bounded.
Thus by the theorem of Arzelá and Ascoli for all $\varepsilon>0$ there is a
finite $\varepsilon$-net $N_{\varepsilon}\subset F_{a,R}$.
Let us denote the elements of
$N_{\varepsilon}=\\{M_{a}f_{1},\dots,M_{a}f_{j}\\}$, where $f_{i}\in K$,
$i=1,\dots,j$ and $j=j(\varepsilon)$, that is for all $f\in K$ there is an
$f_{i}$, such that $|M_{a}f(s)-M_{a}f_{i}(s)|\leq\varepsilon$ if $0\leq s\leq
R$. We show that $\\{f_{i}\\}_{i=1}^{j}$ is an $\varepsilon$-net in $K$ if
$\\{M_{a}f_{i}\\}_{i=1}^{j}=N_{\frac{\varepsilon}{L}}$ is an
$\frac{\varepsilon}{L}$-net in $F_{a,R}$ with a suitable $a$, $L$ and $R$
(will be given later).

Applying again Hölder’s inequality and then Fubini’s theorem

$\|M_{a}f-f\|_{p,(\alpha)}^{p}\leq\int_{0}^{\infty}\left(\int_{0}^{a}\frac{1}{A}|T_{t,\alpha}f(s)-f(s)|t^{2\alpha+1}dt\right)^{p}s^{2\alpha+1}ds$
$\leq\frac{1}{A}\int_{0}^{\infty}\int_{0}^{a}|T_{t,\alpha}f(s)-f(s)|^{p}t^{2\alpha+1}dts^{2\alpha+1}ds$
$=\frac{1}{A}\int_{0}^{a}\int_{0}^{\infty}|T_{t,\alpha}f(s)-f(s)|^{p}s^{2\alpha+1}dst^{2\alpha+1}dt.$

Thus

(39) $\|M_{a}f-f\|_{p,(\alpha)}^{p}\leq\sup_{0\leq t\leq
a}\int_{0}^{\infty}|T_{t,\alpha}f(s)-f(s)|^{p}s^{2\alpha+1}ds.$

Similarly

$\|M_{a}f-M_{a}f_{j}\|_{p,(\alpha)}^{p}\leq\left(\int_{0}^{R}|M_{a}f(s)-M_{a}f_{j}(s)|^{p}s^{2\alpha+1}ds\right)^{\frac{1}{p}}$
$+\left(\int_{R}^{\infty}|M_{a}f(s)-M_{a}f_{j}(s)|^{p}s^{2\alpha+1}ds\right)^{\frac{1}{p}}=I+II.$

As
$II\leq\left(\int_{R}^{\infty}|M_{a}f(s)|^{p}s^{2\alpha+1}ds\right)^{\frac{1}{p}}+\left(\int_{R}^{\infty}|M_{a}f_{j}(s)|^{p}s^{2\alpha+1}ds\right)^{\frac{1}{p}}$,
it is enough to investigate $M_{a}g$, where $g\in K$.

$\left(\int_{R}^{\infty}|M_{a}g(s)|^{p}s^{2\alpha+1}ds\right)^{\frac{1}{p}}\leq\|M_{a}g-g\|_{p,(\alpha)}^{p}+\left(\int_{R}^{\infty}|g(s)|^{p}s^{2\alpha+1}ds\right)^{\frac{1}{p}}.$
$I\leq\sup_{0\leq s\leq R}|M_{a}f(s)-M_{a}f_{i}(s)|B(R),$

where $B(R)=\left(\frac{R^{2\alpha+2}}{2\alpha+2}\right)^{\frac{1}{p}}$. Thus

(40) $\|M_{a}f-M_{a}f_{j}\|_{p,(\alpha)}^{p}\leq B(R)\sup_{0\leq s\leq
R}|M_{a}f(s)-M_{a}f_{i}(s)|$ $+\sup_{g\in
K}\left(\|M_{a}g-g\|_{p,(\alpha)}^{p}+\left(\int_{R}^{\infty}|g(s)|^{p}s^{2\alpha+1}ds\right)^{\frac{1}{p}}\right).$

Let $\varepsilon>0$ be arbitrary and let us choose by (34) $R$ so large that
$\forall f\in K$
$\left(\int_{R}^{\infty}|f(x)|^{p}x^{2\alpha+1}dx\right)^{\frac{1}{p}}<\frac{\varepsilon}{5}$
and by (35) let us choose $a$ so small such that for all $0\leq h\leq a$ and
for all $f\in K$, $t\in[0,R]\hskip
4.0pt\left(\int_{0}^{\infty}|T_{t+h,\alpha}f(s)-T_{t,\alpha}f(s)|^{p}s^{2\alpha+1}ds\right)^{\frac{1}{p}}<\frac{\varepsilon}{5}$.
Now let us take an $\frac{\varepsilon}{5B(R)}$-net,
$N:=N_{\frac{\varepsilon}{5B(R)}}\subset F_{a,R}$. By this choice for all
$g\in K$

$\|g-M_{a}g\|_{p,(\alpha)}\leq\frac{\varepsilon}{5},$

cf. (39). Moreover for an arbitrary $f\in K$ there is a suitable $f_{j}$ with
$M_{a}f_{j}\in N$, such that

$\|M_{a}f-M_{a}f_{j}\|_{p,(\alpha)}\leq
B(R)\frac{\varepsilon}{5B(R)}+\frac{\varepsilon}{5}+\frac{\varepsilon}{5},$

cf. (40). Finally we finish the proof with

$\|f-f_{j}\|_{p,(\alpha)}\leq\|f-M_{a}f\|_{p,(\alpha)}+\|M_{a}f-M_{a}f_{j}\|_{p,(\alpha)}+\|M_{a}f_{j}-f_{j}\|_{p,(\alpha)}\leq\varepsilon.$

Now we define the Bessel-Fourier or Hankel transformation to facilitate
formulating a Pego-type theorem. Let us denote the Hankel trasform of a
function $f$ by

$H_{\alpha}(f)(y)=\hat{f}(y):=c_{\alpha}\int_{0}^{\infty}f(x)j_{\alpha}(xy)x^{2\alpha+1}dx.$

The inverse transformation is

$H_{\alpha}^{[-1]}(\hat{f})(x)=c_{\alpha}\int_{0}^{\infty}\hat{f}(y)j_{\alpha}(xy)x^{2\alpha+1}dy,$

if exists. If $K$ is a set of certain functions, we denote by $\hat{K}$ the
set of the Hankel transforms of the functions in $K$.
Below we need that Hankel transformation fulfils the Hausdorff-Young
inequality, that is

(41) $\|H_{\alpha}(f)\|_{p^{\prime},(\alpha)}\leq
C_{p}\|f\|_{p,(\alpha)},\hskip 4.0pt\hskip 4.0pt\hskip 4.0pt1\leq p\leq 2,$

see [8]. We also use the next property of the Hankel transform.

(42) $H_{\alpha}(T_{t,\alpha}f)(y)=j_{\alpha}(ty)\hat{f}(y);\hskip 4.0pt\hskip
4.0pt\hskip
4.0ptT_{t,\alpha}\hat{f}(y)=H_{\alpha}(j_{\alpha}(t\cdot)f(\cdot))(y),$

see [8]. Furthermore

(43) $\lim_{y\to\infty}\hat{f}(y)=0,$

see [24]. Defining the convolution by

$(f*g)(x)=\int_{0}^{\infty}T_{x,\alpha}f(t)g(t)t^{2\alpha+1}dt,$

we have

(44) $H_{\alpha}(f*g)=\hat{f}\hat{g},$

see [8].

######  Theorem 4.

$1\leq p\leq 2$. Let $K\subset L_{p,\alpha}$. If $K$ is bounded and satisfies
${\bf P_{A}}$ in $L_{p,\alpha}$, $\hat{K}$ satisfies ${\bf P_{B}}$ in
$L_{p^{\prime},\alpha}$, and if $K$ satisfies ${\bf P_{B}}$ in $L_{p,\alpha}$,
$\hat{K}$ satisfies ${\bf P_{A}}$ in $L_{p^{\prime},\alpha}$.

###### Proof.

Let $K$ be bounded by $M$ and $f\in K$. Then by (42) and then (41)

$\|T_{t+h,\alpha}\hat{f}-T_{t,\alpha}\hat{f}\|_{p^{\prime},(\alpha)}=\|H_{\alpha}\left((j_{\alpha}((t+h)(\cdot))-j_{\alpha}(t(\cdot)))f(\cdot)\right)\|_{p^{\prime},(\alpha)}$
$\leq C_{p}\|(j_{\alpha}((t+h)s)-j_{\alpha}(ts))f(s)\|_{p,(\alpha)}\leq
C_{p}\left(\int_{0}^{R}\left|(j_{\alpha}((t+h)s)-j_{\alpha}(ts))f(s)\right|^{p}s^{2\alpha+1}ds\right)^{\frac{1}{p}}$
$+C_{p}\left(\int_{R}^{\infty}\left|(j_{\alpha}((t+h)s)-j_{\alpha}(ts))f(s)\right|^{p}s^{2\alpha+1}ds\right)^{\frac{1}{p}}=C_{p}(I+II).$

Let $\varepsilon$ be arbitrary and we choose $R=R(\varepsilon)$ to
$\frac{\varepsilon}{4}$ according to (34). By (31)

$II\leq
2\left(\int_{R}^{\infty}\left|f(s)\right|^{p}s^{2\alpha+1}ds\right)^{\frac{1}{p}}\leq\frac{\varepsilon}{2}.$

To estimate $I$, let $M_{0}>0$ be arbitrary and assume that $0<t<M_{0}$. Let
us choose $\delta=\delta(R(\varepsilon),M_{0},\varepsilon)<1$ such that

$\delta\leq\frac{(\alpha+1)\varepsilon}{(M_{0}+1)R^{2}M}.$

Then in view of (32) and (31) for all $0\leq h\leq\delta$, by choice of
$\delta$

$I\leq(M_{0}+1)R\frac{1}{2(\alpha+1)}\|j_{\alpha+1}\|_{\infty}\delta
R\|f\|_{p,(\alpha)}\leq\frac{\varepsilon}{2}.$

To prove the second statement, for an arbitrary fixed $\delta>0$ we choose
nonnegative functions $g_{\delta}$ supported on $(0,\delta)$ such that

$\int_{0}^{\infty}g_{\delta}(x)x^{2\alpha+1}dx=1.$

Then in view of (43) we choose $R=R(\delta)$ such that
$|H_{\alpha}(g_{\delta})(y)|\leq\frac{1}{2}$ if $y>R$. Thus, by (44) and (41)

$\left(\int_{R}^{\infty}|\hat{f}(y)|^{p^{\prime}}y^{2\alpha+1}dy\right)^{\frac{1}{p^{\prime}}}\leq
2\left(\int_{R}^{\infty}|H_{\alpha}(f)(y)(1-H_{\alpha}(g_{\delta})(y))|^{p^{\prime}}y^{2\alpha+1}dy\right)^{\frac{1}{p^{\prime}}}$
$\leq
C(p)\|f-f*g_{\delta}\|_{p,(\alpha)}=C(p)\left(\int_{0}^{\infty}\left|\int_{0}^{\infty}(f(x)-T_{x,\alpha}f(t))g_{\delta}(t)t^{2\alpha+1}dt\right|^{p}x^{2\alpha+1}dx\right)^{\frac{1}{p}}$
$\leq
C(p)\int_{0}^{\infty}\left(\int_{0}^{\infty}|f(x)-T_{t,\alpha}f(x)|^{p}x^{2\alpha+1}dx\right)^{\frac{1}{p}}g_{\delta}(t)t^{2\alpha+1}dt$
$\leq C(p)\sup_{0\leq
t\leq\delta}\left(\int_{0}^{\infty}|f(x)-T_{t,\alpha}f(x)|^{p}x^{2\alpha+1}dx\right)^{\frac{1}{p}},$

where in the last but one (Minkowski) inequality the symmetry of translation
is considered. Thus the proof can be finished by choosing an appropriate
$\delta$ to the arbitrary $\varepsilon$ in view of (35), and $g_{\delta}$ and
$R(\delta)$ to $\delta$ as above.

######  Corollary 2.

$\mathrm{(I)}$ A bounded subset $K\subset L_{2,\alpha}$ is precompact if and
only if for all $M_{0}>0$, $t\in[0,M_{0}]$
$\|T_{t+h,\alpha}f-T_{t,\alpha}f\|_{2,(\alpha)}\to 0$, and
$\|T_{t+h,\alpha}\hat{f}-T_{t,\alpha}\hat{f}\|_{2,(\alpha)}\to 0$ uniformly in
$K$.

$\mathrm{(II)}$ A bounded subset $K\subset L_{2,\alpha}$ is precompact if and
only if
$\left(\int_{R}^{\infty}|f(x)|^{2}x^{2\alpha+1}dx\right)^{\frac{1}{2}}\to 0$,
and
$\left(\int_{R}^{\infty}|\hat{f}(y)|^{2}y^{2\alpha+1}dy\right)^{\frac{1}{2}}\to
0$ uniformly in $K$.

######  Corollary 3.

Let $u,v:\mathbb{R}_{+}\to\mathbb{C}$, $u,v\in L_{\infty}$ such that
$\lim_{x\to\infty}u(x),v(x)=0$, and assume that $u$ is continuous. Then the
operator $L:=uH_{\alpha}v$, $f\mapsto uH_{\alpha}(vf)$ is compact on
$L_{2,(\alpha)}$.

###### Proof.

Let $K\subset L_{2,\alpha}$ be a subset bounded by $M$. By the assumption, for
an $\varepsilon>0$ there is an $R>0$ such that $|v(x)|<\varepsilon$ if $x>R$.
Thus
$\left(\int_{R}^{\infty}|(vf)(x)|^{2}x^{2\alpha+1}dx\right)^{\frac{1}{2}}\leq\varepsilon
M$, that is $vK$ is bounded and equivanishing. So by Theorem 4
$H_{\alpha}(vK)$ is $L_{2,\alpha}$-equicontinuous, and by assumption
$LK=uH_{\alpha}(vK)$ is also $L_{2,\alpha}$-equicontinuous and equivanishing.

Remark. (1) In the proof of the first statement of Theorem 4 we can also use
the asymptotics of Bessel functions $|j_{\alpha+1}(u)|\leq
Cu^{-\alpha+\frac{3}{2}}$. Then a larger $\delta$ can be allowed.
(2) Comparing [18] to these special translations, the omission of the
criterion of boundedness needs further investigations in Bessel and Laguerre
cases.

## References

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Department of Analysis,
Budapest University of Technology and Economics

<EMAIL_ADDRESS>


# Reinforced Meta Active Learning

Michael Katz Huawei Technologies
Israel Research Center  Eli Kravchik Huawei Technologies
Israel Research Center

###### Abstract

In stream-based active learning, the learning procedure typically has access
to a stream of unlabeled data instances and must decide for each instance
whether to label it and use it for training or to discard it. There are
numerous active learning strategies which try to minimize the number of
labeled samples required for training in this setting by identifying and
retaining the most informative data samples. Most of these schemes are rule-
based and rely on the notion of _uncertainty_ , which captures how small the
distance of a data sample is from the classifier’s decision boundary.
Recently, there have been some attempts to learn optimal selection strategies
directly from the data, but many of them are still lacking generality for
several reasons: 1) They focus on specific classification setups, 2) They rely
on rule-based metrics, 3) They require offline pre-training of the active
learner on related tasks. In this work we address the above limitations and
present an online stream-based meta active learning method which learns on the
fly an informativeness measure directly from the data, and is applicable to a
general class of classification problems without any need for pretraining of
the active learner on related tasks. The method is based on reinforcement
learning and combines episodic policy search and a contextual bandits approach
which are used to train the active learner in conjunction with training of the
model. We demonstrate on several real datasets that this method learns to
select training samples more efficiently than existing state-of-the-art
methods.

##  1 Introduction

In many machine learning applications, data for training the model arrives in
the form of a sequential stream of unlabeled data samples, where some or all
of these samples are then labeled and used for training. Since annotated data
is typically difficult and expensive to obtain, especially in domains where
reliable labels can only be obtained by experts, active learning (AL)
alleviates this difficulty by focusing the labeling budget only on informative
data instances which enable to train the model most effectively. However, in
some situations, due to practical limitations, e.g. storage constraints at
edge devices, etc., upon the arrival of each data sample, an immediate
decision must be made whether to retain the sample for training or to discard
it. Such a setting is known as _stream-based active learning_ , as opposed to
_pool-based active learning_ , where all the unlabeled data samples are
available to the active learner right from the start. Examples for such
learning scenarios include spam filtering (Sculley,, 2007), human activity
recognition (Adaimi and Thomaz,, 2019), sentiment analysis (Smailović et al.,,
2014), learning from social-media (Pohl et al.,, 2018), collecting data from
fleet vehicles for training perception systems for self-driving cars, and
more. In fact, this scenario is applicable to any learning system which
accumulates large amounts of unlabeled data and wishes to use only a small
representative subset for training, but is unable to store the data to be
selected for labeling at a later time, and must decide on the spot whether to
keep each data instance or not.

Over the years, several rule-based approaches were proposed for the stream-
based AL setting (Settles,, 2009; Lewis and Gale,, 1994; Seung et al.,, 1992;
Sculley,, 2007; Chu et al.,, 2011; Zliobaitė et al.,, 2014; Kottke et al.,,
2015; Rosa and Cesa-Bianchi,, 2017). Most of these approaches attempt to
assign an informativeness measure to each data sample and give preference to
samples with a high measure, also known as uncertainty sampling. Other
approaches maintain a committee of classifiers and select data samples for
which there is a large disagreement over their predictions by all committee
members. More recently, learning-based approaches which learn AL policies
directly from the data were also reported (Woodward and Finn,, 2017; Fang et
al.,, 2017; Kvistad et al.,, 2019; Wassermann et al.,, 2019; Krawczyk and
Cano,, 2019; Cheng et al.,, 2013). Most of these approaches try to harness
reinforcement learning in order to learn optimal active learning policies for
training a classifier. However, many of them either suffer from lack of
generality or are impractical as an online algorithm for AL. For instance,
(Woodward and Finn,, 2017) is applicable only in a specific and non-standard
classification setup known as one-shot learning, (Wassermann et al.,, 2019;
Cheng et al.,, 2013) are based on combinations of rule-based informativeness
measures such as uncertainty sampling, and finally (Fang et al.,, 2017)
requires pre-training of the active learner on a related task before applying
it on the actual task of interest, rendering the approach impractical for any
new task.

In this work we address the limitations of previously reported meta-learning
approaches for AL from several perspectives. First, we do not limit the active
learner to consider any specific function of the model’s predictions for
making a decision to select samples for training. Secondly, we consider a
meta-learning strategy which is applicable to any classification problem, and
finally, our approach is an online approach, which does not require any
offline pre-training. Our main contributions are summarized as follows:

  * •

A novel stream-based meta AL method which adapts itself to the task and
dataset online and learns to select on the fly the most informative data
samples for training a classifier (our method is also suitable for a
regression type of problem).

  * •

The method is based on reinforcement learning and combines episodic policy
optimization and a contextual bandits (CB) approach which are applied
_online_. In particular, the method does not require any offline pre-training
on related tasks and/or datasets.

  * •

An empirical evaluation demonstrating that for a given budget constraint
imposed by the system, this method learns to select training samples more
efficiently than existing uncertainty-based approaches.

##  2 Related work

Most stream-based AL algorithms use rule-based query strategies to evaluate
the informativeness of unlabeled data samples. The most commonly used query
strategy is _uncertainty sampling_ (Lewis and Gale,, 1994), where the active
learner queries the instances for which it is least certain about their most
probable label. The uncertainty can be defined in many ways, e.g., in terms of
the maximum prediction value across classes, the difference between the two
highest prediction values a.k.a. _margin sampling_ , and even the Shannon
entropy of the prediction vector or the mutual information between predictions
and model parameters (Gal et al.,, 2017). Another query strategy is _query-by-
committee_ (QBC) (Seung et al.,, 1992), where a committee of models which are
all trained on the labeled instances represent competing hypotheses. Each
model member is allowed to vote on the labeling of query candidates and the
candidates over which the committee members disagree most are considered more
informative.

Other rule-based approaches reported more recently include (Kottke et al.,,
2015), which defines a probabilistic spatial usefulness measure to model the
true posterior probability of the classifier and uses it to select the best
samples, (Fujii and Kashima,, 2016) which applies pool-based AL approaches
based on adaptive submodularity for the stream-based setting, and (Rosa and
Cesa-Bianchi,, 2017) which analyzes confidence intervals for splitting leaves
in decision trees and uses them for selective sampling in online decision tree
learning.

Some learning-based approaches aimed at learning selection strategies directly
from the data were also recently reported. For example, (Cheng et al.,, 2013)
proposes a reinforcement AL algorithm that adaptively mixes two sample
selection criteria based on the KL divergence measured from classifier change.
In (Woodward and Finn,, 2017) meta learning and reinforcement learning are
combined to learn an active learner for the one-shot classification problem
(Santoro et al.,, 2016). In (Fang et al.,, 2017) reinforcement learning is
also used to train a stream-based active learner agent to select samples for
training a classifier on a Natural Language Processing task. In this work the
agent is first trained to select samples for one task, and then transferred to
select samples for a similar task. This work is further extended in (Kvistad
et al.,, 2019) by introducing a memory-augmented neural architecture. RAL
(Wassermann et al.,, 2019) is a stream-based AL framework modeled as a CB
problem. A committee of expert classifiers, where each member has a decision
weight, is used to vote on whether or not to label an incoming sample. Once a
decision is made, a reward signal which measures the usefulness of the
committee’s decision is used to update the decision weights of the
classifiers. A similar pure bandits approach is also reported in (Krawczyk and
Cano,, 2019).

##  3 A Reinforced Meta Active Learning Framework

###  3.1 Problem Formulation - Stream-based Active Learning

Let
$\mathcal{D}_{0}=\left\\{\left(x_{i}^{0},y_{i}^{0}\right)\right\\}_{i=1}^{N_{0}}$
be an initial training set composed of pairs $(x_{i}^{0},y_{i}^{0})$ of input
feature vectors $x_{i}^{0}\in\mathbb{R}$ and corresponding ground truth labels
${y_{i}^{0}\in\left\\{1,...,C\right\\}}$ independently drawn from some joint
probablity distribution. Next, let
$\mathcal{X}=\left\\{x_{1},x_{2},...,x_{N}\right\\}$ be a data stream of $N$
additional input feature vectors $x_{i}\in\mathbb{R}$, and
$\mathcal{Y}=\left\\{y_{1},y_{2},...,y_{N}\right\\}$ be their corresponding
_unknown_ ground truth labels $y_{i}\in\left\\{1,...,C\right\\}$ randomly
drawn from the same distribution. We further assume the existence of an oracle
annotator $c(x)$ who is able to recover the class $y$ associated with a given
input feature vector $x$. The initial training set $\mathcal{D}_{0}$ and the
data stream $\mathcal{X}$ along with the oracle $c(x)$ are used to train a
classifier model $p\left(y|x;\theta\right)$ parameterized by $\theta$ with a
given training procedure.

Figure 1: Framework of Reinforced Meta Active Learning

An active learner $\mathcal{L}$ for this setting is defined as follows. At
each time step $t$, the active learner computes a classifier $f_{t}\coloneqq
p\left(y|x;\theta_{t}\right)$. At time step $t=0$, the classifier $f_{0}$ is
obtained by training on the initial training set $\mathcal{D}_{0}$. Next, for
$t=1,2,...,N$, in response to a sample $x_{t}$ arriving from the stream, the
active learner does one of the following actions: 1) Compute the label
$y_{t}=c(x_{t})$, add the pair $(x_{t},y_{t})$ to the training set, and re-
train the classifier; or 2) Discard the sample, in which case the classifier
remains unchanged. The active learner may utilize the classifier in order to
make a decision. Denote by $N_{u}$ the total number of samples labeled by the
active learner during the learning process until all $N$ samples from the
stream are exhausted. Then, a budget constraint $b\in(0,1]$ limits the
fraction of total labeled samples such that $N_{u}\leq bN$ (Zliobaitė et al.,,
2014).

Finally, let $\mathcal{U}_{t}$ be the training set accumulated by an active
learner $\mathcal{L}$ up to time $t$. Then, the classifier at step $t$ is
given by $f_{t}=\mathcal{L}\left(f_{t-1},\mathcal{U}_{t-1},x_{t}\right)$. Let
$\textrm{Acc}(t)$ denote the accuracy of the classifier $f_{t}$ measured on a
held out test set. Then, for a budget fraction $\beta\in(0,1]$ we define the
partial budget accuracy metric $\textrm{PBA}(\beta)$ as
$\textrm{Acc}(t_{\beta})$, where $t_{\beta}$ is the smallest $t$ such that
$\left|\mathcal{U}_{t}\right|\geq\beta bN$.

###  3.2 System Setup

The RMAL framework is illustrated in Figure 1. Samples arrive in a stream to a
budget controller, which monitors the number of samples $u_{t}$ selected for
training until time $t$, and only passes on samples for which $u_{t}/t<b$.
Samples for which this condition is not met are discarded. In order to prevent
under consumption, whenever the instantaneous consumption rate $u_{t}/t$ falls
below $b_{min}<b$, an amount of labeled samples are immediately replenished
from the stream until the rate returns to a level of $b$.

A sample $x_{j}$ arriving from the budget controller is classified by the
classifier, who produces a class prediction vector
$\boldsymbol{p}_{j}\in[0,1]^{C}$. The prediction vector is then fed to an AL
agent who makes a decision $a_{j}\in\\{0,1\\}$ whether to select the current
sample for training the classifier or to discard it. If the sample is chosen
for training, the annotator produces the label $y_{j}$ and the pair
$(x_{j},y_{j})$ is added to the training set, and the classifier is re-
trained, otherwise the current step is done, and the system proceeds to
process the next sample.

We take the approach of (Fang et al.,, 2017) and model the sequence of
decisions, whether to label an input sample or not, as a Markov Decision
Process (MDP), which allows the learning of a policy that can dynamically
select instances most informative for training. We represent the MDP as a
tuple
$\left(\mathcal{S},\mathcal{A},p\left(s_{j+1}|s_{j},a_{j}\right),r(s,a)\right)$,
where $\mathcal{S}\subseteq\mathbb{R}^{C}$ is the state space,
$\mathcal{A}=\left\\{0,1\right\\}$ is the action space,
$p\left(s_{j+1}|s_{j},a_{j}\right)$ is the state transition probability, and
$r(s,a)$ is the reward function. Under this framework, we represent the AL
agent as a policy $\pi\left(a|s;\varphi\right)$ parameterized by $\varphi$.
Once the classifier is re-trained, its accuracy is evaluated on a held out
validation set, and used to produce a reward signal $r_{j}$ which is
subsequently used to update the agent. This process repeats until all samples
in the stream are exhausted.

In this framework there are two learning processes taking place: 1) learning
of a classifier; and 2) learning of an agent. In the first process, the
classifier is successively trained on an increasing set of samples chosen from
the data stream by the agent. In the second process, the quality of the
classifiers which were trained based on the agent’s decisions is used to train
the agent. The interplay between these learning processes is further detailed
in the next sections.

##  4 An Alternating Approach

Learning To Sample (LTS) (Shao et al.,, 2019) is a pool-based AL framework
where a sampling model and a boosting model mutually learn from each other in
iterations. The boosting model combines several weak models which are
successively trained on a sequence of increasing datasets, and the sampling
model is a regressor which estimates the uncertainty of unlabeled data
samples.

Input : Initial dataset $\mathcal{D}_{0}$, Stream $\\{x_{1},...,x_{N}\\}$,
Budget $b$, Batch size $T$, Training episodes $E$

Output : Classifier $f$

Initialization: $\pi\leftarrow\textrm{Random}$,
$\mathcal{D}\leftarrow\mathcal{D}_{0}$, $\mathcal{N}\leftarrow\emptyset$,
$u\leftarrow 0$

1 $f_{0}\leftarrow\textnormal{{TrainModel}}\left(\mathcal{D}\right)$

2 $f^{\textrm{PXY}}\leftarrow f_{0}$

3 Pretrain the policy $\pi$

4 for $t=1,2,...,N$ do

5 $s_{t}\leftarrow f_{t-1}(x_{t})$

6 Draw action $a_{t}\sim\textrm{Bernoulli}(\pi(a=1|s_{t}))$

7 if $(u/t<b)$ and $(a_{t}=1)$ then

8 $y_{t}\leftarrow\textrm{query label of }x_{t}$

9 $\mathcal{N}\leftarrow\mathcal{N}\cup\\{(x_{t},y_{t})\\}$

10 $f_{t}\leftarrow\textnormal{{TrainModel}}(\mathcal{D}\cup\mathcal{N})$

11 $u\leftarrow u+1$

12 if $|\mathcal{N}|=T$ then

13
$\pi\leftarrow\textnormal{{UpdateAgent}}\left(\pi,f^{\textrm{PXY}},\mathcal{D},\mathcal{N},E\right)$

14 $\mathcal{D}\leftarrow\mathcal{D}\cup\mathcal{N}$

15 $\mathcal{N}\leftarrow\emptyset$

16 $f^{\textrm{PXY}}\leftarrow f_{t}$

17 end if

18

19 else

20 $f_{t}\leftarrow f_{t-1}$

21 end if

22

23 end for

Algorithm 1 RMAL-AL Algorithm

We borrow the alternating approach of LTS. The AL process continuously
alternates between two interleaved learning phases: a classifier training
phase and an agent training phase. In the classifier training phase, the
classifier is trained and the agent remains fixed, while in the agent training
phase, the classifier remains fixed and the agent is updated. A high level
description of the algorithm is presented in Algorithm 1. We assume the
existence of a training procedure $\texttt{TrainModel}(\mathcal{D})$ for
training the model on a given dataset $\mathcal{D}$. The AL process begins
with training of the classifier on the initial training set $\mathcal{D}_{0}$
and initializing a proxy classifier (lines 1-2). The agent is then pretrained
to mimic uncertainty sampling (we detail this procedure in section 6). Next,
as long as there exists budget, each new sample is classified by the most up-
to-date classifier, and the predictions are used by the agent to decide
whether to use this sample for training. The decision is made stochastically
by drawing from a Bernoulli distribution with parameter $\pi(a=1|s_{t})$
(lines 5-7). Selected samples are annotated and added to the training set
(lines 8-9), and a new classifier is trained on the new training set (line
10). Once an amount of $T$ samples has been accumulated, the agent undergoes
an update phase where it is trained to select samples for training the proxy
classifier (line 13).

Input : Agent $\pi_{\varphi}$, Classifier $f$, Training dataset $\mathcal{D}$,
New labeled samples $\mathcal{N}=\\{(z_{1},y_{1}),...,(z_{T},y_{T})\\}$,
Number of episodes $E$, Validation dataset $\mathcal{V}$,

Learning rate $\alpha$

Output : Updated agent $\pi_{\varphi}$

Initialization: $\mathcal{M}\leftarrow\emptyset$, $Acc_{0}\leftarrow$ accuracy
of $f$ on $\mathcal{V}$

1 for episode = 1, 2, …, $E$ do

2 $f^{\textrm{PXY}}_{0}\leftarrow f$

3 $\textnormal{{Shuffle}}(\mathcal{N})$

4 for $t=1,2,...,T$ do

5 $s_{t}\leftarrow f^{\textrm{PXY}}_{t-1}(z_{t})$

6 Draw action $a_{t}\sim\textrm{Bernoulli}(\pi_{\varphi}(a=1|s_{t})$

7 if $a_{t}=1$ then

8 $\mathcal{M}\leftarrow\mathcal{M}\cup\\{(z_{t},y_{t})\\}$

9
$f^{\textrm{PXY}}_{t}\leftarrow\textnormal{{TrainModel}}(\mathcal{D}\cup\mathcal{M})$

10
$\textrm{Acc}_{t}\leftarrow\textnormal{{Accuracy}}(f^{\textrm{PXY}}_{t},\mathcal{V})$

11
$r_{t}\leftarrow\left(\textrm{Acc}_{t}-\textrm{Acc}_{t-1}\right)/\textrm{Acc}_{t-1}$

12 else

13
$f^{\textrm{CF}}\leftarrow\textnormal{{TrainModel}}(\mathcal{D}\cup\mathcal{M}\cup\\{(z_{t},y_{t})\\})$

14
$\textrm{Acc}^{CF}\leftarrow\textnormal{{Accuracy}}(f^{\textrm{CF}},\mathcal{V})$

15 $r_{t}\leftarrow-(\textrm{Acc}^{CF}-\textrm{Acc}_{t-1})/\textrm{Acc}_{t-1}$

16 $f^{\textrm{PXY}}_{t},\,\textrm{Acc}_{t}\leftarrow
f^{\textrm{PXY}}_{t-1},\,\textrm{Acc}_{t-1}$

17 end if

18

19 end for

20 $\varphi\leftarrow\varphi+\alpha\nabla_{\varphi}J(\varphi)$
$\triangleright$ Compute $\nabla_{\varphi}J(\varphi)$ by eq. 5

21 end for

Algorithm 2 RMAL-AL Algorithm: UpdateAgent (line 13 in Algorithm 1)

The UpdateAgent procedure, presented in Algorithm 2, is based on the REINFORCE
policy search method (Williams,, 1992). We define a training episode of the
agent as a sequence of $T$ steps, during which the agent selects samples for
training a proxy classifier. The set $\mathcal{N}$ consists of the samples
selected in the previous classifier training phase along with their labels. At
the beginning of each episode, the proxy classifier $f^{\textrm{PXY}}_{0}$ is
initialized to the state of the last classifier which was not yet trained on
any of the samples in $\mathcal{N}$, and the set $\mathcal{N}$ is shuffled.
Next, for each sample in $\mathcal{N}$, the proxy classifier is used to
classify the sample, and the agent uses its prediction to decide
stochastically whether or not to select the sample for training the proxy
classifier. To facilitate policy learning we introduce an instantaneous reward
signal which provides feedback on the quality of the actions taken by the
agent.

In each step of the episode, if the agent decides to select the sample for
training ($a=1$), the proxy classifier is re-trained with this sample111No
annotation is required as this sample was already annotated in the previous
training phase of the classifier. and its accuracy on a held-out validation
set is recorded as $\textrm{Acc}_{t}$. In this case, the reward signal $r_{t}$
is defined as the relative accuracy improvement of the model resulting from
selecting the $t^{th}$ sample for training. If, however, the agent does not
select the current sample for training, a mock classifier is trained with this
sample and its validation accuracy is recorded as
$\textrm{Acc}^{\textrm{CF}}$. We refer to this accuracy as counter-factual, as
it is the accuracy which would have been attained by the classifier, had the
agent selected this sample for training. In this case, the reward signal
$r_{t}$ is defined as minus the relative accuracy improvement of the model
resulting from training on this samples. It follows that the instantaneous
reward for a given state-action pair $(s_{t},a_{t})$ is given by:

$r_{t}(s_{t},a_{t})=\begin{cases}\phantom{-}\left(\textrm{Acc}^{\phantom{\textrm{C}}}_{t}-\textrm{Acc}_{t-1}\right)/\textrm{Acc}_{t-1},&a_{t}=1\\\
-\left(\textrm{Acc}_{t}^{CF}-\textrm{Acc}_{t-1}\right)/\textrm{Acc}_{t-1},&a_{t}=0\end{cases}.$
(1)

Our goal is to maximize the cumulative return $R(\tau)=\sum_{t=1}^{T}r_{t}$,
where $\tau$ is a trajectory of states, actions, and rewards
$(s_{1},a_{1},r_{1},s_{2},a_{2},r_{2},...)$ occurring in one episode. Thus, we
wish to maximize the expected return

$J(\varphi)=\mathbb{E}_{\tau\sim\pi_{\varphi}}\left[R(\tau)\right].$ (2)

We use the REINFORCE method (Williams,, 1992; Sutton and Barto,, 2018) to
update $\varphi$ via the gradient

$\nabla_{\varphi}J(\varphi)=\mathbb{E}_{\tau\sim\pi_{\varphi}}\left[\sum_{t=1}^{T}\nabla_{\varphi}\log\pi_{\varphi}(a_{t}|s_{t})R(\tau)\right]$
(3)

which can be empirically approximated by sampling $m$ trajectories following
policy $\pi_{\varphi}$ and averaging the result

$\nabla_{\varphi}J(\varphi)\approx\frac{1}{m}\sum_{k=1}^{m}\sum_{t=1}^{T}\nabla_{\varphi}\log\pi_{\varphi}(a_{t}|s_{t})\left(R(\tau)-b(s_{t})\right).$
(4)

$b(s_{t})$ is any baseline function which estimates the expected return in
state $s_{t}$, and does not depend on the actions. It is used to reduce the
variance of the gradient’s estimation. Finally, we replace the return
$R(\tau)$ in 4 with the discounted future return
$R_{t}=\sum_{t^{\prime}=t}^{T}\gamma^{t^{\prime}-t}r_{t^{\prime}}$, where
$\gamma\in[0,1]$ is a discount rate which controls how much influence on
future rewards we associate to each decision (Sutton and Barto,, 2018). We
discuss considerations for setting $\gamma$ in section 6. The gradient then
becomes

$\nabla_{\varphi}J(\varphi)\approx\frac{1}{m}\sum_{k=1}^{m}\sum_{t=1}^{T}\nabla_{\varphi}\log\pi_{\varphi}(a_{t}|s_{t})\left(R_{t}-b(s_{t})\right)$
(5)

and is used in line 19 of Algorithm 2 to update the agent’s weights.

##  5 A Hybrid Reinforcement Learning Approach

In this section we incorporate into our meta AL framework a CB interpretation
where the agent updates can be made simpler. We first discuss a pure CB
solution and then describe how to combine it with our previous RMAL-AL
algorithm.

###  5.1 A Pure Contextual Bandits Strategy

In contrast to a full reinforcement learning problem, in a CB setting, the
distribution of the next state does not depend on the action taken by the
agent. Thus, while taking an action in any state, the agent need not take
delayed rewards into consideration and does not need to consider a long
episode, rather a single state-action-reward $(s,a,r)$ triplet forms an
episode, and the agent just focuses on selecting an optimal action for the
given state.

One could argue that the assumption of independence between the actions and
the next state does not hold in our AL setting, since the agent’s actions
indirectly affect the model which is trained on the samples selected by the
agent, and the model affects the predictions for the next data sample which
constitute the next state. However, the state in our case is largely dependent
on the sample itself which is the input for the model’s predictions, and the
data sequence is i.i.d. So, the CB approach is a reasonable approximation
which focuses the agent on selecting samples which give a maximal immediate
accuracy improvement. Besides reducing complexity, another motivation for
pursuing this direction is the fact that by continuously updating the agent
online, we are less prone to overfitting on a particular sample batch.

We now give the system in Figure 1 a CB interpretation and describe how we can
train the AL agent. We focus on the differences with respect to the RMAL-AL
scheme. One crucial difference in the contxtual bandits approach is that there
is no distinction between a model training phase and an agent training phase,
rather the two occur concurrently. Whereas in the RMAL-AL scheme, the agent
was trained by running episodes on a batch of already labeled samples, here
there is no notion of episode, rather, the agent is continuously updated based
on the actual reward which comes from the previous agent’s decisions. Thus,
the reward signal used to train the agent does not include the counter factual
component, and is simply defined as
$r_{t}(s_{t},a_{t})=\left(\textrm{Acc}_{t}-\textrm{Acc}_{t-1}\right)/\textrm{Acc}_{t-1}$.

Next, instead of the objective in 2, we wish to maximize the expected
instantaneous reward $r_{t}(s_{t},a_{t})$, so our objective function becomes

$J(\varphi)=\mathbb{E}_{s,a\sim\pi_{\varphi}}\left[r(s,a)\right]$ (6)

and its policy gradient can be approximated by

$\nabla_{\varphi}J(\varphi_{t})\approx\frac{1}{m}\sum_{k=1}^{m}\nabla_{\varphi}\log\pi_{\varphi}(a_{k}|s_{k})\left(r_{k}-b(s_{k})\right)$
(7)

where we average the gradient over $m$ successive labeled samples which were
used to train the classifier. After each $m$ data samples are labeled and used
to train the classifier, we update the agent using the gradient approximation
in 7. The parameter $m$ is a hyper parameter.

Input : Initial dataset $\mathcal{D}_{0}$, Stream $\\{x_{1},...,x_{N}\\}$,
Budget $b$, Validation set $\mathcal{V}$, Batch sizes
$\left\\{T_{1},T_{2},...\right\\}$, Training episodes
$\left\\{E_{1},E_{2},...\right\\}$

Output : Classifier $f$

Initialization: $\pi_{\varphi}\leftarrow\textrm{Random}$,
$\mathcal{D}\leftarrow\mathcal{D}_{0}$, $\mathcal{N}\leftarrow\emptyset$,
$u\leftarrow 0$, $k\leftarrow 1$

1 $f^{\textrm{PXY}}\leftarrow
f_{0}\leftarrow\textnormal{{TrainModel}}(\mathcal{D})$

2 Pretrain the policy $\pi$

3 for $t=1,2,...,N$ do

4 $s_{t}\leftarrow f_{t-1}(x_{t})$

5 Draw action $a_{t}\sim\textrm{Bernoulli}(\pi_{\varphi}(a=1|s_{t}))$

6 if $(u/t<b)$ and $(a_{t}=1)$ then

7 $y_{t}\leftarrow\textrm{query label of }x_{t}$

8 $\mathcal{N}\leftarrow\mathcal{N}\cup\\{(x_{t},y_{t})\\}$

9 $f_{t}\leftarrow\textnormal{{TrainModel}}(\mathcal{D}\cup\mathcal{N})$

10 $u\leftarrow u+1$

11 if $E_{k}=0$ then $\triangleright$ Contextual bandits updates

12 $\textrm{Acc}_{t}\leftarrow$ accuracy of $f_{t}$ on $\mathcal{V}$

13
$r_{t}\leftarrow\left(\textrm{Acc}_{t}-\textrm{Acc}_{t-1}\right)/\textrm{Acc}_{t-1}$

14 if $|\mathcal{N}|>T_{k}$ then

$\varphi\leftarrow\varphi+\alpha\nabla_{\varphi}J(\varphi)$ ;

$\triangleright$ Compute $\nabla_{\varphi}J(\varphi)$ by eq. 7

15 $\mathcal{D}\leftarrow\mathcal{D}\cup\mathcal{N}$,
$\mathcal{N}\leftarrow\emptyset$, $k\leftarrow k+1$

16 end if

17

18 else $\triangleright$ Episodic RL updates

19 if $|\mathcal{N}|>T_{k}$ then

20
$\pi_{\varphi}\leftarrow\textnormal{{UpdateAgent}}\left(\pi_{\varphi},f^{\textrm{PXY}},\mathcal{D},\mathcal{N},E_{k}\right)$

21 $\mathcal{D}\leftarrow\mathcal{D}\cup\mathcal{N}$,
$\mathcal{N}\leftarrow\emptyset$, $k\leftarrow k+1$,
$f^{\textrm{PXY}}\leftarrow f_{t}$

22 end if

23

24 end if

25

26 else

27 $f_{t}\leftarrow f_{t-1}$

28 end if

29

30 end for

Algorithm 3 RMAL-HY Algorithm

###  5.2 Combining Contextual Bandits with RMAL-AL

We now incorporate the CB approach by allowing each batch of selected samples
to be used for updating the agent either as was done in RMAL-AL by the
procedure UpdateAgent, or as in the pure CB approach described above.

In this way, it is possible to balance between two complementary update
strategies. The former can better fit the agent to the current model state and
data statistics but requires more computation and is more prone to
overfitting, while the later is less computationally demanding and has better
chances of generalizing to the remainder of the data stream but its updates
are more noisy.

To better trade-of complexity and policy accuracy, we allow to choose the size
of each sample batch and the number of episodes in that batch. Specifically,
the sequences $\\{T_{i}\\}$ and $\\{E_{i}\\}$, $i=1,2,...$ define the size of
and the number of episodes in each batch, respectively, where a value of
$E_{i}>0$ means that at the end of batch $i$, the agent is updated as in the
RMAL-AL scheme, i.e., based on rewards associated with a proxy classifier
using $E_{i}$ episodes, while a value of ${E_{i}=0}$ means that the agent is
updated as in the CB scheme, i.e., based on rewards computed directly on the
classifier which is actively learned. Thus, earlier batches can be longer with
more training episodes, yielding a more accurate policy, while later batches
can be shorter with fewer episodes allowing quicker policy updates. This
hybrid scheme named RMAL-HY is illustrated in Algorithm 3.

##  6 Experimental Results

We evaluate our algorithms by comparing their performance against two
baselines: (1) random sampling (RND) and (2) variable uncertainty (VU)
proposed in (Zliobaitė et al.,, 2014) which is considered state-of-the-art for
stream-based AL. We experiment with one synthetic dataset (described below)
and 5 real datasets extracted from the UCI Machine Learning Repository and
kaggle whose details are listed in Figure 4.

We divide each dataset into four disjoint parts: an initial training set, a
streaming part, a validation set, and a test set. In each experiment we use
the dataset to train a logistic regressor. The agent’s architecture consists
of a fully connected network with 4 hidden layers of 256, 512, 256, and 256
neurons, a non-linearity of tanh in the 1st layer, ReLU non-linearities in the
remaining layers, and finally a Sigmoid layer at the output. For training the
agent we use SGD with weight decay of $5\text{\times}{10}^{-4}$. The discount
rate $\gamma$ is set close to zero222We noticed little performance improvement
by considering delayed rewards., $m$ is set to 1, the baseline $b(s_{t})$ is
an exponential moving average, and $b_{min}$ is set in the range 0.6-0.9
depending on the value of $b$. For each algorithm, we first train the model on
the initial training set, and then let the AL algorithm select samples for
training the classifier. The validation set is used by the RMAL algorithms to
produce a reward signal for updating the agent as described in sections 4 and
5.

In the case of the real datasets, after training the model on the initial
training set, and prior to processing the data stream, we perform pretraining
of the agent’s policy to mimic a policy based on uncertainty sampling
(Zliobaitė et al.,, 2014). The purpose of this procedure is to let the agent
start with a policy which is known to perform well, and try to improve on it,
rather than start learning an optimal policy from scratch. To this end, we
generate a dataset of random probability vectors and train the agent in a
supervised manner to produce a positive decision if the maximum element of the
input probabililty vector is below a threshold ${T_{unc}=0.6}$. In the
synthetic dataset, we do not perform such pretraining of the agent.

###  6.1 A Synthetic Dataset

In order to demonstrate the effectiveness of our approach in finding good
policies for training the classifier, we test our scheme on a synthetic
dataset which is specifically tailored such that the VU baseline attains a
significant gain w.r.t. the RND baseline. Our target in this test is to show
that our RMAL-HY scheme can learn a policy which attains at least as good a
performance as the VU baseline even without pretraining of the agent.

The synthetic dataset is a combination of an exponential distribution and a
uniform one and is defined as follows. For a dataset of size $N$ and a
constant $\alpha\in\left(0,\frac{1}{2}\right)$ we draw $\frac{(1-\alpha)N}{2}$
2-dimensional vectors from the exponential density
$p(x_{1},x_{2})=e^{-x_{1}-x_{2}},\,(x_{1},x_{2})\in[0,\infty]^{2}$, assign to
them a label $y=0$, and then draw another $\frac{(1-\alpha)N}{2}$
2-dimensional vectors from the density
$p(x_{1},x_{2})=e^{-x_{1}+x_{2}},\,(x_{1},x_{2})\in[0,\infty]\times[-\infty,0]$,
and assign to them a label $y=1$. Next we draw $\frac{\alpha N}{2}$ samples
uniformly distributed in the rectangle $[0,3]\times[0,3]$, label them with
$y=1$, and finally draw $\frac{\alpha N}{2}$ samples uniformly distributed in
the rectangle $[0,3]\times[-3,0]$, and label them with $y=0$. For this
dataset, it can be shown that an optimal decision boundary is the line
$x_{2}=0$, and its accuracy would be equal to $1-\alpha$. A sample of this
data set is plotted in Fig. 2(a) for $N=1200$ and $\alpha=0.2$.

We use RMAL-HY to train a logistic regressor on this dataset in the stream-
based AL setup described in section 3. Fig. 2(b) shows the learning curve
performance for a budget $b=0.1$. The large performance gap between the RND
and VU baselines is a result of this particular dataset. This is explained as
follows. When the RND baseline is employed, in any given set of training
samples, a fraction $\alpha$ of the samples causes the decision boundary to
fluctuate due to the noisy labels.

(a)

(b)

Figure 2: (a) A two-dimensional synthetic dataset for which the VU baseline
achieves superior performance to RND by a large margin. (b) Results of AL
algorithms used for training a logistic regressor on the dataset shown in (a).

However, in the VU baseline, by choosing samples closer to the decision
boundary, the relative proportion of the outlier samples is dramatically
reduced, leading to improved AL performance. As is seen in Figure 2(b), after
some learning period, the RMAL-HY algorithm is able to learn an improved
selection policy which is competitive with both baselines. Note that during
the selection of the 1st batch of 1000 samples, the agent is still in its
random initial state, so its performance coincides with the RND baseline.
However, at the end of this batch, the 1st training phase of the agent is
carried out, and the resulting policy learns to select more informative
samples for training the classifier. In subsequent batches, the cheap online
updates are enough to maintain the advantage of the policy even as more data
continues to arrive.

Figure 4 illustrates how the policies evolve during the AL process. Starting
with the blue curve, which corresponds to a randomly initialized agent, these
policies adapt to the underlying data distribution and progress towards more
and more peaky distributions, which give increasing preference to data samples
with prediction values closer to 1/2, i.e., samples with a higher level of
uncertainty. Each curve is an average of corresponding policy updates in 100
trials.

Figure 3: Evolution of policy updates exhibiting an increasing preference of
uncertain data samples.

Dataset |  Features |  Classes |  Samples
---|---|---|---
Magic |  10 |  2 |  19,020
Activity |  8 |  4 |  75,128
Covertype |  54 |  7 |  581,012333In our experiments we use a subset of size 80,000.
Phishing |  30 |  2 |  11,054
Web Club Acceptance |  23 |  2 |  20,000
Figure 4: A list of real datasets used for evaluating the RMAL-AL and RMAL-HY
algorithms.

###  6.2 Real Datasets

To further demonstrate the applicability of our scheme, we evaluate our
algorithms on 5 real datasets with various number of attributes and classes,
detailed in Figure 4. We perform our experiments for low budget values, e.g.,
5%, 10%, 20%, which are typical for an AL setting. For each budget value, we
consider several values of $\beta$, e.g., 25%, 50%, and 100%, in order to
evaluate the performance of the algorithm during various stages of the AL
process. We repeat each experiment for 20-100 trials, depending on the size of
the dataset.

Figure 5 shows learning curves for all considered AL algorithms when used to
train a logistic regressor on 4 real datasets with a budget $b=0.05$. As can
be seen, for very low budget values, our method outperforms the baselines
throughout the active learning.

Accuracy results of all experiments are summarized in Table 1, were we report
the average partial budget accuracy metric $\textrm{PBA}(\beta)$, defined in
section 3.1. The largest accuracy for each setting is marked with boldface
letters. In all experiments, RMAL-AL was run with 50 episodes in the 1st
batch, 20 episodes in the 2nd batch, and 10 episodes for the remaining batches
with a learning rate of $0.1$. In RMAL-HY there were typically 100 episodes in
the 1st batch. Results show a consistent advantage of the RMAL algorithms over
both baselines across all considered datasets, across various budget values,
and for different stages in the AL process.

(a) Magic

(b) Covertype

(c) Phishing

(d) Web Club

Figure 5: Results of AL algorithms used for training a logistic regressor on several real datasets. Dataset |  $b[\%]$ | RMAL-HY  | RMAL-AL  | VU  | RND
---|---|---|---|---|---
|  $\beta[\%]$ |  $\beta[\%]$ |  $\beta[\%]$ | $\beta[\%]$
|  25 |  50 |  100 |  25 |  50 |  100 |  25 |  50 |  100 |  25 |  50 |  100
Magic |  5 |  78.27 |  78.86 |  79.37 |  78.27 |  78.78 |  79.35 |  78.05 |  78.54 |  78.91 |  77.76 |  78.18 |  78.38
10 |  78.78 |  79.27 |  79.60 |  78.76 |  79.27 |  79.64 |  78.58 |  78.93 |  79.37 |  78.15 |  78.34 |  78.55
20 |  79.13 |  79.56 |  79.71 |  79.11 |  79.61 |  79.72 |  78.96 |  79.37 |  79.47 |  78.38 |  78.46 |  78.69
Activity |  5 |  89.68 |  90.18 |  90.88 |  89.67 |  90.19 |  90.95 |  88.94 |  89.95 |  90.61 |  83.73 |  87.81 |  89.71
10 |  90.13 |  90.85 |  91.54 |  90.10 |  90.92 |  91.96 |  89.83 |  90.49 |  91.52 |  87.77 |  89.61 |  90.70
20 |  90.73 |  91.13 |  91.93 |  90.74 |  91.82 |  91.88 |  90.56 |  91.54 |  91.81 |  89.64 |  90.65 |  91.76
Covertype |  5 |  68.81 |  69.25 |  69.56 |  68.81 |  69.23 |  69.61 |  68.36 |  69.08 |  69.52 |  67.83 |  68.60 |  69.18
10 |  69.31 |  69.75 |  70.12 |  69.31 |  69.74 |  70.07 |  69.06 |  69.57 |  69.94 |  68.69 |  69.41 |  69.86
20 |  69.68 |  70.00 |  70.77 |  69.65 |  70.15 |  70.97 |  69.63 |  70.17 |  70.88 |  69.21 |  69.93 |  70.75
Phishing |  5 |  92.06 |  92.55 |  92.92 |  92.06 |  92.56 |  92.95 |  91.38 |  91.84 |  92.27 |  90.92 |  91.37 |  91.90
10 |  92.33 |  92.70 |  92.82 |  92.33 |  92.78 |  92.85 |  91.88 |  92.41 |  92.68 |  91.34 |  91.81 |  92.28
20 |  92.41 |  92.70 |  92.88 |  92.39 |  92.72 |  92.90 |  92.44 |  92.65 |  92.75 |  91.88 |  92.33 |  92.55
Web Club Acceptance |  5 |  77.23 |  77.78 |  77.98 |  77.23 |  77.76 |  78.01 |  76.91 |  77.50 |  77.85 |  77.04 |  77.55 |  77.74
10 |  77.55 |  77.85 |  78.05 |  77.55 |  77.82 |  78.05 |  77.49 |  77.83 |  78.04 |  77.43 |  77.83 |  78.00
20 |  77.91 |  77.97 |  78.04 |  77.89 |  78.00 |  78.06 |  77.80 |  78.00 |  78.09 |  77.70 |  77.88 |  77.98
Table 1: Performance comparison on real datasets.

##  7 Conclusions and Future Work

In this work we present novel data-driven AL methods for selecting samples for
training a classifier in a stream-based setting. The proposed methods, which
combine several notions of reinforcement learning, learn to optimally select
samples based on their model’s predictions without any apriori rule-based
functional restrictions or assumptions. Furthermore, these methods are
applicable to a wide range of classification problems and are applied online
without any need for offline training on any related task. We demonstrate on
both a synthetic dataset and real datasets the efficacy of these methods with
respect to state-of-the-art.

Although not pursued in this work, streaming data may sometimes exhibit
changes in the underlying data distribution over time also known as concept
drift. One future research direction would be to extend the meta active
learner to adapt to these changes during the active learning process. For
example, one could adapt the reward mechanism to account for changing
statistics and thereby induce an appropriate change in the active learner.

Another direction would be to alleviate the need to retrain the classifier
after every selected sample. Allowing to train the model in batches may open
the door for higher dimensional and large scale datasets and tasks, e.g,.
image datasets which are typically trained in batches.

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