The Non-coincidence of Ordinary and Peano Derivatives

Zoltán Buczolich^{∗}, Department of Analysis, Eötvös Loránd University, Rákóczi út 5, 1088 Budapest, Hungary

Clifford E. Weil, Mathematics Department, Michigan State University, East Lansing, MI 48824-1027, USA

Dedicated to the memory of Jan Mařík

Abstract Let f : H ⊂ ℝ → ℝ be k times differentiable in both the usual (iterative) and Peano senses. We investigate when the usual derivatives and the corresponding Peano derivatives are different and the nature of the set where they are different.

1 Introduction

Professor Jan Mařík, whose death in January of 1994 ended an outstanding career, made significant contributions to several areas of mathematics including extensions of differentiable functions. (See [3].) For his enormous contributions to analysis and for our genuine affection for him, we dedicate this paper to his memory.

This paper is motivated by the following question. Assume that H ⊂ ℝ is perfect and for a function f : H → ℝ both the kth ordinary derivative, f^{(k)}, and the kth Peano derivative, f_{k}, exist at all points of H. How large can the set E_{k} of those points x in H be where f^{(k)}(x) and f_{k}(x) are different?

For k = 1 the ordinary and Peano derivatives are the same. It follows from Theorem 2 of this paper that for a given perfect set H the set E_{2} is countable. Theorem 3 implies that if, in addition, we assume that the third ordinary and Peano derivatives exist on H, then E_{2} is scattered. On the other hand Example 2 shows that for n ≥ 3, E_{n} = H is also possible for some perfect sets H.

In our Theorems for higher values of k we must impose coincidence assumptions about lower order Peano derivatives of the ordinary derivatives of f in order to obtain results. Actually the non-coincidence sets we consider are noncoincidence sets of “exact order” k, while the non-coincidence set E_{k} considered in the original question is of “order” less or equal k. (We give more explanation of this heuristic background in a remark following Theorem 6.)

∗This author was supported by Grants FKFP B-07/1997 and Hungarian National Foundation for Scientific Research Grant No. T 016094

1991 Mathematics Subject Classification: Primary 26A24

Keywords: Peano derivatives, nowhere dense perfect sets, porosity.

In Section 4 the concept of γ-gap porosity is introduced. In Theorem 5 it is proved that if the set H is γ_{n}-gap porous at the points x_{n} ∈ H, then there is a k ≥ 2 times ordinary and Peano differentiable function such that x_{n} ∈ E_{k} for all n. Theorem 6 shows that γ-gap porosity, in a certain sense, is also a necessary condition.

2 Definitions and Other Preliminaries

Throughout this paper H will denote a perfect subset of ℝ, k will be a fixed element of ℕ, i and j will denote nonnegative integers and f : H → ℝ. The usual or iterative kth derivative of f will be denoted by f^{(k)}. For example if

x ∈ H, then f′(x) = lim_{\begin{matrix}y→x \\ y∈H \end{matrix}} \frac{f(y) − f(x)}{y − x}. Next the corresponding Peano derivative

is defined.

Definition 1. Let f : B ⊂ ℝ → ℝ let k ∈ ℕ and let x ∈ B. Then f is k times Peano differentiable at x means that there are numbers f_{j}(x) for j = 1, 2, … k and there is a function ϵ : B → ℝ such that lim_{\begin{matrix}y→x \\ y∈B \end{matrix}} ϵ(y) = 0, and for each y ∈ B

f(y) = f(x) + ∑_{j=1}^{k} \frac{f_{j}(x)}{j!}(y − x)^{j} + ϵ(y)(y − x)^{k}.

If x is an isolated point of B, then the numbers f_{1}(x), f_{2}(x), …, f_{k}(x) are completely arbitrary. Otherwise they are unique if they exist. Examining the above sum it is obvious that setting f(x) = f_{0}(x) will be useful as will f(x) = f^{(0)}(x). The reader unfamiliar with the notion of Peano derivatives is directed to [4]. The major conditions imposed on the sets studied in this paper are motivated by the work done in [1]. The specific theorem is as follows. (See page 395 of [1].)

Theorem 1. Let H ⊂ ℝ be closed, let k ∈ ℕ and let f : H → ℝ be k times differentiable in both the usual sense and in the Peano sense on H. Suppose for each i, j ∈ ℕ ∪ {0} with i + j ≤ k we have that f^{(i)} is j times Peano differentiable on H and that (f^{(i)})_{j} = f^{(i+j)} on H. Then there is a function F : ℝ → ℝ which is k times Peano differentiable on ℝ such that F_{j} = f_{j} on H for each j = 0, 1, 2, …, k.

Simply stated, the purpose of this paper is to investigate the equality (f^{(i)})_{j} = f^{(i+j)} where f is defined on a nowhere dense set, H. If H is an interval, then the existence of f^{(i)} implies the existence of f_{i} and the equality f^{(i)} = f_{i}. Consequently the equality under study holds. However for a nowhere dense set it is possible for f_{k}(x) and f^{(k)}(x) to both exist but to be different.

Since the hypotheses of Theorem 1 are used often in this paper, we introduce the following useful notation. Let

P D_{k}(H) = {f : H → ℝ; f is k times differentiable in both the usual sense

and in the Peano sense}

and

NP D_{k}(H) = {f ∈ P D_{k}(H); i + j ≤ k and x ∈ H imply
(f^{(i)})_{j}(x) exists}.

From Theorem 2 it follows that if the condition (f^{(i)})_{j} = f^{(i+j)} holds on H for all i+j ≤ k, with the exception i = 0, j = k, and k is even, then the set f_{k} ≠ f^{(k)} is countable. If we have the additional information that f ∈ P D_{k+1}(H), then in Theorem 3 we show that the previous exceptional set is scattered. However, for odd k’s in Example 2 it is shown that there are non-empty perfect sets, H and functions, f which satisfy the assumptions of Theorem 2 and f^{(k)} ≠ f_{k} everywhere on H.

3 Non-coincidence Sets

We begin with a very simple but illustrative example.

Example 1. Let P = {p_{1}, p_{2}, … } be a countable set in ℝ with no isolated points, let {k_{n}} be a sequence in ℕ with k_{n} ≥ 2 for each n ∈ ℕ and let {α_{n}} be a sequence in ℝ. Then there is a function f : P → ℝ which is infinitely differentiable in the usual sense and in the Peano sense on P such that f^{(k)} ≡ 0 on P for all k ∈ ℕ and for each n ∈ ℕ we have f_{k}(p_{n}) = 0 if k ≠ k_{n} while f_{k_{n}}(p_{n}) = α_{n}.

Let f(p_{1}) = 0 and set g_{1}(x) = f(p_{1})+ \frac{α_{1}}{k_{1}!}(x−p_{1})^{k_{1}}. Let a_{1} = −∞ and b_{1} = +∞.

Let n ∈ ℕ with n ≥ 2 and suppose for j = 1, 2, …, n − 1, f(p_{j}) has been defined and set g_{j}(x) = f(p_{j}) + \frac{α_{j}}{k_{j}!}(x − p_{j})^{k_{j}}. Also suppose that for j =

1, 2, …, n − 1 numbers a_{j}, b_{j} ∉ P have been selected so that p_{j} ∈ (a_{j}, b_{j}) and for i = 1, 2, …, (j−1) either (a_{j}, b_{j})∩(a_{i}, b_{i}) = ∅, or (a_{j}, b_{j}) ⊂ (a_{i}, b_{i}) and in the

latter case; i.e., p_{j} ∈ (a_{i}, b_{i}), |g_{i}(x) − g_{j}(x)| < exp(−\frac{1}{|x − p_{i}|}). To define f(p_{n}),

let j_{n} = max{j ∈ {1, 2, …, n − 1}; p_{n} ∈ (a_{j}, b_{j})}. (Since (a_{1}, b_{1}) = (−∞, ∞), j_{n} is defined.) Set f(p_{n}) = g_{j_{n}}(p_{n}) and then let g_{n}(x) = f(p_{n}) + \frac{α_{n}}{k_{n}!}(x − p_{n})^{k_{n}}.

To define a_{n} and b_{n} first select a closed subinterval I of (a_{j_{n}}, b_{j_{n}}) with p_{n} in its interior. Let i < j_{n} with p_{n} ∈ (a_{i}, b_{i}). By the induction hypotheses,

(a_{j_{n}}, b_{j_{n}}) ⊂ (a_{i}, b_{i}) and |g_{i}(x) − g_{j_{n}}(x)| < exp(−\frac{1}{|x − p_{i}|}) for x ∈ (a_{j_{n}}, b_{j_{n}}). It

follows that there is an ϵ > 0 such that for all i < j_{n} with p_{n} ∈ (a_{i}, b_{i}) and

for all x ∈ I we have ϵ ≤ exp(−\frac{1}{|x − p_{i}|}) − |g_{i}(x) − g_{j_{n}}(x)|. Because each g_{j} is

continuous and since g_{n}(p_{n}) − g_{j_{n}}(p_{n}) = 0, it is not difficult to see that there are a_{n}, b_{n} ∉ P with p_{n} ∈ (a_{n}, b_{n}) ⊂ I such that

|g_{j_{n}}(x) − g_{n}(x)| < exp(−\frac{1}{|x − p_{j_{n}}|}) and |g_{n}(x) − g_{j_{n}}(x)| < ϵ.

To complete the induction step we need only consider the case i < n with p_{n} ∈ (a_{i}, b_{i}). By definition i ≤ j_{n}. If i = j_{n}, then for x ∈ (a_{n}, b_{n}) we have

|g_{j_{n}}(x) − g(x)| < exp(−\frac{1}{|x − p_{j_{n}}|}). If i < j_{n}, then

|g_{i}(x) − g_{n}(x)| ≤|g_{i}(x) − g_{j_{n}}(x)| + |g_{j_{n}}(x) − g_{n}(x)|
<|g_{i}(x) − g_{j_{n}}(x)| + ϵ
<|g_{i}(x) − g_{j_{n}}(x)| + exp(−\frac{1}{|x − p_{j_{n}}|}) − |g_{i}(x) − g_{j_{n}}(x)|.

To show that the function f has the desired properties, fix i ∈ ℕ and let n ∈ ℕ with n > i and p_{n} ∈ (a_{i}, b_{i}). Then j_{n} ≥ i and by definition f(p_{n}) = g_{j_{n}}(p_{n}). Thus

|f(p_{n}) − g_{i}(p_{n})| = |g_{j_{n}}(p_{n}) − g_{i}(p_{n})| < exp(−\frac{1}{|x − p_{j_{n}}|}).

By the definition of g_{i} and since k_{i} ≥ 2, this estimate proves that f′(p_{i}) = f_{1}(p_{i}) = 0 (and hence that f^{(k)} ≡ 0 on P) and that f_{k}(p_{i}) = 0 if k ≠ k_{i} while f_{k_{i}}(p_{i}) = α_{i}.

The next example shows that for some perfect sets H for any integer larger than 2 the corresponding usual and Peano derivatives may exist and be different everywhere on H.

Example 2. There is a perfect set H and, for each m ∈ ℕ with m ≥ 3, a function f ∈ NP D_{k}(H) for all k ∈ ℕ such that for all k ∈ ℕ with k ≥ 2 we have f^{(k)} = 0 on H and f_{k} = 0 on H except for k = m while f_{m} = m! on H.

Remark 1. If m is odd then in the above Example we have f′ = f_{1} = 0 on H which implies that (f^{(i)})_{j} = f^{(i+j)} holds except for i = 0 and j = m. When m is even (f^{(i)})_{j} = f^{(i+j)} is always satisfied when i ≥ 2, if i = 0, or 1 it is satisfied for all j’s with the exception of j = m − i.

Set ℓ_{0} = 1 and for each n ∈ ℕ let ℓ_{n} = 10^{−n^{n}}. In fact, any sequence satisfying the following three properties can be chosen for our construction.

(i) ℓ_{n−1} − 2ℓ_{n} > \frac{1}{2}ℓ_{n−1} holds for n = 2, 3, ... .

(ii) For m, n_{0} ∈ ℕ, m ≥ 3 letting σ_{n_{0}} = ∑_{n=n_{0}+1}^{∞} ℓ_{n−1}^{m−1} ≥ σ′_{n_{0}} = ∑_{n=n_{0}+1}^{∞} ℓ_{n−1}^{m} we

have lim_{n_{0}→∞} ℓ_{n_{0}−1}^{−n}σ_{n_{0}} = 0 for all n ∈ ℕ.

(iii) \frac{ℓ_{n}}{ℓ_{n−1}} < ℓ_{n−1}^{n−1} for n = 2, 3, ... .

Observe that the ℓ_{n}’s we chose satisfy properties (i)–(iii).

Set I_{0,1} = [0, 1], I_{1,1} = [0, ℓ_{1}], and I_{1,2} = [1 − ℓ_{1}, 1]. Suppose I_{n,j} = [a, b] has been defined for n ∈ ℕ and for j ∈ N_{n} = {1, 2, …, 2^{n}}. Then I_{n+1,2j−1} = [a, a + ℓ_{n+1}] and I_{n+1,2j} = [b − ℓ_{n+1}, b] defines I_{n+1,j} for j ∈ N_{n+1}. Let H = ⋂_{n∈ℕ} ⋃_{j=1}^{2^{n}} I_{n,j}. For each x ∈ H and for each n ∈ ℕ let j_{n,x} be that integer in N_{n} such that x ∈ I_{n,j}. For each n ∈ ℕ and for each j ∈ N_{n} let p_{n,j}(x) = α_{n,j}(x − a_{n,j}) + β_{n,j} where a_{n,j} is the left endpoint of I_{n,j} and the constants α_{n,j} and β_{n,j} will be defined later depending on whether m is odd or even in a way that they will satisfy

|α_{n,j}| ≤ ℓ_{n−1}^{m−1} and |β_{n,j}| ≤ ℓ_{n−1}^{m}.  (1)

For x ∈ H set f(x) = ∑_{n∈ℕ} p_{n,j_{n,x}}(x).

Note that for x, y ∈ [0, 1]

p_{n,j_{n,y}}(y) − p_{n,j_{n,x}}(x) = p_{n,j_{n,y}}(y) − p_{n,j_{n,x}}(y) + α_{n,j_{n,x}}(y − x).  (2)

For x, y ∈ H with x ≠ y let n_{0}(x, y) = min{n ∈ ℕ; j_{n,x} ≠ j_{n,y}}. Consequently, for x, y ∈ H with x ≠ y (denoting n_{0}(x, y) by n_{0}) from (2) it follows that

\frac{f(y) − f(x)}{y − x} = ∑_{n<n_{0}} α_{n,j_{n,x}} + \frac{∑_{n≥n_{0}} p_{n,j_{n,y}}(y) − p_{n,j_{n,x}}(x)}{y − x}.

Thus using (1) we have |p_{n,j_{n,x}}(x)| ≤ 2ℓ_{n−1}^{m}. Since |y − x| > ℓ_{n_{0}−1} − 2ℓ_{n_{0}} > ℓ_{n_{0}−1}/2 by (i) and (ii) it follows that the second term above tends to 0 as y → x (The term n = n_{0} must be dealt with separately, but clearly it is no more than 8ℓ_{n_{0}−1}^{m−1} which tends to 0 because m ≥ 3.) and hence f′(x) = ∑_{n∈ℕ} α_{n,j_{n,x}}.

Similarly for x, y ∈ H with x ≠ y

f(y) − f(x) − f′(x)(y − x) − (y − x)^{m} =( ∑_{n≥n_{0}} p_{n,j_{n,y}}(y) − p_{n,j_{n,x}}(y))− (y − x)^{m}
=(p_{n_{0},j_{n_{0},y}}(y) − p_{n_{0},j_{n_{0},x}}(y) − (y − x)^{m}) +

∑_{n>n_{0}} p_{n,j_{n,y}}(y) − p_{n,j_{n,x}}(y) = T (x, y) + S(x, y).

Since |y − x| > ℓ_{n_{0}−1} − 2ℓ_{n_{0}}, and n_{0} = n_{0}(x, y) → ∞ as y → x, conditions (i) and (ii) imply that lim_{y→x} \frac{S(x,y)}{|y−x|^{n}} = 0 for all n > m.

First let m be odd. Then put α_{n,j} = 0 and β_{n,j} = \begin{cases}0 & if j is odd \\ ℓ_{n−1}^{m} & if j is even. \end{cases}

Let x, y ∈ H with x ≠ y. If x < y, then x ∈ I_{n_{0},2j−1} and y ∈ I_{n_{0},2j}. Thus T (x, y) = ℓ_{n_{0}−1}^{m} − (y − x)^{m}. Since ℓ_{n_{0}−1} − 2ℓ_{n_{0}} < y − x < ℓ_{n_{0}−1}, by condition (iii)

0 < ℓ_{n_{0}−1}^{m} − (y − x)^{m} < ℓ_{n_{0}−1}^{m} − (ℓ_{n_{0}−1} − 2ℓ_{n_{0}})^{m} = ℓ_{n_{0}−1}^{m}[1 − (1 − 2\frac{ℓ_{n_{0}}}{ℓ_{n_{0}−1}} )^{m}]

< ℓ_{n_{0}−1}^{m}[1 − (1 − m · 2\frac{ℓ_{n_{0}}}{ℓ_{n_{0}−1}} )] < ℓ_{n_{0}−1}^{m}(2mℓ_{n_{0}−1}^{n_{0}−1})

= 2mℓ_{n_{0}−1}^{m+n_{0}−1}.

So by condition (i) for n > m

lim_{y→x+} \frac{|T (x, y)|}{|y − x|^{n}} ≤ lim_{y→x+} 2^{n} \frac{2mℓ_{n_{0}(x,y)−1}^{m+n_{0}(x,y)−1}}{ℓ_{n_{0}(x,y)−1}^{n}} = 0.

If y < x, then y ∈ I_{n_{0},2j−1} and x ∈ I_{n_{0},2j}. Thus, since m is odd, T (x, y) = −ℓ_{n_{0}−1}^{m} + (x − y)^{m}. Now as above

0 > (x − y)^{m} − ℓ_{n_{0}−1}^{m} > (ℓ_{n_{0}−1} − 2ℓ_{n_{0}})^{m} − ℓ_{n_{0}−1}^{m}

and hence |T (x, y)| < ℓ_{n_{0}−1}^{m} − (ℓ_{n_{0}−1} − 2ℓ_{n_{0}})^{m}. So by the same argument as

above, lim_{y→x−} \frac{|T (x, y)|}{|y − x|^{n}} = 0 for n > m.

Now, let m be even. Then put β_{n,j} = 0 and α_{n,j} = \begin{cases}0 & if j is odd \\ ℓ_{n−1}^{m−1} & if j is even. \end{cases}

If x < y, then x ∈ I_{n_{0},2j−1} and y ∈ I_{n_{0},2j}. Thus

|T (x, y)| = |ℓ_{n_{0}−1}^{m−1}(y−a_{n,j_{n_{0},y}})−(y−x)^{m}| < |ℓ_{n_{0}−1}^{m−1}ℓ_{n_{0}}−(y−x)^{m}| < ℓ_{n_{0}−1}^{m}−(y−x)^{m}

and proceeding as in the first part of the previous case yields lim_{y→x+} \frac{|T (x, y)|}{|y − x|^{n}} =

0. If y < x, then y ∈ I_{n_{0},2j−1} and x ∈ I_{n_{0},2j}. Thus since m is even,

|T (x, y)|=| − ℓ_{n_{0}−1}^{m−1}(y − a_{n_{0},j_{n_{0},x}})−(y − x)^{m} =|ℓ_{n_{0}−1}^{m−1}(a_{n_{0},j_{n_{0},x}} − y) − (x − y)^{m}|

< |(ℓ_{n_{0}−1} − 2ℓ_{n_{0}})^{m} − ℓ_{n_{0}−1}^{m}| = ℓ_{n_{0}−1}^{m} − (ℓ_{n_{0}−1} − 2ℓ_{n_{0}})^{m}

and proceeding as in the second part of the first case yields lim_{y→x−} \frac{|T (x, y)|}{|y − x|^{n}} =

0.

Therefore in all cases if n > m, then

lim_{y→x} \frac{f(y) − f(x) − (y − x)f′(x) − (y − x)^{m}}{(y − x)^{n}} = 0.

Consequently if m is odd and m ≥ 3, then f′(x) = 0 (because each α_{n,j} = 0) for all x ∈ H and hence f^{(k)} = 0 on H for all k ≥ 1. Moreover f_{j}(x) = 0 if j ≠ m while f_{m}(x) = m! for all x ∈ H.

On the other hand if m ≥ 4 is even, then m′ = m − 1 ≥ 3 is odd and letting β′_{n,j} = α_{n,j} we have f′(x) = ∑_{n∈ℕ} β′_{n,j_{n,x}}; that is, our earlier argument for odd m’s shows that (f′)′(x) = f^{(2)}(x) = 0 for all x ∈ H and hence f^{(k+1)}(x) = 0 for all k ≥ 1. Therefore f′′(x) = 0 on H and hence f^{(k)} = 0 on H for all k ≥ 2. Moreover, as above, f_{j}(x) = 0 if j ≠ m while f_{m}(x) = m! and we also have (f′)_{j}(x) = 0 if j ≠ m − 1, while (f′)_{m−1}(x) = (m − 1)! for all x ∈ H.

The preceding example shows in particular that for n ≥ 3 the nth ordinary and the nth Peano derivatives can both exist and be different everywhere on H for some perfect sets, H. The case n = 2 proves to be quite different as the next theorem demonstrates. For example it shows that the second ordinary and the second Peano derivatives can differ only on a countable set.

Theorem 2. Let H ⊂ ℝ be perfect, let k ∈ ℕ with k ≥ 2 and let f ∈ P D_{k}(H). Suppose 0 ≤ i < k with k − i even and put

E_{i} = {x ∈ H; if i′ + j′ < k or if i′ + j′ = k and i′ > i, then
(f^{(i′)})_{j′}(x) exists and = f^{(i′+j′)}(x) and
(f^{(i)})_{k−i}(x) ≠ f^{(k)}(x)}.

Then E_{i} is countable.

Proof. For rational numbers α and β with α > β and for n ∈ ℕ let

E_{n,i}^{α,β} = {x ∈ H; (f^{(i′)})_{j′}(x) = f^{(i′+j′)}(x) for i′ + j′ < k,

(f^{(i′)})_{k−i′}(x) = f^{(k)}(x) for i < i′ < k

|f^{(i)}(y) − ∑_{j=0}^{k−i} \frac{(f^{(i)})_{j}(x)}{j!}(y − x)^{j}| < \frac{α − β}{2(k − i)!}|y − x|^{k−i}

for y ∈ H, |y − x| < \frac{1}{n}, and f^{(k)}(x) > α > β > (f^{(i)})_{k−i}(x)\}.

The theorem will be proved if it can be shown that E_{n,i}^{α,β} is an isolated set. So suppose to the contrary that x is a non-isolated point of E_{n,i}^{α,β}. Let ϵ > 0 and select y ∈ E_{n,i}^{α,β} with |y − x| < \frac{1}{n} such that for i < i′ < k

|f^{(i′)}(y) − ∑_{j=0}^{k−i′} \frac{(f^{(i′)})_{j}(x)}{j!}(y − x)^{j}| < ϵ|y − x|^{k−i′}.  (3)

Since both x and y belong to E_{n,i}^{α,β},

|f^{(i)}(y) − ∑_{j=0}^{k−i} \frac{(f^{(i)})_{j}(x)}{j!}(y − x)^{j}|< \frac{α − β}{2(k − i)!}|y − x|^{k−i}  (4)

and

|f^{(i)}(x) − ∑_{j=0}^{k−i} \frac{(f^{(i)})_{j}(y)}{j!}(x − y)^{j}| < \frac{α − β}{2(k − i)!}|y − x|^{k−i}.  (5)

For u ∈ H put

g(u) = f(u) − ∑_{j=0}^{k} \frac{f^{(j)}(x)}{j!}(u − x)^{j} = f(u) − h(u)

where h is a polynomial of degree no more than k. Then g^{(j)}(x) = 0 for j = 0, 1, …, k and for 0 ≤ i′ < k

g^{(i′)}(u) = f^{(i′)}(u) − ∑_{j=0}^{k−i′} \frac{f^{(i′+j)}(x)}{j!}(u − x)^{j}.

Using the assumptions in the definition of E_{n,i}^{α,β} and (3) for i < i′ < k we have

|g^{(i′)}(y)| < ϵ|y − x|^{k−i′}.  (6)

Again using the definition of E_{n,i}^{α,β} and (4) we infer

|g^{(i)}(y) + \frac{f^{(k)}(x) − (f^{(i)})_{k−i}(x)}{(k − i)!}(y − x)^{k−i}| < \frac{α − β}{2(k − i)!}|y − x|^{k−i}. (7)

Since h is a polynomial of degree no more than k, by (5) we have

\frac{α − β}{2(k − i)!}|y − x|^{k−i} >|g^{(i)}(x) + h^{(i)}(x) − ∑_{j=0}^{k−i} \frac{(g^{(i)})_{j}(y) + (h^{(i)})_{j}(y)}{j!}(x − y)^{j}|

=|g^{(i)}(x) − ∑_{j=0}^{k−i} \frac{(g^{(i)})_{j}(y)}{j!}(x − y)^{j}|

(using (6) and that (g^{(i)})_{j}(y) = g^{(i+j)}(y) for 1 < j < k − i)

>|0 − g^{(i)}(y) − \frac{(g^{(i)})_{k−i}(y)}{(k − i)!}(x − y)^{k−i}|

− ϵ ∑_{j=1}^{k−i−1} \frac{|y − x|^{k−i−j}}{j!} |x − y|^{j}

>| − g^{(i)}(y) − \frac{(g^{(i)})_{k−i}(y)}{(k − i)!}(x − y)^{k−i}| − ϵ(k − i)|x − y|^{k−i}.

Since (g^{(i)})_{k−i}(y) = (f^{(i)})_{k−i}(y) − f^{(k)}(x), we obtain

\frac{α − β}{2(k − i)!}|y − x|^{k−i} + ϵ(k − i)|x − y|^{k−i} >

| − g^{(i)}(y) − \frac{(f^{(i)})_{k−i}(y) − f^{(k)}(x)}{(k − i)!}(x − y)^{k−i}|.

Using (7) we have

\frac{α − β}{2(k − i)!}|y − x|^{k−i} > |g^{(i)}(y) − \frac{(f^{(i)})_{k−i}(x) − f^{(k)}(x)}{(k − i)!}(y − x)^{k−i}|.

Adding the two preceding inequalities together, canceling and keeping in mind that k − i is even we obtain

(α − β) + ϵ(k − i)(k − i)! > |2f^{(k)}(x) − (f^{(i)})_{k−i}(y) − (f^{(i)})_{k−i}(x)| = A. (8)

On the other hand x, y ∈ E_{n,i}^{α,β} implies

f^{(k)}(x) > α > β > (f^{(i)})_{k−i}(x) and β > (f^{(i)})_{k−i}(y).

Hence A > 2(α − β) and this contradicts (8) when ϵ is small. Thus E_{n,i}^{α,β} is countable.

In a similar fashion the set resulting from E_{n,i}^{α,β} by reversing the inequality between α and β in the definition is also clearly countable. This observation concludes the proof.

The next theorem shows that if i = 0 the set defined in Theorem 2, besides being countable, is scattered when f is k + 1 times differentiable in both senses.

Theorem 3. Let k ∈ ℕ be even and let f ∈ P D_{k+1}(H). Set

E = {x ∈ H;if i + j < k or if i + j = k and 0 < i, then(f^{(i)})_{j}(x) exists and
= f^{(i+j)}(x) and f_{k}(x) ≠ f^{(k)}(x)}.

Then E is nowhere dense in each ∅ ≠ F ⊂ H with F perfect.

Proof. By Theorem 2, E is countable. Let ∅ ≠ F ⊂ H be perfect. Suppose there is an interval I_{1} such that F ∩ I_{1} ≠ ∅ and F ∩ E is dense in I_{1} ∩ F. Since f_{k+1} is a Baire one function, there is an interval I_{2} ⊂ I_{1} and an M ∈ (0, ∞) such that I_{2} ∩ F ≠ ∅ and |f_{k+1}(x)| ≤ M for each x ∈ I_{2} ∩ F. By the Baire Category Theorem there is a δ > 0 such that

K_{δ} = {x ∈ I_{2} ∩ F; y ∈ H and |x − y| < δ implies

|f(y) − ∑_{j=0}^{k+1} \frac{f_{j}(x)}{j!}(y − x)^{j}| < |y − x|^{k+1}\}

is of the second category in I_{2} ∩ F. Also there is an interval I_{3} ⊂ I_{2} with I_{3} ∩ F ≠ ∅ such that K_{δ} is of the second category in every subportion of I_{3} ∩ F. Since F ∩ E is dense in I_{3} ∩ F, we may select x ∈ I_{3} ∩ F ∩ E. Let ϵ > 0. For y ∈ E, y sufficiently close to x

|f(y) − ∑_{j=0}^{k} \frac{f_{j}(x)}{j!}(y − x)^{j}| < ϵ|y − x|^{k}  (9)

and for 0 < i < k

|f^{(i)}(y) − ∑_{j=0}^{k−i} \frac{(f^{(i)})_{j}(x)}{j!}(y − x)^{j}| < ϵ|y − x|^{k−i}.  (10)

Since y ∈ E, (10) may be rewritten as

|f^{(i)}(y) − ∑_{j=0}^{k−i} \frac{f^{(i+j)}(x)}{j!}(y − x)^{j}| < ϵ|y − x|^{k−i}.  (11)

As in the proof of Theorem 2 for u ∈ H let

g(u) = f(u) − ∑_{j=0}^{k} \frac{f^{(j)}(x)}{j!}(u − x)^{j}.

Then g^{(j)}(x) = 0 for j = 0, 1, …, k. Since x ∈ E, (9) implies for y ∈ E sufficiently close to x

|g(y) + \frac{f^{(k)}(x) − f_{k}(x)}{k!}(x − y)^{k}| < ϵ|y − x|^{k}.  (12)

Moreover for 0 < i ≤ k

g^{(i)}(y) = f^{(i)}(y) − ∑_{j=0}^{k−i} \frac{f^{(i+j)}(x)}{j!}(y − x)^{j}.

So (11) and the assumption that x ∈ E imply |g^{(i)}(y)| < ϵ|y − x|^{k−i}. Also |g^{(k)}(y)| = |f^{(k)}(y) − f^{(k)}(x)| < ϵ for y sufficiently close to x since f^{(k+1)}(x) exists. Because g −f is a polynomial of degree at most k, for y ∈ K_{δ} sufficiently close to x we have

|g(x) − ∑_{j=0}^{k+1} \frac{g^{(j)}(y)}{j!}(x − y)^{j}| = |f(x) − ∑_{j=0}^{k+1} \frac{f^{(j)}(y)}{j!}(x − y)^{j}| < |x − y|^{k+1}.

Thus by (12)

|x − y|^{k+1} >|g(x) − ∑_{j=0}^{k+1} \frac{g^{(j)}(y)}{j!}(x − y)^{j}|

≥|\frac{f^{(k)}(x) − f_{k}(x)}{k!}(x − y)^{k}| − |g(y) + \frac{f^{(k)}(x) − f_{k}(x)}{k!}(x − y)^{k}|

− ∑_{j=1}^{k} \frac{|g^{(j)}(y)|}{j!}|x − y|^{j} − |\frac{f^{(k+1)}(y)}{(k + 1)!}||x − y|^{k+1}

≥|\frac{f^{(k)}(x) − f_{k}(x)}{k!}||x − y|^{k} − ϵ|x − y|^{k} − ϵ|x − y|^{k} ∑_{j=0}^{k} \frac{1}{j!}

− \frac{M}{(k + 1)!}|x − y|^{k+1}.

Dividing by |x − y|^{k}, using that y can be chosen arbitrarily close to x and that ϵ was arbitrary we obtain an inequality which contradicts f_{k}(x) ≠ f^{(k)}(x).

4 A Porosity Condition

In this section we introduce a condition on the set H sufficient for the existence of a function in NP D_{k}(H) for which f^{(k)} = f_{k} fails to hold on a dense subset of H. We show that in some sense the condition is necessary.

Definition 2. Let H ⊂ ℝ, let 0 < γ < 1 and let x ∈ H. Then H is γ-gap porous at x means there exist sequences a_{1} < a_{2} < ... < a_{ℓ} < ... < x < ... < b_{ℓ} < ... < b_{2} < b_{1} such that [a_{2ℓ+1}, a_{2ℓ+2}] ∩ H = ∅, [b_{2ℓ+2}, b_{2ℓ+1}] ∩ H = ∅, for ℓ ∈ ℕ ∪ {0}

γ|x − a_{2ℓ+1}| ≤ |a_{2ℓ+2} − a_{2ℓ+1}|, γ|b_{2ℓ+1} − x| ≤ |b_{2ℓ+1} − b_{2ℓ+2}|

and

lim_{ℓ→∞} \frac{|a_{2ℓ+1} − a_{2ℓ}|}{|a_{2ℓ+2} − a_{2ℓ+1}|} = 0, lim_{ℓ→∞} \frac{|b_{2ℓ} − b_{2ℓ+1}|}{|b_{2ℓ+1} − b_{2ℓ+2}|} = 0.

The first condition asserts that [a_{2ℓ+1}, a_{2ℓ+2}] is at least a fixed portion of the interval [a_{2ℓ+1}, x] while the second condition can be shown to be equivalent to stating that the length of [a_{2ℓ}, a_{2ℓ+1}] divided by the length of [a_{2ℓ}, x] tends to 0. Analogous statement can be made concerning the sequence {b_{ℓ}}. These remarks are expressed in a very useful way in the following proposition.

Proposition 4. Let H ⊂ ℝ be perfect, let 0 < γ < 1 and let x ∈ H. The set H is γ-gap porous at x if and only if for each ϵ > 0 there is a δ > 0 such that if y_{1}, y_{2} ∈ H with |y_{i} − x| < δ and either y_{1} < y_{2} < x or x < y_{2} < y_{1}, then we

have \frac{|y_{2} − y_{1}|}{|x − y_{1}|} ≥ γ or \frac{|y_{2} − y_{1}|}{|x − y_{1}|} ≤ ϵ.

The proof of the proposition is standard and hence is omitted.

Theorem 5. Let H be a perfect set, let k ∈ ℕ with k ≥ 2 and for each n ∈ ℕ let x_{n} ∈ H. Suppose for each n ∈ ℕ there is a γ_{n} ∈ (0, 1) such that H is γ_{n}-gap porous at x_{n}. Then there is an f ∈ NP D_{k}(H) such that for each n ∈ ℕ we have f^{(k)}(x_{n}) ≠ f_{k}(x_{n}). In addition, for all n we can also insist that i+j < k implies (f^{(i)})_{j}(x_{n}) = f^{(i+j)}(x_{n}) and 0 < i < k implies (f^{(i)})_{k−i}(x_{n}) = f^{(k)}(x_{n}).

Before proving this theorem we remark that given a γ ∈ (0, 1) it is easy to construct a perfect set H and a dense subset {x_{n}; n ∈ ℕ} of H such that H is γ-porous at each x_{n}. Then Theorem 5 provides a function, f, which is in NP D_{k}(H). Since {x_{n}; n ∈ ℕ} is not scattered, by Theorem 3 f cannot belong to P D_{k+1}(H).

Proof. For each n ∈ ℕ let a_{n,ℓ} ↗ x_{n}, b_{n,ℓ} ↘ x_{n} such that for each ℓ ∈ ℕ we have

γ_{n}|x_{n} − a_{n,2ℓ+1}| ≤ |a_{n,2ℓ+2} − a_{n,2ℓ+1}|, γ_{n}|b_{n,2ℓ+1} − x_{n}| ≤ |b_{n,2ℓ+1} − b_{n,2ℓ+2}|,
lim_{ℓ→∞} \frac{|a_{n,2ℓ+1} − a_{n,2ℓ}|}{|a_{n,2ℓ+2} − a_{n,2ℓ+1}|} = 0 and lim_{ℓ→∞} \frac{|b_{n,2ℓ} − b_{n,2ℓ+1}|}{|b_{n,2ℓ+1} − b_{n,2ℓ+2}|} = 0.

Further assume, as we may, that |x_{n} − a_{n,1}| ≤ 1 and |b_{n,1} − x_{n}| ≤ 1 for each n ∈ ℕ. For ℓ ∈ ℕ ∪ {0} set I_{a,n,ℓ} = [a_{n,2ℓ+1}, a_{n,2ℓ+2}], I_{b,n,ℓ} = [b_{n,2ℓ+2}, b_{n,2ℓ+1}], J_{a,n,0} = (−∞, a_{n,1}), J_{b,n,0} = (b_{n,1}, ∞) and for ℓ ∈ ℕ set J_{a,n,ℓ} = (a_{n,2ℓ}, a_{n,2ℓ+1}) and J_{b,n,ℓ} = (b_{n,2ℓ+1}, b_{n,2ℓ}). Since H is γ_{n}-gap porous at x_{n}, for ℓ ∈ ℕ ∪ {0}, I_{a,n,ℓ}∩H = ∅, I_{b,n,ℓ}∩H = ∅, γ_{n}|x_{n}−a_{n,2ℓ+1}| ≤ |I_{a,n,ℓ}|, γ_{n}|b_{n,2ℓ+1}−x_{n}| ≤ |I_{b,n,ℓ}|, and

lim_{ℓ→∞} \frac{|J_{a,n,ℓ}|}{|I_{a,n,ℓ}|} + \frac{|J_{b,n,ℓ}|}{|I_{b,n,ℓ}|} = 0.

For each n ∈ ℕ let α_{n} = \frac{γ_{n}^{k}}{2^{n}}. Then ∑_{n∈ℕ} \frac{α_{n}}{γ_{n}^{k}} < ∞. Fix n ∈ ℕ. We define

f(n) : ℝ → ℝ as follows. First put f(n)(x_{n}) = 0. For ℓ ∈ ℕ ∪ {0} the function f(n) is constant on J_{a,n,ℓ} and on J_{b,n,ℓ} and is linear on I_{a,n,ℓ} and on I_{b,n,ℓ}. In addition

f(n)(a_{n,2ℓ+2}) − f(n)(a_{n,2ℓ+1}) = α_{n}((a_{n,2ℓ+2} − x_{n})^{k} − (a_{n,2ℓ+1} − x_{n})^{k})

and

f(n)(b_{n,2ℓ+1}) − f(n)(b_{n,2ℓ+2}) = α_{n}((b_{n,2ℓ+1} − x_{n})^{k} − (b_{n,2ℓ+2} − x_{n})^{k}).

Finally assume that f(n) is continuous at x_{n}. Hence f(n) is continuous everywhere. Since |x_{n} − a_{n,2ℓ+1}| ≤ |I_{a,n,ℓ}|/γ_{n} and since |b_{n,2ℓ+1} − x_{n}| ≤ |I_{b,n,ℓ}|/γ_{n},

|f(n)(a_{n,2ℓ+2}) − f(n)(a_{n,2ℓ+1})| ≤ α_{n}|x_{n} − a_{n,2ℓ+1}|^{k} ≤ \frac{α_{n}}{γ_{n}^{k}} |I_{a,n,ℓ}|^{k}

and

|f(n)(b_{n,2ℓ+1}) − f(n)(b_{n,2ℓ+2})| ≤ α_{n}|b_{n,2ℓ+1} − x_{n}|^{k} ≤ \frac{α_{n}}{γ_{n}^{k}} |I_{b,n,ℓ}|^{k}.

Let x, y ∈ ∪_{ℓ∈ℕ∪{0}}(J_{a,n,ℓ} ∪ J_{b,n,ℓ}) with x < y. Let h = y − x and let {L_{i} = [c_{i}, d_{i}]; i ∈ N } denote the set of intervals I_{a,n,ℓ} and I_{b,n,ℓ} contained in

(x, y) where N ⊂ ℕ. Put h_{i} = |L_{i}|. Then ∑_{i∈N} \frac{h_{i}}{h} ≤ 1 and since 0 ≤ \frac{h_{i}}{h} ≤ 1

implies 0 ≤ \frac{h_{i}^{k}}{h^{k}} ≤ \frac{h_{i}}{h}, we have ∑_{i∈N} (\frac{h_{i}}{h} )^{k} ≤ 1; that is, ∑_{i∈N} h_{i}^{k} ≤ h^{k}. Because f(n) is continuous at x_{n} and constant on each J_{a,n,ℓ} and on each J_{b,n,ℓ},

|f(n)(y) − f(n)(x)| ≤ \frac{α_{n}}{γ_{n}^{k}} (y − x)^{k}.  (13)

Since H ⊂ ∪_{ℓ∈ℕ∪{0}}(J_{a,n,ℓ} ∪ J_{b,n,ℓ}), the above inequality holds for x, y ∈ H.

For x ∈ H put f(x) = ∑_{n∈ℕ} f(n)(x). Since f(n) is constant on J_{a,n,ℓ} and on J_{b,n,ℓ} and since |x_{n} − a_{n,1}| ≤ 1 and |b_{n,1} − x_{n}| ≤ 1, from (13) it follows that |f(n)(x)| ≤ \frac{α_{n}}{γ_{n}^{k}} for all x ∈ H. Hence the sum defining f converges for all x ∈ H.

Let x ∈ H \ {x_{n}; n ∈ ℕ}. Then for each n ∈ ℕ there is an ℓ_{n} ∈ ℕ ∪ {0} such that x ∈ J_{a,n,ℓ_{n}} ∪ J_{b,n,ℓ_{n}}. Let N_{0} ∈ ℕ. Since f(n) is constant on J_{a,n,ℓ} and on J_{b,n,ℓ}, there is an open interval U such that x ∈ U and f(n) is constant on U for n ∈ {1, 2, …, N_{0} − 1}. Then for y ∈ U by (13) we have

|f(y) − f(x)| = | ∑_{n≥N_{0}}(f(n)(x) − f(n)(y))| ≤ ∑_{n≥N_{0}} \frac{α_{n}}{γ_{n}^{k}} |y − x|^{k}.

Since ∑_{n∈ℕ} \frac{α_{n}}{γ_{n}^{k}} < ∞, by definition it follows that f is k times Peano differen-

tiable at x and that f_{j}(x) = 0 for all 1 ≤ j ≤ k. In particular since k ≥ 2, we have f′(x) = f_{1}(x) = 0.

To complete the proof it suffices to show that for each n ∈ ℕ, f is k times Peano differentiable at x_{n} with f_{j}(x_{n}) = 0 for 1 ≤ j < k and f_{k}(x_{n}) = α_{n}k! because then f′ = 0 on H and hence f^{(k)} = 0 on H while f_{k}(x_{n}) = α_{n}k! ≠ 0. First we show that for each n ∈ ℕ if f(n) is k times Peano differentiable at x_{n}, then so is f and indeed with the same Peano derivatives. To this end fix n_{0} ∈ ℕ and choose N_{0} > n_{0}. Then there is an open interval U such that x_{n_{0}} ∈ U and f(n) is constant on U for all n ≤ N_{0} − 1 with n ≠ n_{0}. Then for y ∈ U

f(y) − f(x_{n_{0}}) = f(n_{0})(y) − f(n_{0})(x_{n_{0}}) + ∑_{n≥N_{0}}(f(n)(y) − f(n)(x_{n_{0}})).

Thus

|f(y) − f(x_{n_{0}}) − (f(n_{0})(y) − f(n_{0})(x_{n_{0}}))| ≤ ∑_{n≥N_{0}} \frac{α_{n}}{γ_{n}^{k}} |y − x_{n_{0}}|^{k}.

Therefore f′(x_{n_{0}}) = f′(n_{0})(x_{n_{0}}) and for 1 ≤ j ≤ k we have f_{j}(x_{n_{0}}) = f_{j}(n_{0})(x_{n_{0}}) whenever the right-hand side exists.

Finally it will be shown that for each n ∈ ℕ, f(n) is k times Peano differentiable at x_{n} with f(n)_{j}(x_{n}) = 0 for 1 ≤ j < k and f(n)_{k}(x_{n}) = α_{n}k!. Now fix n ∈ ℕ and let ϵ > 0. Because 0 < γ_{n} < 1, there is K ∈ ℕ such that (1−γ_{n})^{K} < ϵ and there exists ℓ_{0} ∈ ℕ such that ℓ ≥ ℓ_{0} implies

|J_{a,n,ℓ}| < |I_{a,n,ℓ}|\frac{ϵ}{K} and |J_{b,n,ℓ}| < |I_{b,n,ℓ}|\frac{ϵ}{K}

Let x < x_{n} with x ∈ H so close to x_{n} that if x ∈ J_{a,n,ℓ′}, then ℓ′ ≥ ℓ_{0}. Set t = |x_{n} − a_{n,2ℓ′+1}|; the distance between x_{n} and the right endpoint of J_{a,n,ℓ′}. Since |J_{a,n,ℓ′}| < |I_{a,n,ℓ′}|\frac{ϵ}{K} < t\frac{ϵ}{K}, we conclude that t < |x_{n} − x| ≤ |J_{a,n,ℓ′}| + t < |I_{a,n,ℓ′}|\frac{ϵ}{K} + t < t(1 + \frac{ϵ}{K} ). Since |x_{n} − a_{n,2ℓ′+1}| ≤ |I_{a,n,ℓ′}|/γ_{n}, we have

|x_{n} − a_{n,2ℓ′+2}| = |x_{n} − a_{n,2ℓ′+1} − (a_{n,2ℓ′+2} − a_{n,2ℓ′+1})| < t − γ_{n}t = (1 − γ_{n})t.

Moreover

|x_{n} − a_{n,2ℓ′+4}| ≤ (1 − γ_{n})|x_{n} − a_{n,2ℓ′+3}| < (1 − γ_{n})|x_{n} − a_{n,2ℓ′+2}| < (1 − γ_{n})^{2}t

and in general

|x_{n} − a_{n,2(ℓ′+K)}| ≤ (1 − γ_{n})^{K}t < ϵt.

In the interval (a_{n,2ℓ′+1}, a_{n,2(ℓ′+K)}) there are K intervals I_{a,n,ℓ}; namely I_{a,n,ℓ′}, I_{a,n,ℓ′+1}, … I_{a,n,ℓ′+K−1}. Since |J_{a,n,ℓ}| < |I_{a,n,ℓ}|\frac{ϵ}{K} for all ℓ = ℓ′+1, ℓ′+2, …, ℓ′+ K − 1, and since for these same values of ℓ we have |I_{a,n,ℓ}| < t, it follows that |J_{a,n,ℓ}| < \frac{ϵ}{K} t and consequently

∑_{ℓ=ℓ′+1}^{ℓ′+K−1} |J_{a,n,ℓ}| < (K − 1)\frac{ϵ}{K} t < ϵt.

The function f(n) changes by α_{n}((a_{n,2ℓ+2} − x_{n})^{k} − (a_{n,2ℓ+1} − x_{n})^{k}) on I_{a,n,ℓ} = [a_{n,2ℓ+1}, a_{n,2ℓ+2}], later it will also be useful to keep in mind that the sign of this

change equals that of (−1)^{k+1}. From |x_{n}−a_{n,2(ℓ′+K)}| < ϵt and ∑_{ℓ=ℓ′+1}^{ℓ′+K−1} |J_{a,n,ℓ}| < ϵt

it is easy to see that (Recall that t = |x_{n} − a_{n,2ℓ′+1}|.)

α_{n}t^{k} > (−1)^{k+1}(f(n)(a_{n,2(ℓ′+K)}) − f(n)(a_{n,2ℓ′+1}))

= (−1)^{k+1} ∑_{ℓ=ℓ′}^{ℓ′+K−1} f(n)(a_{n,2ℓ+2}) − f(n)(a_{n,2ℓ+1})

= (−1)^{k} ∑_{ℓ=ℓ′}^{ℓ′+K−1} α_{n}((a_{n,2ℓ+1} − x_{n})^{k} − (a_{n,2ℓ+2} − x_{n})^{k}) > α_{n}((1 − 2ϵ)t)^{k}.

From (13) and from the continuity of f(n) it follows that

|f(n)(a_{n,2(ℓ′+K)}) − f(n)(x_{n})| ≤ \frac{α_{n}}{γ_{n}^{k}} |a_{n,2(ℓ′+K)} − x_{n}|^{k} ≤ \frac{α_{n}}{γ_{n}^{k}} ϵ^{k}t^{k}.

Since f(n) is constant on J_{a,n,ℓ′}, we have f(n)(x) = f(n)(a_{n,2ℓ′+1}); hence

α_{n}t^{k} + \frac{α_{n}}{γ_{n}^{k}} ϵ^{k}t^{k} > (−1)^{k}(f(n)(x) − f(n)(x_{n})) > α_{n}(1 − 2ϵ)^{k}t^{k} − \frac{α_{n}}{γ_{n}^{k}} ϵ^{k}t^{k}.

Using the above inequality, t < |x_{n} − x| < (1 + \frac{ϵ}{K})t, and the fact that the above argument is valid for any ϵ > 0 one can easily verify that

\frac{f(n)(x) − f(n)(x_{n})}{(x − x_{n})^{k}} → α_{n} as x → x_{n}, x < x_{n}, and x ∈ H.

A similar argument is valid when x > x_{n}. This implies f_{j}(n)(x_{n}) = 0 for 0 ≤ j < k and f_{k}(n)(x_{n}) = α_{n}k! ≠ 0. Therefore for each n_{0} ∈ ℕ we have f′(x_{n_{0}}) = f′(n_{0})(x_{n_{0}}) = 0, f_{j}(x_{n_{0}}) = f_{j}(n_{0})(x_{n_{0}}) = 0 for 1 ≤ j < k, and f_{k}(x_{n_{0}}) = f_{k}(n_{0})(x_{n_{0}}) = α_{n_{0}}k! ≠ 0 = f^{(k)}(x_{n_{0}}).

Theorem 6. Let H be a non empty perfect set and let f ∈ NP D_{k}(H). Assume that f_{j} = f^{(j)} on H for all 2 ≤ j ≤ k − 1 and let

E = {x ∈ H; if i + j < k, or if i + j = k and i > 0, then
(f^{(i)})_{j}(x) = f^{(i+j)}(x) and f_{k}(x) ≠ f^{(k)}(x)}.

Suppose E is dense in H. Then there is a (non empty) portion I ∩ H of H such that for each x ∈ I ∩ E there is a γ ∈ (0, 1) such that H is γ-gap porous at x.

The above theorem seems to be too restrictive, but if E is not dense in H, then we can still obtain some information about its size. It is clear that the union of dense in itself sets is dense in itself. Let E_{0} denote a maximal dense in itself subset of E and H_{0} be the closure of E_{0}. Then E \H_{0} is a scattered set, and our theorem is applicable to H_{0}. Here we also point out that the assumptions in the definition of the set E are not unnatural either. If f_{k}(x) ≠ f^{(k)}(x) and x ∉ E, then there is a k′ < k and 0 < i < k such that for g = f^{(i)} the point x belongs to a set of non-coincidence, E′, which is defined analogously to E by using g and k′. This means that one should think of E as an “exact” non-coincidence set of order k.

Theorems 5 and 6 imply that E_{2} can be dense in H if and only if the set of gap porosity points is dense in H. Furthermore the set {f^{(k)} ≠ f_{k}} can be dense in H for some f ∈ NP D_{k}(H) if and only if the set of gap porosity points of H is dense in H.

Proof. By Baire Category Theorem there is a portion I ∩ H, a δ_{0} > 0, an M′ ∈ (0, ∞) and a set F ⊂ I ∩ H dense in I ∩ H such that for each x ∈ I ∩ H we have |f_{k}(x)| ≤ M′ and for each x ∈ F and each y ∈ H with |y − x| < δ_{0}

|f(y) − ∑_{j=0}^{k} \frac{f_{j}(x)}{j!}(y − x)^{j}| ≤ |y − x|^{k}.  (14)

Let p ∈ I ∩ E. As in the proof of Theorem 2 for x ∈ H we let

g(x) = f(x) − ∑_{j=0}^{k} \frac{f^{(j)}(p)}{j!}(x − p)^{j}.

Then g^{(j)}(p) = g_{j}(p) = 0 for all j = 0, 1, …, k − 1, g^{(k)}(p) = 0 and g_{k}(p) = f_{k}(p) − f^{(k)}(p) ≠ 0. In general for each y ∈ H we have g_{k}(y) = f_{k}(y) − f^{(k)}(p) and hence |g_{k}(y)| ≤ M′ + |f^{(k)}(p)| = M for all y ∈ I ∩ H. Moreover since g − f is a polynomial of degree no more than k, for x, y ∈ H

g(y) − ∑_{j=0}^{k} \frac{g_{j}(x)}{j!}(y − x)^{j} = f(y) − ∑_{j=0}^{k} \frac{f_{j}(x)}{j!}(y − x)^{j}.

Thus by (14) for x ∈ F and y ∈ H with |y − x| < δ_{0}

|g(y) − ∑_{j=0}^{k} \frac{g_{j}(x)}{j!}(y − x)^{j}| < |y − x|^{k}.  (15)

Let A = (\frac{|g_{k}(p)|}{2(M + k!)})^{\frac{1}{k−1}} and set γ = \frac{A}{1+A}. We will now show that H is γ-gap

porous at p using Proposition 4.

Let ϵ > 0. By hypothesis, g_{j}(x) = g^{(j)}(x) for each x ∈ H and 1 ≤ j < k and since p ∈ E, if i + j < k or if i + j = k and i > 0, then (g^{(i)})_{j}(p) = g^{(i+j)}(p). Thus there is a 0 < δ_{1} < δ_{0} such that y ∈ H, |y − p| < δ_{1} and 1 ≤ i < k imply

|g_{i}(y) − ∑_{j=0}^{k−i} \frac{g^{(i+j)}(p)}{j!}(y − p)^{j}| < ϵ_{i}|y − p|^{k−i}

where ϵ_{i} = \frac{i!|g_{k}(p)|}{k!}(\frac{1−γ}{γ} )^{i−1}. Since g^{(j)}(p) = 0, for j = 0, 1, …, k and |y−p| <

δ_{1} we obtain

|g_{i}(y)| < ϵ_{i}|y − p|^{k−i}.  (16)

Since g_{j}(p) = 0 for j = 0, 1, …, k − 1, there is a 0 < δ_{2} < δ_{1} such that y ∈ H and |y − p| < δ_{2} implies

|g(y) − \frac{g_{k}(p)}{k!}(y − p)^{k}| < ϵ_{k}|y − p|^{k}  (17)

where ϵ_{k} = \frac{ϵ(1 − γ)^{k−1}|g_{k}(p)|}{(1 + (1 − γ)^{k−1})2k!}. Let x, y ∈ H with |p − y| < δ_{2}. Further

suppose y < x < p. (The argument for p < x < y is similar.) In addition assume that x ∈ F. By (15), (16) and (17)

|\frac{g_{k}(p)}{k!}||(y − p)^{k} − (x − p)^{k}| ≤|\frac{g_{k}(p)}{k!}(y − p)^{k} − g(y)| + |g(y) − g(x)|

+ |g(x) − \frac{g_{k}(p)}{k!}(x − p)^{k}|

<ϵ_{k}|y − p|^{k} + |g(y) − ∑_{j=0}^{k} \frac{g_{j}(x)}{j!}(y − x)^{j}|

+ ∑_{j=1}^{k} \frac{|g_{j}(x)|}{j!}|y − x|^{j} + ϵ_{k}|x − p|^{k}

<ϵ_{k}(p − y)^{k} + |y − x|^{k} + ∑_{j=1}^{k−1} \frac{ϵ_{j}|x − p|^{k−j}|y − x|^{j}}{j!}

+ \frac{M}{k!} |y − x|^{k} + ϵ_{k}(p − x)^{k}.

Since

|(y − p)^{k} − (x − p)^{k}| = |y − x||∑_{j=1}^{k}(y − p)^{k−j}(x − p)^{j−1}|

= (x − y) ∑_{j=1}^{k}(p − y)^{k−j}(p − x)^{j−1},

|\frac{g_{k}(p)}{k!}|(x − y) ∑_{j=1}^{k}(p − y)^{k−j}(p − x)^{j−1}  (18)

<ϵ_{k}((p − y)^{k} + (p − x)^{k}) + ∑_{j=1}^{k−1} \frac{ϵ_{j}(p − x)^{k−j}(x − y)^{j}}{j!} + (\frac{M + k!}{k!})(x − y)^{k}.

Suppose ϵ < \frac{x−y}{p−y} < γ. Since p − y = p − x + x − y < p − x + γ(p − y), (1 − γ)(p − y) < (p − x). So for j = 1, 2, …, k − 1

ϵ_{j}(p − x)^{k−j}(x − y)^{j} < ϵ_{j}(x − y)(p − y)^{k−j}(p − y)^{j−1}γ^{j−1}
< ϵ_{j}(x − y)(p − y)^{k−j}(p − x)^{j−1}(\frac{γ}{1 − γ} )^{j−1}  (19)

= \frac{|g_{k}(p)|}{k!}j!(x − y)(p − y)^{k−j}(p − x)^{j−1}.

Also

ϵ_{k}((p − y)^{k} + (p − x)^{k}) + (\frac{M + k!}{k!})(x − y)^{k}

<ϵ_{k}(x − y)(\frac{(p − x)^{k−1}}{ϵ(1 − γ)^{k−1}} + \frac{(p − x)^{k−1}}{ϵ} )

+ (x − y)(\frac{M + k!}{k!})(\frac{γ}{1 − γ} )^{k−1}(p − x)^{k−1}

=\frac{|g_{k}(p)|}{2k!}(x − y)(p − x)^{k−1} + \frac{|g_{k}(p)|}{2k!}(x − y)(p − x)^{k−1}
=\frac{|g_{k}(p)|}{k!}(x − y)(p − x)^{k−1}.

Summing inequality (19) multiplied by 1/j! for j = 1, 2, …, k − 1, and adding to it the above estimate contradicts inequality (18). Thus \frac{x−y}{p−y} ≤ ϵ or \frac{x−y}{p−y} ≥ γ. Since F is dense in I ∩ H, we may assume that x ∈ I ∩ H. So by Proposition 4 the set H is γ-gap porous at p.

References

[1] H. Fejzić, J. Mařík, and C. E. Weil, Extending Peano derivatives, Math. Bohemica, 119 (1994), 387–406.

[2] V. Jarník, Sur l’extension du domaine de definition des fonctions d’une variable, qui laisse intacte la derivabilité de la fonction, Bull international de l’Acad Sci de Boheme, (1923).

[3] J. Mařík, Derivatives and closed sets, Acta. Math. Acad. Sci. Hungar., 43 (1998), 25–29.

[4] Clifford E. Weil, The Peano notion of higher order differentiation, Math. Japonica, 42 (1995), 587–600.