The first example of a nowhere approximately differentiable function from the Banach space C[0, 1] is due to Jarn´ik (see [77]). Moreover, he showed that such functions are typical, i.e., they constitute a set whose complement is of first category in C[0, 1].
In this chapter we present one precise construction of a function acting from R into R, which is nowhere approximately differentiable. This con- struction is due to Mal´y [140] (cf. also [39] and [40]). It is not difficult and, at the same time, is rather vivid from the geometrical point of view.
We begin with some preliminary notions and facts.
Let λ denote the standard Lebesgue measure on R and let X be a λ- measurable subset of R. We recall that a point x ∈ R is said to be a density point for (of) X if
limh→0, h>0 λ(X ∩ [x − h, x + h])/2h = 1.
According to the classical Lebesgue theorem (see, e.g., the Introduction), almost all points of X are its density points.
Exercise 1. Let (tn)n∈Nbe a sequence of strictly positive real numbers,
such that
limn→+∞ tn = 0, limn→+∞ tn/tn+1= 1.
Let X be a Lebesgue measurable subset of R and let x ∈ R. Prove that the following two assertions are equivalent:
(1) x is a density point of X;
(2) limn→+∞ λ(X ∩ [x − tn, x + tn])/2tn= 1.
Exercise 2. Let X be a Lebesgue measurable subset of R and let x ∈ R. Show that the following two assertions are equivalent:
(1) x is a density point of X;
(2) limh→0+, k→0+ λ(X ∩ [x − h, x + k])/(h + k) = 1.
The notion of a density point turned out to be rather deep and fruitful not only for real analysis but also for general topology, probability theory
and some other domains of mathematics. For example, by use of this no- tion the important concept of the density topology on R was introduced and investigated by several authors (Pauc, Goffman, Waterman, Nishiura, Neugebauer, Tall, and others). This topology was studied, with its further generalizations, from different points of view (see, e.g., [66], [162], [171] and [219]). We shall deal with the density topology (and with some of its natural analogues) in our considerations below.
Now, let f : R → R be a function and let x ∈ R. We recall that f is said to be approximately continuous at x if there exists a λ-measurable set X such that:
(1) x is a density point of X;
(2) the function f |(X ∪ {x}) is continuous at x.
The next two exercises show that all Lebesgue measurable functions can be described in terms of approximate continuity.
Exercise 3. Let g : R → R be a function, let ε be a fixed strictly positive real number and suppose that, for any λ-measurable set X with λ(X) > 0, there exists a λ-measurable set Y ⊂ X with λ(Y ) > 0, such that
(∀x ∈ Y )(∀y ∈ Y )(|g(x) − g(y)| < ε).
Demonstrate that there exists a λ-measurable function h : R → R for which we have
(∀x ∈ R)(|g(x) − h(x)| < ε).
Infer from this fact that if the given function g satisfies the above condition for any ε > 0, then g is measurable in the Lebesgue sense.
Exercise 4. Let f : R → R be a function. By applying the result of Exercise 3 and utilizing the classical Luzin theorem on the structure of Lebesgue measurable functions (see, e.g., [154] orChapter 3of this book), show that the following two assertions are equivalent:
(a) the function f is measurable in the Lebesgue sense;
(b) for almost all (with respect to λ) points x ∈ R, the function f is approximately continuous at x.
Exercise 5. Let f : R → R be a locally bounded Lebesgue measurable function and let
F (x) = Z x
0
f (t)dt (x ∈ R).
Prove that, for any point x ∈ R at which the function f is approximately continuous, we have F0(x) = f (x).
nowhere approximately differentiable functions 139
Let now f : R → R be a function and let x ∈ R. We say that f is approximately differentiable at x if there exist a Lebesgue measurable set Y ⊂ R, for which x is a density point, and a limit
limy→x, y∈Y, y6=x
f (y) − f (x) y − x .
This limit is denoted by fap0 (x) and is called an approximate derivative of
f at the point x.
Exercise 6. Demonstrate that if a function f : R → R is approxi- mately differentiable at x ∈ R, then f is also approximately continuous at x.
Exercise 7. Check that an approximate derivative of a function f : R → R
at a point x ∈ R is uniquely determined, i.e., it does not depend on the choice of a Lebesgue measurable set Y for which x is a density point and for which the corresponding limit exists (applyExercise 29 from the Intro- duction).
Check also that the family of all functions acting from R into R and approximately differentiable at x forms a vector space over R.
Exercise 8. Show that if a function f : R → R is differentiable (in the usual sense) at a point x ∈ R, then f is approximately differentiable at x and f0
ap(x) = f0(x). Give an example showing that the converse assertion
is not true.
For our purposes below, we need two simple auxiliary propositions. Lemma 1. Let f : R → R be a function, let x be a point of R and suppose that f is approximately differentiable at x. Then, for any real number M1> fap0 (x), we have
limh→0+λ({y ∈ [x − h, x + h] \ {x} : (f (y) − f (x))/(y − x) ≥ M1})/2h = 0.
Similarly, for any real number M2< fap0 (x), we have
limh→0+λ({y ∈ [x − h, x + h] \ {x} : (f (y) − f (x))/(y − x) ≤ M2})/2h = 0.
Proof. Because the argument in both cases is comletely analogous, we shall consider only the case of M1. There exists a λ-measurable set X such
that x is a density point of X and
Fix ε > 0 for which
fap0 (x) + ε < M1.
Then there exists a real number δ > 0 such that, for any strictly positive real h < δ, we have
(∀y ∈ X ∩ [x − h, x + h] \ {x})((f (y) − f (x))/(y − x) ≤ fap0 (x) + ε). But, if δ > 0 is sufficiently small, then
λ(X ∩ [x − h, x + h])/2h ≥ 1 − ε for all strictly positive reals h < δ. So we obtain the relation
λ({y ∈ [x − h, x + h] \ {x} : (f (y) − f (x))/(y − x) ≥ M1})/2h ≤ ε,
and the lemma is proved.
Actually, in our further considerations we need only the following aux- iliary assertion, which is an immediate consequence ofLemma 1.
Lemma 2. Let f : R → R be a function, let x be a point of R and suppose that, for every strictly positive real number M , the relation
liminfh→0+
λ({y ∈ [x − h, x + h] \ {x} : |f (y) − f (x)|/|y − x| ≥ M })
2h > 0
holds true. Then f is not approximately differentiable at x. In particular, suppose that two sequences
{hk : k ∈ N}, {Mk : k ∈ N}
of real numbers are given, satisfying the following conditions: (1) hk> 0 and Mk> 0 for all integers k ≥ 0;
(2) limk→+∞hk = 0 and limk→+∞Mk= +∞;
(3) the lower limit
liminfk→+∞λ({y ∈ [x − hk, x + hk] \ {x} :
|f (y) − f (x)|/|y − x| ≥ Mk})/2hk
is strictly positive.
Then we can assert that our function f is not approximately differen- tiable at the point x.
nowhere approximately differentiable functions 141
After these simple preliminary remarks, we are able to begin the con- struction of a nowhere approximately differentiable function.
First of all, let us put
f1(0/9) = 0, f1(1/9) = 1/3, f1(2/9) = 0, f1(3/9) = 1/3,
f1(4/9) = 2/3, f1(5/9) = 1/3, f1(6/9) = 2/3, f1(7/9) = 3/3,
f1(8/9) = 2/3, f1(9/9) = 3/3
and extend (uniquely) this partial function to a continuous function f1 : [0, 1] → [0, 1]
in such a way that f1becomes affine on each segment [k/9, (k + 1)/9] where
k = 0, 1, ..., 8. We shall start with this function f1. In our further construc-
tion, we also need an analogous function g acting from the segment [0, 9] into the segment [0, 3]. Namely, we put
g(x) = 3f1(x/9) (x ∈ [0, 9]).
Obviously, g is continuous and affine on each segment [k, k + 1] where k = 0, 1, ..., 8. Also, another function similar to g will be useful in our construction. Namely, we denote by g∗ the function from [0, 9] into [0, 3],
whose graph is symmetric with the graph of g, with respect to the straight line
{(x, y) ∈ R × R : y = 3/2}. In other words, we put
g∗(x) = 3 − g(x)
for all x ∈ [0, 9]. Suppose now that, for a natural number n ≥ 1, the function fn : [0, 1] → [0, 1]
has already been defined, such that: (a) fn is continuous;
(b) for each segment of the form [k/9n, (k + 1)/9n], where
k ∈ {0, 1, ..., 9n− 1},
the function fn is affine on it and the image of this segment with respect to
fn is some segment of the form [j/3n, (j + 1)/3n], where
Let us construct a function
fn+1 : [0, 1] → [0, 1].
For this purpose, it suffices to define fn+1for any segment [k/9n, (k + 1)/9n]
where k ∈ {0, 1, ..., 9n− 1}. Here only two cases are possible.
1. fn is increasing on [k/9n, (k + 1)/9n]. In this case, let us consider the
following two sets of points of the plane:
{(0, 0), (0, 3), (9, 3), (9, 0)}, {(k/9n, f
n(k/9n)), (k/9n, fn((k + 1)/9n)),
((k + 1)/9n, fn((k + 1)/9n)), ((k + 1)/9n, fn(k/9n))}.
Because we have the vertices of two rectangles, there exists a unique affine transformation
h : R2→ R2
satisfying the conditions
h(0, 0) = (k/9n, fn(k/9n)), h(0, 3) = (k/9n, fn((k + 1)/9n)),
h(9, 3) = ((k + 1)/9n, fn((k + 1)/9n)), h(9, 0) = ((k + 1)/9n, fn(k/9n)).
Let the graph of the restriction of fn+1 to the segment [k/9n, (k + 1)/9n]
coincide with the image of the graph of g with respect to h.
2. fn is decreasing on [k/9n, (k + 1)/9n]. In this case, let us consider the
following two sets of points of the plane:
{(0, 0), (0, 3), (9, 3), (9, 0)}, {(k/9n, f
n((k + 1)/9n)), (k/9n, fn(k/9n)),
((k + 1)/9n, fn(k/9n)), ((k + 1)/9n, fn((k + 1)/9n))}.
Here we also have the vertices of two rectangles, so there exists a unique affine transformation
h∗ : R2→ R2
satisfying the relations
h∗(0, 0) = (k/9n, fn((k + 1)/9n)), h∗(0, 3) = (k/9n, fn(k/9n)),
h∗(9, 3) = ((k + 1)/9n, fn(k/9n)), h∗(9, 0) = ((k + 1)/9n, fn((k + 1)/9n)).
Let the graph of the restriction of fn+1 to the segment [k/9n, (k + 1)/9n]
nowhere approximately differentiable functions 143
The function fn+1has thus been determined. From the above construc-
tion immediately follows that the corresponding analogues of the conditions (a) and (b) hold true for fn+1, too. In other words, fn+1is continuous and,
for each segment of the form [k/9n+1, (k + 1)/9n+1], where
k ∈ {0, 1, ..., 9n+1− 1},
the function fn+1is affine on it and the image of this segment with respect
to fn+1 is some segment of the form [j/3n+1, (j + 1)/3n+1], where
j ∈ {0, 1, ..., 3n+1− 1}. Moreover, our construction shows that
(∀x ∈ [0, 1])(|fn+1(x) − fn(x)| ≤ 1/3n).
In addition, let
[u, v] = [k/9n, (k + 1)/9n]
be an arbitrary segment on which fn is affine. Then it is not hard to check
that
fn+1([u, (2u + v)/3]) = fn([u, (2u + v)/3]),
fn+1([(2u + v)/3, (2v + u)/3]) = fn([(2u + v)/3, (2v + u)/3]),
fn+1([(u + 2v)/3, v]) = fn([(u + 2v)/3, v]).
Proceeding in this way, we come to the sequence of functions (f1, f2, . . . , fn, . . . )
uniformly convergent to some continuous function f also acting from [0, 1] into [0, 1]. We assert that f is nowhere approximately differentiable on the segment [0, 1]. In order to demonstrate this fact, let us take an arbitrary point x ∈ [0, 1] and fix a natural number n ≥ 1.
Clearly, there exists a number k ∈ {0, 1, ..., 9n− 1} such that
x ∈ [k/9n, (k + 1)/9n]. Therefore, we have
fn(x) ∈ [j/3n, (j + 1)/3n]
for some number j ∈ {0, 1, ..., 3n− 1}. For the sake of simplicity, denote
From the remarks made above it immediately follows that, for all natural numbers m > n, we have
fm(x) ∈ [p, q]
and, consequently, f (x) ∈ [p, q], too. Further, we may assume without loss of generality that fn is increasing on [u, v] (the case when fn is decreasing
on [u, v] can be considered completely analogously). Suppose first that f (x) ≤ (p + q)/2 and put
D1= [(2v + u)/3, v].
Then, for each point y ∈ D1, we may write
f (y) ∈ [(2q + p)/3, q]. Hence, we get
(f (y) − f (x))/(y − x) ≥ ((2q + p)/3 − (p + q)/2)/(v − u) = (1/6)(3n). Suppose now that f (x) ≥ (p + q)/2 and denote
D2= [u, (2u + v)/3].
In this case, for any point y ∈ D2, we may write
f (y) ∈ [p, (2p + q)/3]. Hence, we get
(f (x) − f (y))/(x − y) ≥ ((p + q)/2 − (2p + q)/3)/(v − u) = (1/6)(3n). Thus, in the both cases, we have
λ({y ∈ [x − 1/9n, x + 1/9n] \ {x} : |f (y) − f (x)|/|y − x| ≥ (1/6)(3n)}) ≥ (1/3)(1/9n)
or, equivalently,
λ({y ∈ [x − 1/9n, x + 1/9n] \ {x} :
|f (y) − f (x)|/|y − x| ≥ (1/6)(3n)}) ≥ (1/6)λ([x − 1/9n, x + 1/9n]).
The latter relation immediately yields that our function f is not approx- imately differentiable at x (see Lemma 2 and the comments after this lemma).
nowhere approximately differentiable functions 145
Remark 1. The function f constructed above has a number of other interesting properties (for more information concerning f , see [140] and [40]).
Now, starting with an arbitrary continuous nowhere approximately dif- ferentiable function acting from [0, 1] into [0, 1], we can easily obtain an analogous function for R. We thus come to the following classical result (first obtained by Jarn´ik in 1934).
Theorem 1. There exist continuous bounded functions acting from R into R, which are nowhere approximately differentiable.
Remark 2. Actually, Jarn´ik proved that almost all (in the sense of the Baire category) functions from the Banach space C[0, 1] are nowhere ap- proximately differentiable. Clearly, this result generalizes the corresponding result of Banach and Mazurkiewicz for the usual differentiability (seeTheo- rem 2 from the Introduction). Further investigations showed that analogous statements hold true for many kinds of generalized derivatives. The main tool for obtaining such statements is the notion of porosity of a subset X of R at a given point x ∈ R. However, this interesting topic is out of the scope of our book. So we only refer the reader to the fundamental paper [27] where several category analogues of Theorem 1 for generalized derivatives are discussed from this point of view.
InChapter 15of our book we give an application of a nowhere approxi- mately differentiable function to the question concerning some relationships between the sup-measurability and weak sup-measurability of functions act- ing from R × R into R.
Because the concept of an approximate derivative relies essentially on the notion of a density point, it is reasonable to introduce here the so-called density topology on R and to consider briefly some elementary properties of this topology.
Exercise 9. For any Lebesgue measurable subset X of R, let us denote d(X) = {x ∈ R : x is a density point f or X}.
Further, denote by Td the family of all those Lebesgue measurable sets
Y ⊂ R, for which Y ⊂ d(Y ). Show that:
(a) Td is a topology on R strictly extending the standard Euclidean
topology of R;
(b) the topological space (R, Td) is a Baire space and satisfies the Suslin
condition (i.e., no nonempty open set in (R, Td) is of first category and each
(c) every first category set in (R, Td) is nowhere dense and closed (in
particular, the family of all subsets of (R, Td) having the Baire property
coincides with the Borel σ-algebra of (R, Td));
(d) a set X ⊂ R is Lebesgue measurable if and only if X has the Baire property in (R, Td);
(e) a set X ⊂ R is of Lebesgue measure zero if and only if X is a first category subset of (R, Td);
(f) the space (R, Td) is not separable.
The above-mentioned topology Td is usually called the density topology
on R. In a similar way, the density topology can be introduced for the Euclidean space Rn (n ≥ 2) equipped with the n-dimensional Lebesgue
measure λn.
Exercise 10. Let f : R → R be a function and let x ∈ R. Prove that the following two assertions are equivalent:
(a) f is approximately continuous at x;
(b) f regarded as a mapping from (R, Td) into R is continuous at x.
Exercise 11. By starting with the result of the previous exercise, show that the topological space (R, Td) is connected. For this purpose, suppose
to the contrary that there exists a partition {A, B} of R into two nonempty sets A ∈ Td and B ∈ Td. Then define a function
f : R → R
by putting f (x) = 1 for all x ∈ A, and f (x) = −1 for all x ∈ B. Obviously, f is a bounded continuous mapping from (R, Td) into R and hence, according
to Exercise 10, f is approximately continuous at each point of R. Further, define
F (x) = Z x
0
f (t)dt (x ∈ R).
By applyingExercise 5of this chapter, demonstrate that the function F is differentiable everywhere on R and
F0(x) = 1 ∨ F0(x) = −1
for each x ∈ R. This yields a contradiction with the Darboux property of any derivative.
One of the most interesting facts concerning the density topology states that (R, Td) is a completely regular topological space (see, for instance,
[162] and [219]). This property of Td implies some nontrivial consequences
nowhere approximately differentiable functions 147
of everywhere differentiable nowhere monotone functions by applying the above-mentioned fact (note that this approach is due to Goffman [65]).
Exercise 12. Consider any two disjoint countable sets A = {an : n ∈ N} ⊂ R,
B = {bn : n ∈ N} ⊂ R,
each of which is everywhere dense in R. Taking into accountExercise 10and the fact that (R, Td) is completely regular, we can find, for any n ∈ N, an
approximately continuous function fn : R → [0, 1] satisfying the relations
0 ≤ fn(x) ≤ 1 (x ∈ R),
fn(an) = 1, (∀x ∈ B)(fn(x) = 0).
Analogously, for any n ∈ N, there exists an approximately continuous func- tion gn : R → [0, 1] such that
0 ≤ gn(x) ≤ 1 (x ∈ R),
gn(bn) = 1, (∀x ∈ A)(gn(x) = 0).
Now, define a function h : R → R by the formula h = X
n∈N
(1/2n)(fn− gn).
Check that:
(a) h is bounded and approximately continuous; (b) h(a) > 0 for all a ∈ A;
(c) h(b) < 0 for all b ∈ B.
Let H denote an indefinite integral of h. Show that:
(i) H is everywhere differentiable on R and H0(x) = h(x) for each x ∈ R; (ii) H is nowhere monotone.
We thus see that with the aid of the density topology on R it is possible to give another proof of the existence of everywhere differentiable nowhere monotone functions acting from R into R (cf. the proof presented inChap- ter 5).
Remark 3. The density topology on R can be regarded as a very par- ticular case of the so-called von Neumann topology. Let (E, S, µ) be a space with a complete probability measure (or, more generally, with a complete nonzero σ-finite measure). Then, in conformity with a deep theorem of von
Neumann and Maharam (see, e.g., [138], [162], [171], [222]), there exists a topology T = T (µ) on E such that:
(1) (E, T ) is a Baire space satisfying the Suslin condition;
(2) the family of all subsets of (E, T ) having the Baire property coincides with the σ-algebra S;
(3) a set X ⊂ E is of µ-measure zero if and only if X is of first category in (E, T ).
We say that T = T (µ) is a von Neumann topology associated with the given measure space (E, S, µ). Note that T , in general, is not unique. This fact is not so surprising, because the proof of the existence of T is essentially based on the Axiom of Choice. There are many nontrivial applications of a von Neumann topology in various branches of contemporary mathematics (for instance, some applications to the general theory of stochastic processes can be found in [171]).
Remark 4. For the real line R, an interesting analogue of the den- sity topology, formulated in terms of category and the Baire property, was introduced and considered by Wilczy´nski in [231]. Wilczy´nski’s topology was then investigated by several authors. An extensive survey devoted to properties of this topology and to functions continuous with respect to it is contained in [40] (see also the list of references presented in the same work).
Remark 5. There are some invariant extensions of the Lebesgue mea- sure λ, for which an analogue of the classical Lebesgue theorem on density points does not hold. For example, there exist a measure µ on R and a µ-measurable set X ⊂ R, such that:
(1) µ is an extension of λ;
(2) µ is invariant under the group of all isometric transformations of R; (3) there is only one µ-density point for X, i.e., there exists a unique point x ∈ R for which we have
limh→0+
µ(X ∩ [x − h, x + h]) 2h = 1.
A more detailed account of the measure µ and its other extraordinary properties can be found in [97].