# On the nth quantum derivative

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```					                        On the nth quantum derivative

ıos-Collantes-de-Ter´n
J. Marshall Ash, Stefan Catoiu, and Ricardo R´                   a

Abstract. The nth quantum derivative Dn f (x) of the real-valued function
f is deﬁned for each real non-zero x as
n
n
(−1)k   k q
q (k−1)k/2 f   q n−k x
k=0
lim                                              ,
q→1           q (n−1)n/2 (q − 1)n xn
where n q is the q-binomial coeﬃcient. If the nth Peano derivative exists at x,
k
which is to say that if f can be approximated by an nth degree polynomial at
the point x, then it is not hard to see that Dn f (x) must also exist at that point.
Consideration of the function |1 − x| at x = 1 shows that the second quantum
derivative is more general than the second Peano derivative. However, we can
show that the existence of the nth quantum derivative at each point of a set
necessarily implies the existence of the nth Peano derivative at almost every
point of that set.

1. Introduction
Let f be a real valued function of a real variable. The ﬁrst quantum diﬀerence
is deﬁned as
∆1 (q, x)   f (qx) − f (x)
D1 (q, x) =                =                ,
qx − x         qx − x
and for n = 2, 3, ..., the higher quantum diﬀerences are deﬁned inductively by
Dn−1 (q, qx) − Dn−1 (q, x)
Dn (q, x) =                                 .
qx − x
For n = 1, 2, ..., the nth diﬀerence quotient can be written
∆n (q, x)
(1.1)                         Dn (q, x) =          (n−1)n/2 (q − 1)n
q                      xn

1991 Mathematics Subject Classiﬁcation. Primary 26A24; Secondary 26A27.
Key words and phrases. Quantum derivative, q-derivative.
The ﬁrst author’s research was partially supported by NSF grant DMS 9707011 and a grant
from the Faculty and Development Program of the College of Liberal Arts and Sciences, DePaul
University.
The third author’s research was partially supported by DGICYT Grant PB96-1327,the Junta
ıa                          o     a
de Andaluc´ and the Patronato Fundaci´n C´mara Universidad de Sevilla.
1
2     J. MARSHALL ASH, STEFAN CATOIU, AND RICARDO R´                   ´
IOS-COLLANTES-DE-TERAN

with
n
k     n (k−1)k/2
(1.2)                  ∆n (q, x) =                  (−1)          q      f q n−k x ,
k q
k=0
n
−1
where the q-binomial coeﬃcients are deﬁned by setting [n]q = qq−1 if n = 1, 2, ...,

 1                  if n = 0
[n]q ! =                                       ,
[1]q [2]q ... [n]q if n = 1, 2, ...


and
n                    [n]q !
=                         for n ≥ k ≥ 0.
k   q       [k]q ! [n − k]q !
For each positive integer n, and each non-zero real number x,the nth quantum
derivative of f is deﬁned to be
Dn f (x) = lim Dn (q, x) .
q→1

From equation (1.2) it is clear that the nth quantum derivative is an analogue of
the nth forward Riemann derivative
n
k n
(−1)      k     f (x + (n − k) h)
k=0
lim                        .
h→0        hn
There is also a symmetric Riemann derivative based on n + 1 uniformly spaced
points distributed symmetrically about x. The analogue of this is the nth symmetric
quantum derivative,
n
k n    (k−1)k/2
(−1)     k qq          f     q n/2−k x
S                     k=0
Dn f (x) = lim                                           n               .
q→1                    q (n−1)n/2 (q − 1) xn
The simple substitution qx = x + h shows that D1 f (x) is identical to the ordinary
ﬁrst derivative. The second quantum derivative D2 f (x) is given by
f (q 2 x)−f (qx)       f (qx)−f (x)
q 2 x−qx       −       qx−x                    f q 2 x − (1 + q) f (qx) + qf (x)
D2 f (x) = lim                                            = lim        2         .
q→1                qx − x                           q (q − 1) x2
q→1

Perhaps the most important notion of higher order diﬀerentiation is that of
the nth Peano derivative. The function f (x) has an nth Peano derivative if f can
be approximated to nth order by an nth degree polynomial at the point x. More
explicitly, there must exist constants f0 (x) , f1 (x) , ..., fn (x) such that
hn
f (x + h) = f0 (x) + f1 (x) h + ... + fn (x)       + o (hn )
n!
as h tends to 0. That the existence of the ordinary nth derivative at a point x implies
the existence of fn (x) is the content of a strong version of Taylor’s Theorem that
was proved by Peano in 1884. (See page 246 of [AGV] for details.) If n = 0, f0 (x)
exists and coincides with f (x) exactly when f is continuous at x, and if n = 1,
the deﬁnitions of f1 (x) and the ordinary ﬁrst derivative are the same. But if
n ≥ 2, the existence of the nth Peano derivative at each point of a set of positive
Lebesgue measure does not imply the existence of the ordinary nth derivative at
ON THE NTH QUANTUM DERIVATIVE                              3

almost every point of that set.[O] Thus if one accepts the centrality of the notion
of nth degree polynomial approximation(and there are many physical reasons for
doing so), when n ≥ 2, it is the nth Peano derivative and not the ordinary nth
derivative that is the benchmark for testing generalized diﬀerentiation schemes
against. For example, each of the nth forward Riemann derivative and the nth
symmetric Riemann derivative is equivalent almost everywhere to the nth Peano
derivative.[MZ], [A]
If the nth Peano derivative fn exists at x, then it is easy to see that Dn f (x)
must also exist and that then Dn f (x) = fn (x). Consideration of the function
|x − 1| at x = 1 shows that the second quantum derivative is more general than
the second Peano derivative. However, the nth quantum derivative is essentially
equivalent to the nth Peano derivative. Explicitly, we show that the existence of
the nth quantum derivative at each point of a Lebesgue measurable set necessarily
implies the existence of the nth Peano derivative at almost every point of that set.
It will be clear from the proof (see especially the Sliding Lemma below) that the
S
same result holds true for the nth symmetric quantum derivative Dn f (x).
The nth quantum derivative has proven to be useful in the study of hypergeo-
metric series[GR]. One defect that it has is that it cannot be deﬁned at the point
x = 0. The nth Riemann derivative and its generalizations discussed in [A] involve
sums of the form h−n an f (x + bn h) , where the coeﬃcients {an } are constants.
The setting here is more complicated and more delicate techniques are needed since
now the coeﬃcients are no longer constant. For example, if we let h = qx − x, the
deﬁnition of D2 f (x) becomes

x                    h2       2x + h
lim h−2       f    x + 2h +          −          f (x + h) + f (x) .
h→0       x+h                   x         x+h

One technical point that arises in the proofs of two Lemmas below is the ques-
tion of the measurability of certain sets. Much light is shed on this question by
H. Fejzic and C. E. Weil who give an example that depends on the existence of a
measurable set S for which S + S is not measurable.[FW] K. Ciesielski has shown
us how such a set S can be constructed.[C] See his theorem and his proof in the
appendix at the end of this paper.

2. Results
Proposition 1. If the nth Peano derivative fn (x) exists and x = 0, then the
nth quantum derivative Dn f (x) also exists and Dn f (x) = fn (x) .

Proof. The existence of the nth Peano derivative at x means that there are
constants f0 (x) = f (x) , f1 (x) , f2 (x) , ..., fn (x) such that

n
fj (x) j
f (x + h) =               h + o (hn ) .
j=0
j!
4   J. MARSHALL ASH, STEFAN CATOIU, AND RICARDO R´                   ´
IOS-COLLANTES-DE-TERAN

We have

∆n (q, x)
Dn (q, x) =       (n−1)n/2 (q − 1)n
q                                 xn
n
1                                 k       n (k)
=                                         (−1)             q 2 f x + q n−k − 1 x
q (n) (q − 1)n xn
2                                               k q
k=0
n                           n
1           fj (x) j         k n      k                                             j
=      n                        x      (−1)      q (2) q n−k − 1                                      + o (1) ,
q ( ) (q − 1)n xn
2                    j!               k q
j=0          k=0

so it suﬃces to prove that

n
k       n (k−1)k/2 n−k                          j
(2.1)                   (−1)               q      q   −1                           = 0, j = 0, 1, ..., n − 1
k q
k=0

and

n
1                                    k       n (k−1)k/2 n−k                 n
n            (−1)                q      q   −1                  = [n]q !,
q (n−1)n/2 (q − 1)            k=0
k q

where [n]q ! satisﬁes

[n]q ! = (1 + q) 1 + q + q 2 · · · (1 + q + ... + q n−1 ), n = 1, 2, ...,

and

lim [n]q ! = n!.
q→1

Let

k
(−1) q (k−1)k/2 n
Bk =          (n−1)n/2 (q − 1)n k
and ak = q n−k − 1, k = 0, 1, ..., n.
q                                          q

It would suﬃce to know that Xk = Bk , where the n dimensional vector X is the
solution to the Vandermonde system of equations


n                        0                  j = 0, 1, ..., n − 1
aj Xk =
k                                                         .
[n]q !           j=n

k=0
ON THE NTH QUANTUM DERIVATIVE                                         5

Indeed, by Cramer’s rule, the unique solution to this system is given by
[n]q !                   [n]q !                      [n]q !
Xk =                 =         n−k − 1) − (q n−j − 1))
=       n−k − q n−j )
(ak − aj )       ((q                              (q
j=k                     j=k                                                  j=k
[n]q !
=
(q n−k − q n−j ) ·            (q n−k − q n−j )
j<k                         j>k
[n]q !
=                k                                             n−k
k
(q n−k )           (1 − q i ) q 1+2+...+(n−k−1)                  (q i − 1)
i=1                                           i=1
[n]q !
=                                       k                 n−k
k
q (n−k)(n+k−1)/2 (−1)                     (q i − 1)         (q i − 1)
i=1                i=1
k                                                             k
(−1) q k(k−1)/2                       [n]q !                   (−1) q k(k−1)/2 n
=                     n                                        =                     n    .
q n(n−1)/2 (q − 1)         k
q i −1
n−k
q i −1       q n(n−1)/2 (q − 1) k q
q−1            q−1
i=1               i=1

Corollary 1. If the ordinary nth derivative f (n) (x) exists and x = 0, then
the nth quantum derivative Dn f (x) also exists and Dn f (x) = f (n) (x) .
Proof. By Peano’s strong version of Taylor’s Theorem, the existence of f (n) (x)
is suﬃcient for the existence of fn (x) as well as for the equation fn (x) = f (n) (x) .

Corollary 2. For n ≥ 2,the nth quantum derivative may exist at every point
of a set of positive measure without the nth derivative existing at any point of that
set.
Proof. For n ≥ 2,there is a function having an nth Peano derivative and not
having the ordinary nth derivative at each point of a set of positive measure.[O],
Theorem 5
Proposition 2. There is a measurable function f such that D2 f (1) exists, but
f is discontinuous at x = 1.
Remark 1. Referring back to the discussion of the Peano derivative in the
introduction, we see that this means that f not only has no second Peano derivative
at x = 1, but that f has neither a ﬁrst, nor even a zeroth Peano derivative at x = 1.
1
Proof. For x ∈ [2, 4) deﬁne f by f (x) = 4−x . Since every point in the interval
n
(1, 2) is uniquely of the form q 1/2 for some q ∈ [2, 4) and some positive integer
n,we may extend f to the interval (1, 2) by setting
n
q 1/2 − 1
n
(2.2)                                f q 1/2         f (q)
=
q−1
for each q ∈ [2, 4) and each positive integer n. Finally deﬁne f to be 0 on (−∞, 1] ∪
(4, ∞) . Fix q and note that equation (2.2) may be rewritten as
n
f q 1/2 − 0   f (q) − 0
(2.3)                                            =           ,
q 1/2n − 1     q−1
6   J. MARSHALL ASH, STEFAN CATOIU, AND RICARDO R´                   ´
IOS-COLLANTES-DE-TERAN

n               n         n
which shows that the triangles (1, 0)(q, 0)(q, f (q)) and (1, 0)(q 1/2 , 0)(q 1/2 , f (q 1/2 ))
n        n
are similar, so that all of the points (q 1/2 , f (q 1/2 )) : n = 0, 1, ... lie on a line
passing through (1, 0) = (1, f (1)) . Since limx→4− f (x) = ∞, lim supx→1+ f (x) =
∞ and f is not continuous at x = 1. However, if p ≤ 1, f (1) = f (p) = f p2 = 0,
m
while if p ∈ (1, 2) , then p = q 1/2 for some q ∈ [2, 4) and some positive integer m
and
f p2 1 − (1 + p) f (p1) + pf (1)
         m−1

 f q 1/2      −0      q 1/2m
+1 f q  1/2m
− 0  1/2m−1
=      1/2m−1 − 1
−           m         m
q      −1 ,
 q                       q 1/2 + 1 q 1/2 − 1 

where the quantity in curly brackets is 0 by virtue of equations 2.3. Thus
f p2 1 − (1 + p) f (q1) + pf (1)                                 0
D2 f (1) = lim                            2                      = lim                  2    = 0.
p→1               p (p − 1)        12                    p→1   p (p − 1)

We will also need another generalized q-derivative. We will call this one
˜                                                                ˜
Dn f (x) . It will be given by means of an nth diﬀerence ∆n (q, x; f ) which eval-
2    4       2n−1
uates the function f at the n + 1 points x, qx, q x, q x, ..., q      x.Let
˜ 1 (q, x; f ) = f (qx) − f (x) ,
∆
and for n ≥ 2 let
(2.4)          ˜              ˜                             ˜
∆n (q, x; f ) = ∆n−1 (q 2 , x; f ) − λn−1 (q) ∆n−1 (q, x; f ),
where λ1 (q) = q + 1, and, when n ≥ 2,
n−1              n−1                  n−1                   n−1            n−2
(2.5)      λn (q) = q 2         +1    q2         +q        q2         + q 2 ... q 2           + q2          .
We also deﬁne diﬀerence quotients for each n by setting
˜               f (qx) − f (x)
D1 (q, x; f ) =
qx − x
and for n ≥ 2,
˜                   ˜
Dn−1 q 2 , x; f − Dn−1 (q, x; f )
(2.6)              ˜
Dn (q, x; f ) = n                                   .
q 2n−1 x − qx

Finally for x = 0 and each n ≥ 1, we deﬁne the generalized derivative
˜             ˜
Dn f (x) = lim Dn (q, x; f ) .
q→1

Lemma 1. If the nth Peano derivative fn (x) exists and x = 0, then the nth
˜                        ˜
generalized derivative Dn f (x) also exists and Dn f (x) = fn (x) .
Proof. First we establish the simple connection
˜
n!∆n (q, x; f )
˜
(2.7) Dn (q, x; f ) = 2n−1                                         , n = 2, 3, ....
n−1
q     − 1 q2      − q ... q 2n−1 − q 2n−2 xn
To do this, we simplify notation by letting y n = f (q n x) , n = 0, 1, ..., Pn (q, y) =
n
˜                             ˜
∆n (q, x; f ) and Qn (q, y) = Dn (q, x; f ) x . In this notation, everything that we will
n!
need to prove for this lemma will become an identity between polynomials in y
ON THE NTH QUANTUM DERIVATIVE                                                                7

whose coeﬃcients are rational functions of q. For example, equations (2.4), (2.6),
and (2.7) become
(2.8)                  Pn (q, y) = Pn−1 q 2 , y 2 − λn−1 (q) Pn−1 (q, y) ,

Qn−1 q 2 , y 2 − Qn−1 (q, y)
(2.9)                      Qn (q, y) =                                   ,
q 2n−1 − q
and
Pn (q, y)
(2.10)      Qn (q, y) =                                                          , n = 2, 3, ....
q 2n−1 − 1          q 2n−1 − q ... q 2n−1 − q 2n−2
So to prove equations (2.7) it is enough to establish equations (2.10). If n = 2,
2
−(q+1)y+q
both sides are equal to y 2 −1)(q2 −q) . Assuming the equation for n − 1, let µn (q) be
(q
the denominator of Qn (q, y) in equation (2.10) and calculate
Pn (q, y) = Pn−1 q 2 , y 2 − λn−1 (q) Pn−1 (q, y)
= µn−1 q 2 Qn−1 q 2 , y 2 − λn−1 (q) µn−1 (q) Qn−1 (q, y)
= µn−1 q 2     Qn−1 q 2 , y 2 − Qn−1 (q, y)
n−1
= µn−1 q 2      q2         − q Qn (q, y) = µn (q) Qn (q, y) ,
since
λn−1 (q) µn−1 (q) =
n−2               n−2                    n−2          n−4            n−2          n−3
= q2          +1     q2         + q ... q 2            − q2             q2         − q2
n−1               n−1                        n−1          n−2
= q2          −1     q2         − q 2 ... q 2              − q2         = µn−1 q 2
and
n−1                 n−1                  n−1                      n−1          n−2             n−1
µn−1 q 2    q2         − q = q2            −1        q2         − q 2 ... q 2            − q2              q2         −q
= µn (q) .
k                       k
Suppose that x = 0 and fn (x) exists. Since q j − 1                                = O (q − 1)              , we may
n                  k                k                      n
write f q j x = f x + q j − 1 x = k=0 fk (x) x q j − 1 + o ((q − 1) ) . Thus
k!
˜
Dn (q, x; f ) is equal to
                                                 
n         n
1                                            xk 2j−1     k
n
A0 (q) f (x) +     Aj (q)   fk (x)    q    − 1  + o (1) ,
(q − 1) xn                     j=1
k!
k=0

where the Aj are bounded rational functions of q when q is near 1. Interchanging
the order of summation shows that the limit as q → 1 of this expression will be
˜                                             ˜
fn (x) if Dn q, x; xj = 0 for j = 0, 1, ..., n − 1, and Dn (q, x; xn ) = n!. Converting
to polynomial language, it suﬃces to show that
(2.11)                        Qn q, q j = 0 for j = 0, 1, ..., n − 1
and
(2.12)                                      Qn (q, q n ) = 1.
8   J. MARSHALL ASH, STEFAN CATOIU, AND RICARDO R´                   ´
IOS-COLLANTES-DE-TERAN

We begin by proving relation (2.12). For k ≥ 1, let Q−1,k (q) = 0, Q0,k (q) = 1, and
R0,k = 1. For m ≥ 0, let Rm,1 = 1.For m, k ≥ 1, let

m
k1 −1              kl −1
Qm,k (q) = 1 +                                q2           · · · q2           ,
l=1 k≥k1 ≥...≥kl ≥1

and for m ≥ 1, k ≥ 2, let

m
k1 −1              kl −1
Rm,k (q) = 1 +                                q2           · · · q2           .
l=1 k≥k1 ≥...≥kl ≥2

So we have the following properties

k
s−1
(a) Qm,k (q) = 1 +              q2         Qm−1,s (q)         for all m ≥ 0, k ≥ 1
s=1

(b) Qm,k (q) − Rm,k (q) = q · Qm−1,k (q)                      for all m ≥ 0, k ≥ 1

k
s−1
(c) Rm,k (q) = 1 +              q2         Rm−1,s (q)         for all m ≥ 1, k ≥ 2
s=2

(d) Qm,k (q 2 ) = Rm,k+1 (q)                                 for all m ≥ 0, k ≥ 1.

We now establish that for all n and for all m = 0, 1, . . . n − 1

(2.13)                             Qn−m (q, q n ) = Qm,n−m (q).

We ﬁx n and do induction backwards on the variable m. For m = n − 1,

n−1              n−1
k1 −1              kl −1
Q1 (q, q n ) = 1 +          ql = 1 +                           q2           · · · q2               = Qn−1,1 (q).
l=1              l=1 1≥k1 ≥...≥kl ≥1

We suppose that equation (2.13) holds for 0 < m ≤ n − 1. We have to prove

Qn−m+1 (q, q n ) = Qm−1,n−m+1 (q).

Applying equation (2.9) and the induction hypothesis, we have to prove

n−m
(q 2         − q)Qm−1,n−m+1 (q) = Qm,n−m (q 2 ) − Qm,n−m (q).
ON THE NTH QUANTUM DERIVATIVE                                                  9

Using the properties (a) – (d), we compute
Qm,n−m (q 2 ) − Qm,n−m (q) = Rm,n−m+1 (q) − Qm,n−m (q)

n−m+1                                n−m
2s−1                             s−1
=              q          Rm−1,s (q) −         q2         Qm−1,s (q)
s=2                                 s=1

n−m
n−m                                          s−1
= q2         Rm−1,n−m+1 (q) −                  q2         (Qm−1,s (q) − Rm−1,s (q)) − qQm−1,1 (q)
s=2

n−m+1
n−m                                                     s−1
= q2         Qm−1,n−m+1 (q) − q                           q2         Qm−2,s (q) + Qm−1,1 (q)
s=2

n−m
= (q 2         − q)Qm−1,n−m+1 (q).
In particular, equation (2.12) holds.
We can now prove relation (2.11). If n = 1, Q1 q, q 0 = 1−1 = 0. We assume
q−1
the property true for n−1. Let j < n. If j ≤ n−2, then by the induction hypothesis
Qn−1 (q, q j ) = 0. This means that y−q j divides Qn−1 (q, y) . But then y 2 −q 2j divides
Qn−1 (q 2 , y 2 ) so that Qn−1 (q 2 , q 2j ) = 0. But then
Qn−1 (q 2 , q 2j ) − Qn−1 q, q j    0−0
Qn (q, q j ) =                                          = 2n−1    = 0.
q 2n−1 − q            q     −q
In the remaining case, j = n − 1. Then by equation (2.12), Qn−1 q, q n−1 = 1. As
above, this implies that Qn−1 (q 2 , q 2n−2 ) = 1 also. Again we have
Qn−1 (q 2 , q 2n−2 ) − Qn−1 q, q n−1    1−1
Qn (q, q n−1 ) =                             2n−1 − q
= 2n−1    = 0.
q                     q     −q

˜
Lemma 2. If x = 0 and fn−1 (x) and Dn f (x) exist, then fn (x) also exists and
˜ n f (x) = fn (x).
D
˜
Proof. Fix x = 0. We may assume that f (x) , f (x) , ..., fn−1 (x) and Dn f (x)
are all 0. Let ε > 0 be given. Then if q ∈ (1/2, 2) is suﬃciently close to 1, abbrevi-
˜               ˜
ating ∆n (q, x; f ) to ∆n (q) , we have

˜                   ˜                   n   n
∆n−1 q 2 − λn−1 (q) ∆n−1 (q) < ε |q − 1| |x| .
i                                        i−1              j
Iterate this relation by replacing q by q 1/2 and then multiplying by                             j=0   λn−1 (q 1/2 ),
n
i = 1, 2, ..., M. Letting d = ε |x| , we get

˜                   ˜                   n
(2.14)                ∆n−1 q 2 − λn−1 (q) ∆n−1 (q) < d |q − 1|
1   n
˜                  1
˜       1               1 − q2                                         n
|λn−1 (q)| ∆n−1 (q) − λn−1 (q 2 )∆n−1 (q 2 ) < dλn−1 (q)                                         |1 − q|
1−q
10 J. MARSHALL ASH, STEFAN CATOIU, AND RICARDO R´                   ´
IOS-COLLANTES-DE-TERAN

1           1          1           1
˜                     ˜
λn−1 (q) λn−1 (q 2 ) ∆n−1 (q 2 ) − λ(q 22 )∆n−1 (q 22 ) <
1       n
1 − q 221                                            n
dλn−1 (q) λn−1 (q )         2                               |1 − q|
1−q
...
1                   1               1                                                1                          1
˜                          ˜
λn−1 (q) λn−1 (q 2 ) · · · λn−1 (q 2M −1 ) ∆n−1 (q 2M −1 ) − λ(q 22M )∆n−1 (q 22M ) <

1        n
1                    1            1 − q 22M                              n
d λn−1 (q) λn−1 (q ) · · · λn−1 (q  2                  2M −1        )                                |1 − q| .
1−q
Adding all these inequalities, it follows from the triangle inequality that
1                  1                                                    1
(2.15)             ˜                                                 ˜
∆n−1 q 2 − λn−1 (q) λn−1 (q 2 ) · · · λn−1 (q 2M )∆n−1 (q 2M )
M                                                                                   1   n
n                                        1                       1         1 − q 2i
< d |1 − q|                        λn−1 (q) λn−1 (q ) · · · λn−1 (q
2                     2i−1      )                          .
i=0
1−q
The ration test shows that the last sum is uniformly bounded,
1        n
1
1                               1−q 2i+1
λn−1 (q) λn−1 (q 2 ) · · · λn−1 (q 2i )                       1−q
lim sup sup                                                                                              1    n
i→∞ q∈[ 1 ,2]                                      1
1                                                             1−q 2i
2     λn−1 (q) λn−1 (q 2 ) · · · λn−1 (q 2i−1 )                                       1−q

1
λn−1 (q 2i )             2n−1  1
= lim sup sup                                        n     =        = < 1.
1
i→∞ q∈[ 1 ,2] 1 + q 2i+1                                   2n   2
2

So we may rewrite inequality (2.15) as
                                                
M
j                   1
˜                                                   ˜                    n       n
(2.16)         ∆n−1 q 2 −                          λn−1 (q 1/2 ) ∆n−1 (q 2M ) ≤ εC |x| |1 − q| .
j=0
+1
1−q 2
Now applying the identity 1 + q 2 =                                          to deﬁnition (2.5) and then using the
1−q 2
M    aj            a0
identity     j=0 aj+1      =   aM +1 ,          we calculate
n−1−j     n−3                         n−2
−2i )2−j+1
M                           M                               1 − q2                  i=0          1 − q (2
1/2j                     (2n−2 −1)2−j
λn−1 (q          )=            q                                                  n−3
j=0                         j=0                                 1 − q 2n−2−j            i=0      1 − q (2n−2 −2i )2−j
n−1                              n−1
n−3                           −2i+1
1 − q2                i=0       1 − q2
=q     (2n−1 −2)(1−2−M −1 )                                                                                               .
n−3
1 − q 2n−2−M                i=0       1 − q (2n−2 −2i )2−M
From this it follows that for ﬁxed q, as M → ∞ there are constants c = c (n, q) and
d = d (n, q) so that

M
c                                                j                  d
(2.17)                                      n−1     >         λn−1 (q 1/2 ) >                                n−1
1                                                                   1
1−q       2M                      j=0                            1 − q 2M
ON THE NTH QUANTUM DERIVATIVE                                        11

Since fn−1 (x) = 0, from Lemma 1 we have
              
M
j                  1
lim                              ˜
λn−1 (q 1/2 ) ∆n−1 (q 2M ) = fn−1 (x) = 0,
M →∞
j=0

˜                       n
so that letting M → ∞ in inequality (2.16) shows that ∆n−1 (q 2 ) = o (|q − 1| ) ,
or, equivalently,
˜                           n
∆n−1 (q, x; f ) = o (|q − 1| ) .
From this we similarly deduce
˜                           n    ˜                           n
∆n−2 (q, x; f ) = o (|q − 1| ) , ∆n−3 (q, x; f ) = o (|q − 1| ) , ...
˜                         n                             n
and ﬁnally ∆1 (q, x; f ) = o (|q − 1| ) , i. e. f (qx) = o (|q − 1| ) . This proves the
lemma.
Lemma 3. For each n and q, there are constants C0 , C1 , ..., C2n−1 −n such that
2n−1 −n
(2.18)                             ˜
Dn (q, x; f ) =                    Ci Dn q, q i x .
i=0
For ﬁxed n, these constants are uniformly bounded in q near q = 1.
Proof. Fix n. We will show that there are polynomials in q, C0 , C1 , ..., C2n−1 −n
such that
2n−1 −n
(2.19)                             ˜
∆n (q, x; f ) =                    Ci ∆n q, q i x .
i=0

This is suﬃcient, because it will then follow from deﬁnitions (1.1) and (2.7) and
the fact that
n−1          j−1          j−1        n−1
−2j−1
q2         − q2         = q2         q2                  −1        c (j) (q − 1) as q → 1
a(q)
that there is a rational function               b(q)       with a (1) = 0 and b (1) = 0 such that identity
a(q)
(2.18) holds with Cj (q) =            (q) for all j. The polynomials a (q) and Cj (q) are
b(q) Cj
uniformly bounded above in q near q = 1 and since b (1) = 0 the polynomial b (q)
is uniformly bounded below in q near q = 1.
To prove equation (2.19), write ∆n (q, x) = f (q n x) + an−1 (q) f q n−1 x +
... + a0 (q) f (x) as p (q, y) = y n + an−1 (q) y n−1 + ... + a0 (q), using once more the
polynomial notation used in the proof of Lemma 1 above. Then for each nonnegative
˜
integer i, we have the identiﬁcation of ∆n q, q i x with y i p(q, y). Recalling that ∆n
was identiﬁed with a polynomial Pn of degree 2n−1 in y, we see that the lemma
n−1
amounts to proving that Pn (y) = (C0 + C1 y + ... + C2n−1 −n y 2 −n )p (y) . By the
division algorithm, there are polynomials Q and R with deg (Q) = 2n−1 − n and
deg (R) < deg (p) = n so that
R = Pn − Qp.
Note that the coeﬃcients of Q are polynomials in q since the deﬁnitions (1.2), (2.4)
and (2.5) show the coeﬃcients of Pn and p to be polynomials in q and p is monic.
But Pn (q j ) = 0, for j = 0, 1, ..., n − 1 by equations (2.11) and (2.10), and that
j
p(q ) = 0, for j = 0, 1, ..., n − 1 follows easily from equations (2.1). Since R is
12 J. MARSHALL ASH, STEFAN CATOIU, AND RICARDO R´                   ´
IOS-COLLANTES-DE-TERAN

a polynomial of degree smaller than n with n distinct zeros, R must be 0 which
establishes equation (2.19). Lemma 3 is proved.

Lemma 4 (The Sliding Lemma). Let a0 , . . . , aN be distinct real numbers and
A0 (q), . . . , AN (q) bounded measurable functions such that a0 = 0 and |A0 (q)| >
C > 0 when q is near 1. Let b be a non-negative real number. If
N
(2.20)                             Ai (q)f (q ai x) = O((q − 1)b )
i=0

for all x ∈ E, then for any real number a,
N
(2.21)                            Ai (q)f (q ai −a x) = O((q − 1)b )
i=0

for almost every x ∈ E. If “O” is replaced by “o” in the hypothesis, then the
conclusion also holds with “o” in place of “O”.

Proof. We begin with some measure theoretic preliminaries. Let a = 0 and
I = (p, q) with 0 < p < q < +∞. The Mean Value Theorem guarantees that there
exits a constant C(I) such that for all x, y ∈ I,
|a−1|
q
(2.22)           C(I)|x − y| ≤ |xa − y a | ≤                          C(I)|x − y|,
p
where C(I) = inf z∈I |a|z a−1 . Let E ⊂ I and denote outer Lebesgue measure by
m∗ . We have
+∞                          +∞
m∗ (E a ) = inf            pn − q n ; E a ⊂             (pn , qn ) ⊂ I a
n=1                         n=1
+∞                         +∞
= inf             pn − q n ; E ⊂          (pn , qn )1/a ⊂ I
n=1                        n=1
+∞                          +∞
= inf                    a
|pa − qn | ; E ⊂
n                          (pn , qn ) ⊂ I     ,
n=1                         n=1

so by inequalities (2.22)
(2.23)                  C(I)m∗ (E) ≤ m∗ (E a )
|a−1|
q
(2.24)                                    ≤                   C(I)m∗ (E).
p
Let G be a cover of E. This means that E is contained in the Lebesgue mea-
surable set G and for every measurable set A, m∗ (E ∩ A) = |G ∩ A| . We now show
that Ga is a cover of E a . It is clear that Ga is measurable and E a ⊂ Ga . So it’s
enough to prove that m∗ (E a ) ≥ |Ga |. Let pk,n = p + k(q − p)/n for n a natural
number and k = 0, 1, ..., n and Ik,n = (pk−1,n , pk,n ), It follows that
n
m∗ (E a ) =           m∗ ((E ∩ Ik,n )a ),
k=1
ON THE NTH QUANTUM DERIVATIVE                                       13

whence by inequality (2.23)
n
m∗ (E a ) ≥                  C(Ik,n )m∗ (E ∩ Ik,n ).
k=1
Since G is a cover of E
n
m∗ (E a ) ≥               C(Ik,n ) |G ∩ Ik,n | ,
k=1

so by inequality (2.24)
n                        |a−1|
pk−1,n
m∗ (E a ) ≥                                          |(G ∩ Ik,n )a | .
pk,n
k=1
Bound the fractions from below and then let n → ∞,
|a−1|
p
m∗ (E a ) ≥                                     |Ga |
p + q−p
n
≥ |Ga | .
Now let a, b be real numbers, with a not zero, and Iq the interval with endpoints
q and 2q − 1. Let x = 0 be a point of outer density of E and deﬁne Aq = {p ∈
1/a
Iq : q b pa x ∈ E} = q −b x−1 E              ∩ Iq . We next prove that for all ε > 0, there exits
δ > 0 such that:
(2.25)                 m∗ (Aq ) > |Iq |(1 − ε) if 0 < |q − 1| < δ.
We can suppose that x = 1 and E ⊂ (1/2, 3/2). Let G ⊂ (1/2, 3/2) be a cover of
E. As we showed above, Ga is a cover of E a , so that

m∗ (Aq ) =                  χG (q b pa )dp,
Iq

where χG is the characteristic function of G. Letting s = q b pa ,
1
m∗ (Aq ) =                 χG (s)s1/a−1 ds
|a|q b/a qb Iq
a

k(q)
≥                       χG (s)ds
|a|            a
q b Iq

where k(q) = inf s∈qb Iq s1/a−1 → 1 as q → 1. Since x is a point of density of G,
a

a
k(q)                                    |Iq |
χG (s)ds                         |q − 1| = |Iq |.
|a|      q b Iq
a                         |a|
This establishes relation (2.25).
Passing to the proof, let:
N
Ej =   x∈E:            Ai (q)f (q ai x) ≤ j|q − 1|b whenever |q − 1| ≤ 1/j
i=0

c
(The following argument, modeled on work of Fejzi´ and Weil avoids the question
of whether the sets Ej are Lebesgue measurable.[FW]. There is a strongly related
example in [FW] that depends on the existence of a measurable set S for which S+S
is not measurable. To see how such a set S can be constructed, see the Theorem of
14 J. MARSHALL ASH, STEFAN CATOIU, AND RICARDO R´                   ´
IOS-COLLANTES-DE-TERAN

K. Ciesielski in the appendix at the end of this paper.[C]) Then E = j Ej . For
each positive integer j, the subset Nj of Ej consisting of those points of Ej that
are not points of outer density of Ej is measurable and has measure 0, regardless of
whether Ej is itself measurable.[M], p. 290 Thus if a property holds at every point
of outer density of every Ej , then the subset of E where that property fails to hold
∞
is a subset of j=1 Nj , and thus has measure 0. So, ﬁxing j and letting x = 0 be a
point of outer density of Ej , it suﬃces to prove that relation (2.21) holds at x. We
deﬁne
A∗ = {p ∈ Iq : q ak −a p−a0 x ∈ Ej }, k = 0, . . . , N
k

and
B ∗ = {p ∈ Iq : q −a pa          −a0
x ∈ Ej }, = 1, . . . , N.
Also deﬁne
N
Ak = {p ∈ Iq :                Ai (p)f (pai [q ak −a p−a0 x]) ≤ j2b |q − 1|b }, k = 0, ..., N
i=0
and
N
B = {p ∈ Iq :                    Ai (q)f (q ai [q −a pa   −a0
x]) ≤ j|q − 1|b }, = 1, ..., N.
i=0
−1
Let ε ∈ (0, (2N + 1) ). By relation (2.25), there exits δ ∈ (0, 1/ (2j)) such that if
0 < |q − 1| < δ then m∗ (A∗ ) > |q − 1|(1 − ε) and m∗ (B ∗ ) > |q − 1|(1 − ε). Using
k
that |p − 1| < 1/j, it follows that A∗ ⊂ Ak . Similarly, but using that |q − 1| < 1/j,
k
it’s easy to prove that B ∗ ⊂ B . Therefore

m (          Ak ) ∩ (          B)      > |q − 1|(1 − (2N + 1)ε) > 0.
k

Let p ∈ (      k   Ak ) ∩ (         B ). Then
N                                               N
A0 (p)           Ak (q)f (q ak −a x) = A0 (p)                     Ak (q)f (q ak [q −a pa0 −a0 x]) =
k=0                                             k=0

N              N                                                   N             N
A (p)           Ak (q)f (q ak [q −a pa         −a0
x]) −         A (p)         Ak (q)f (q ak [q −a pal −a0 x]) ≤
=0           k=0                                                   =1           k=0
N                   N                                                   N               N
|Ak (q)|            A (p)f (pa [q ak −a p−a0 x]) +                      |A (p)|         Ak (q)f (q ak [q −a pal −a0 x])
k=0                  =0                                                  =1             k=0
b             b                                  b
≤ sup Ak           ∞   (N + 1) j2 |q − 1| + sup A                        ∞   N j |q − 1| .
k

Dividing by |A0 (p)|, we ﬁnish the proof.
Lemma 5. Let a0 , . . . , aN be distinct real numbers and A0 (q), . . . , AN (q) bounded
measurable functions such that a0 = 0 and |A0 (q)| > C > 0 when q is near 1. If
N
(2.26)                                                Ai (q)f (q ai x) = O(1)
i=0

for all x ∈ E, then f (x) is bounded in a neighborhood of almost every point of E.
ON THE NTH QUANTUM DERIVATIVE                                         15

Proof. For each positive integer j, let
N
1
Fj =      x∈E:             Ai (q)f (q ai x) ≤ j for all 0 < |q − 1| <
i=0
j
∞
and Ej = Fj ∩ {x ∈ R : |f (x)| < j} . Then E = j=1 Ej . As in the proof of the
Sliding Lemma, we avoid the question of whether Ej is measurable. Working at
a point x = 0 of outer density of Ej and making use of a cover of Ej , just as in
the proof of the Sliding Lemma, we produce for each q suﬃciently close to 1, a
corresponding number p such that
(2.27)     all of the points p−a0 q a0 x and pai −a0 q a0 x, i = 1, 2, ..., N are in Ej .
Thus all y suﬃciently close to x can be written in the form y = q a0 x where there
is a number p such that condition (2.27) holds. For such a y we have
N                                       N
A0 (p) f (y) =          Ai (p)f (pai p−a0 q a0 x )    −         Ai (p) f (pai −a0 q a0 x) .
i=0                                     i=1

By deﬁnition of Ej , every term in curly brackets is bounded by j. Since A0 (p) is
bounded below, we may divide through by it and then apply the triangle inequality
to get a bound for |f (y)| .

Lemma 6. If for every x in the measurable set E,
˜
lim sup Dn (q, x; f ) < ∞,
q→1

then
(2.28)                                   ˜
lim sup Dn−1 (q, x; f ) < ∞
q→1

for almost every x ∈ E.
Proof. Fix x = 0, x ∈ E such that f is bounded in a neighborhood of x. So
there is a constant K such that for all q suﬃciently close to 1 there hold

(2.29)                      ˜
Dn (q, x; f ) ≤ K and |f (qx)| ≤ K.

Because of Lemma 5, it suﬃces to show that inequality (2.28) holds at x. From
(2.29) we have
˜                   ˜                    n   n
∆n−1 q 2 − λn−1 (q) ∆n−1 (q) ≤ KC |q − 1| |x| .

Repeating the iteration process that was used in the proof of Lemma 2 leads to
                
M
j              M                         n    n
˜
∆n−1 q 2 −                           ˜
λn−1 (q 1/2 ) ∆n−1 q 1/2           ≤ KC |q − 1| |x| .
j=0

˜
But ∆n−1 q 2      ≤ C K, so
                            
M
j               M
                        ˜
λn−1 (q 1/2 ) ∆n−1 q 1/2
j=0
16 J. MARSHALL ASH, STEFAN CATOIU, AND RICARDO R´                   ´
IOS-COLLANTES-DE-TERAN

is uniformly bounded as q varies between b and b1/2 for some ﬁxed b close to 1.
From inequalities (2.17), it follows that
M
˜
∆n−1 q 1/2
n−1
1 − q 1/2M
M
˜
and, as a consequence of equation (2.7), also Dn−1 q 1/2          is uniformly bounded
1/2
as q varies between b and b       for some ﬁxed b close to 1. As M varies through
1/2M
the positive integers, q      takes on all values between b and 1. To prove relation
(2.28), make two appropriate choices of b, one larger than 1 and one less than 1.
Lemma 7. If for every x ∈ E, we have
tn−1             tn
f (x + t) = f (x) + tf1 (x) + ... +            fn−1 (x) + ω (x, t) ,
(n − 1)!           n!
where ω (x, t) = O (1) as t → 0, then the Peano derivative fn (x) exists for almost
every x ∈ E.[MZ], Lemma 7
Lemma 8. If x = 0, fn−1 (x) exists, and
˜
lim sup Dn (q, x; f ) < ∞,
q→1

then
tn−1             tn
f (x + t) = f (x) + tf1 (x) + ... +            fn−1 (x) + ω (x, t)
(n − 1)!           n!
where ω (t) is bounded as t → 0.
This lemma follows by exactly the same argument as was used to prove Lemma
2.
Theorem 1. If for every x in the measurable set E,
(2.30)                        lim sup |Dn (q, x)| < ∞,
q→1

then the nth Peano derivative exists for almost every x ∈ E.
Corollary 3. If Dn f (x) exists for every x in the measurable set E, then the
nth Peano derivative exists at almost every point of E.
S
Corollary 4. If Dn f (x) exists for every x in the measurable set E, then the
nth Peano derivative exists at almost every point of E.
The ﬁrst corollary clearly follows from Theorem 1, while the second follows
from the Sliding Lemma and Theorem 1, so we pass to the proof of Theorem 1.
Proof. Because of Lemma 3 and the Sliding Lemma, we may replace condition
(2.30) by
˜
lim sup Dn (q, x) < ∞ for almost every x ∈ E.
q→1

Now apply Lemma 6 n − 1 times to see that
(2.31)                   ˜
lim sup Dk (q, x; f ) < ∞, k = n − 1, n − 2, ..., 1,
q→1
ON THE NTH QUANTUM DERIVATIVE                                  17

holds almost everywhere on E. In particular,

˜                       f (qx) − f (x)
lim sup D1 (q, x; f ) = lim sup                <∞
q→1                     q→1     qx − x
holds almost everywhere on E. But the change of variable h = qx − x shows that
this is equivalent to
f (x + h) − f (x)
lim sup                     < ∞,
h→0           h
so f (x) exists almost everywhere on E by a theorem of Denjoy.[D], p. 270 of [S]
Proceed inductively. First use the existence of f (x) , condition (2.31) with
k = 2, Lemma 8, and 7 to establish the existence of f2 (x) almost everywhere on E.
Then use the existence of f2 (x) , condition (2.31) with k = 3, Lemma 8, and 7 to
establish the existence of f3 (x) almost everywhere on E. After n − 2 similar steps,
Theorem 1 is proved.

3. Appendix
Theorem 2 (K. Ciesielski). There is a Lebesgue measurable set S such that
S + S is not Lebesgue measurable.[C]
Proof. Let K = {.a1 a2 ... : every ai ∈ {0, 1, 3, 4}}, where .a1 a2 ... is a base 5
decimal expansion, be the Cantor set obtained by removing the open middle ﬁfth
of [0, 1] , then removing the open middle ﬁfth of each of the 4 intervals [0, 1/5] ,
2
[1/5, 2/5] , [3/5, 4/5] and [4/5, 1] , et cetera. Then |[0, 1] \ K| = 1/5 + 4 (1/5) +
2       3
4 (1/5) + ... = 1, so |K| = 0. Every x = α.x1 x2 ... ∈ (0, 2) has c representations
of the form y + z where y = .y1 y2 ... and z = .z1 z2 ... are in K, and c is the
cardinality of the continuum R. Indeed, when α = 0, there are two possible cases.
First, if inﬁnitely many xi = 0, each such xi may be written as yi + zi , with
yi , zi ∈ {0, 1, 3, 4}in two diﬀerent ways, while yi = zi = 0 otherwise. Second,
if n is the largest index with xn = 0, for each i > n, choose yi to be either 1
or 3 and set zi = 4 − yi , while if 1 ≤ i ≤ n − 1, there are yi , zi ∈ {0, 1, 3, 4}
such that .x1 x2 ...xn − 5−n = .y1 y2 ...yn + .z1 z2 ...zn . In all cases, there are 2ℵ0 = c
representations of x. If α = 0, then x = 1.444...4xm xm+1 ..., where xm < 4. This case
is treated similarly, since there exist ym , zm ∈ {0, 1, 3, 4} such that 1.444...4xm =
.444...4ym + .444...4zm and as above, there exist c distributions of the tail of x
between y and z.
It is enough to show that there is a set S ⊂ K so that S is linearly independent
over the rational numbers Q and S +S is a Bernstein set, that is, both S +S and its
complement in [0, 2] have nonempty intersection with every perfect subset of [0, 2] ;
because then S, a subset of a measure 0 set will also have measure 0 and hence be
measurable, but S + S is not measurable, since neither S + S nor its complement
can have positive measure (since such sets contain perfect subsets).
To construct S, we will work in R/2Z, which we will identify with [0, 2). Note,
however, that since S will be contained in [0, 1] , the “+” sign in S +S coincides with
ordinary addition with possibly the single exception of 1 + 1 = 0. Let {Pξ : ξ < c}
be a well ordering of all the perfect subsets of [0, 2) . (We know that there are
exactly c perfect sets, each of cardinality c.[AW]) For each x ∈ [0, 2) , let K (x) =
{u ∈ S : u + v = x for some v ∈ S} . Pick p1 = 0 ∈ P1 . Since # (K (p1 )) = c, there
is a1 ∈ K (p1 ) \ p1 , where · denotes the Q-span in R/2Z of a set ·. Then
18 J. MARSHALL ASH, STEFAN CATOIU, AND RICARDO R´                   ´
IOS-COLLANTES-DE-TERAN

S1 = {a1 , b1 = p1 − a1 } is a linearly independent set and S1 + S1 intersects P1 .
Having chosen Sω ⊂ K for each ordinal ω < ξ, such that for each α ≤ ω, there
is a pair of elements in Sω whose sum is in Pα , set Sξ = ∪ω<ξ Sω ∪ {aξ , bξ }where
aξ and bξ are chosen as follows. Since # (Pξ ) > # ∪ω<ξ Sω and # (K (pξ )) >
# ∪ω<ξ Sω ∪ {pξ } , we can ﬁnd pξ ∈ Pξ independent from ∪ω<ξ Sω and then ﬁnd
aξ ∈ K (pξ ) independent from ∪ω<ξ Sω ∪ {pξ } . Let bξ = pξ − aξ . It is clear that Sξ is
a linearly independent set and Sξ + Sξ intersects Pω for all ω ≤ ξ. Then S = ∪ξ<c Sξ
c
is linearly independent and intersects every Pα . It remains to show that (S + S)
also has nonempty intersection with every perfect subset of [0, 2). Fix any s ∈ S.
The perfect subsets of [0, 2) can also be well ordered as {s + Pξ : ξ < c} . For each
α, we must ﬁnd a point in s + Pα which is not in S + S. From the construction of S,
we know that aα + bα ∈ (S + S) ∩ Pα , so that the point s + aα + bα ∈ s + Pα . But if
s + aα + bα ∈ S + S, then s + aα + bα − u − v = 0, where all ﬁve of s, aα , bα , u, and
v ∈ S. But a sum of an odd number of independent elements can not be zero.

References
[A]   J. M. Ash, Generalizations of the Riemann derivative, Trans. Amer. Math. Soc.,
126(1967), 181-199.
a
[AGV] J. M. Ash, A. E. Gatto, and S. V´gi, A multidimensional Taylor’s theorem with minimal
hypotheses, Colloq. Math., 60-61(1990), 245-252.
[AW] J. M. Ash and G. Wang, Sets of uniqueness of the power of the continuum, Proceedings
of A. Razmadze Mathematical Institute, 122(2000), 15-19.
[C]   K. Ciesielski, Measure zero sets whose algebraic sum is non-measurable, Real Analysis
Exchange, 26(2000/01), 919–922.
[D]                 e                        e e
A. Denjoy, M´moire sur les nombres d´riv´s, J. de Math., 7(1915), 105–240.
[FW] H. Fejzic and C. E. Weil, Repairing the proof of a classical diﬀerentiation result, Real
Analysis Exchange, 19(1993/94), 639–643.
[GR] G. Gasper and M. Rahman, Basic hypergeometric series. Encyclopedia of Mathematics
and its Applications, 35. Cambridge Univ. Press, Cambridge, 1990.
[M]   M. E. Munroe, Introduction to Measure and Integration, Addison–Wesley, Reading, MA,
1953.
[MZ] J. Marcinkiewicz and A. Zygmund, On the diﬀerentiability of functions and summability
of trigonometric series, Fund. Math. 26(1936), 1–43.
[O]   H. W. Oliver, The exact Peano derivative, Trans. Amer. Math. Soc., 76(1954), 444–456.
[S]   S. Saks, Theory of the integral, Dover, New York, 1964.

Department of Mathematics, DePaul University, Chicago, IL 60614
URL: http://www.depaul.edu/~mash/

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