Zoltán Buczolich, Eötvös Loránd University, Department of Analysis, Budapest, Múzeum krt 6-8, H-1088, Hungary.

Second Peano Derivatives are not Extendable

Introduction. A well-known theorem of V.Jarník [J] (see also [PL] Th. 5.5.2.) states that if a real function is defined and differentiable on a perfect subset of the real line, then this function can be extended to be a function defined and differentiable everywhere.

C. E. Weil in a personal communication asked what can be said about the extensions of Peano differentiable functions. Since the first Peano derivatives coincide with the ordinary derivatives, this question is interesting for the second and higher Peano derivatives. In this paper we show that in general there is no extension theorem for the second Peano derivative.

Definition. Suppose that F ⊂ R is perfect. We say that f : F → R is k times Peano differentiable at x ∈ F if there exist finite numbers f_{(0)}(x) = f(x), f_{(1)}(x),...,f_{(k)}(x) such that

|f(x + h) − ∑_{i=0}^{k} \frac{f_{(i)}(x)}{i!}h^{i}| = o(h^{k}),

when x + h ∈ F and h → 0. The set of k times Peano differentiable functions on F will be denoted by D^{k}(F).

Theorem. There exists a non-empty perfect set F and a function f : F → R such that f ∈ D^{2}(F) and there does not exist g : R → R such that g|_{F} = f and g ∈ D^{2}(R).

Research supported by the Hungarian National Foundation for Scientific Research, Grant No. 1807.

Proof. Put d_{n} := (n + 1)^{−(n+1)}. Denote by S the set of infinite zero-one sequences. For every n ∈ N and s ∈ S denote by s|n the first n terms of s. Put I_{∅} := [0, 1]. Suppose that for an n ∈ N and s ∈ S we have defined I_{s|n−1} = [a_{s|n−1}, b_{s|n−1}]. If the n’th term of s equals zero (or one), then we put I_{s|n} := [a_{s|n−1}, a_{s|n−1}+d_{n}] (or I_{s|n} := [b_{s|n−1}−d_{n}, b_{s|n−1}] respectively). Put X(s) := ∩_{n=1}^{∞}I_{s|n} and F := ∪{X(s) : s ∈ S}. It is obvious that the mapping s ↦ X(s) is one-to-one and the set F is non-empty and perfect. Thus for x ∈ F denote by S(x) ∈ S the inverse of x; that is, X(S(x)) = x. Denote by N(s) ∈ {0} ∪ N ∪ {+∞} the number of the ones in the sequence s ∈ S. Suppose that x ∈ F. If i ≤ N(S(x)) is a natural number then choose x_{i} ∈ F such that S(x_{i})|n = S(x)|n if n denotes the place in the sequence S(x) where the i’th term one appeares, and the remaining terms of S(x_{i}) equal zero. For i > N(S(x)) put x_{i} = x. We also put x_{0} = 0. It is clear that x_{i} ≥ x_{i−1} for every x ∈ F and i = 1, 2, ... . The function f claimed in our theorem will be defined on F by

f(x) := ∑_{i=1}^{∞} \frac{1}{2^{i}} (x_{i} − x_{i−1})(x − x_{i}).

First we show that f ∈ D^{2}(F). Namely, we shall show that for every x ∈ F we have f_{(1)}(x) = ∑_{i=1}^{∞} 2^{−i}(x_{i} − x_{i−1}) and f_{(2)}(x) = 0. Suppose that x, y ∈ F, S(x)|m = S(y)|m, S(x)|(m+1) ≠ S(y)|(m+1). Denote by j the number of the ones in S(x)|m. By the choice of m it is obvious that the (m + 1)’th terms of S(x) and S(y) are different. Therefore by the definition of F we have that d_{m}/2 < |x − y| ≤ d_{m}. We also have that

x_{i} = y_{i} for i = 1, ..., j,

|x − x_{i}| < d_{m}, |y − y_{i}| < d_{m} for i ≥ j,
|x_{i} − x_{i−1}| < d_{m}, |y_{i} − y_{i−1}| < d_{m} for i ≥ j + 1.

Using these estimations we obtain

| ∑_{i=1}^{∞} 2^{−i}(x_{i} − x_{i−1})(x − x_{i}) + ∑_{i=1}^{∞} 2^{−i}(x_{i} − x_{i−1})(y − x)−

∑_{i=1}^{∞} 2^{−i}(y_{i} − y_{i−1})(y − y_{i})| ≤

| ∑_{i=1}^{j} 2^{−i}(x_{i} − x_{i−1})(x − x_{i}) + ∑_{i=1}^{j} 2^{−i}(x_{i} − x_{i−1})(y − x)−

∑_{i=1}^{j} 2^{−i}(x_{i} − x_{i−1})(y − x_{i})|+

| ∑_{i=j+1}^{∞} 2^{−i}(x_{i} − x_{i−1})(x − x_{i}) + ∑_{i=j+1}^{∞} 2^{−i}(x_{i} − x_{i−1})(y − x)−

∑_{i=j+1}^{∞} 2^{−i}(y_{i} − y_{i−1})(y − y_{i})| ≤

0 + | ∑_{i=j+1}^{∞} \frac{d_{m}}{2^{i}} d_{m}| + | ∑_{i=j+1}^{∞} \frac{d_{m}}{2^{i}} | · |y − x| + | ∑_{i=j+1}^{∞} \frac{d_{m}}{2^{i}} d_{m}| ≤ d_{m}^{2} · 3 · 2^{−j}.

Therefore

\frac{|f(y) − f(x) − f_{(1)}(x)(y − x) − \frac{f_{(2)}(x)}{2!}(y − x)^{2}|}{|y − x|^{2}} ≤ \frac{d_{m}^{2} · 3 · 2^{−j}}{2^{−2} · d_{m}^{2}} ≤ 2^{−j+4}

If N(S(x)) = ∞, then y → x implies j → ∞ in the above computations and hence f is twice Peano differentiable at x.

For N(S(x)) < ∞ we have to modify the above computations a little. We can obviously suppose that y is so close to x that m > n(N(S(x)), S(x)). Then j = N(S(x)) and we also have x_{i} − x_{i−1} = 0 for i ≥ j + 1. This implies that

| ∑_{i=j+1}^{∞} 2^{−i}(x_{i} − x_{i−1})(x − x_{i}) + ∑_{i=j+1}^{∞} 2^{−i}(x_{i} − x_{i−1})(y − x)−

∑_{i=j+1}^{∞} 2^{−i}(y_{i} − y_{i−1})(y − y_{i})| = | ∑_{i=j+1}^{∞} 2^{−i}(y_{i} − y_{i−1})(y − y_{i})|.

From j = N(S(x)) < ∞ it also follows that S(x) has a zero at the (m + 1)’th place and S(y) has a one there. Thus |y − y_{i}| ≤ d_{m+1} for i ≥ j + 1.

Hence

| ∑_{i=j+1}^{∞} 2^{−i}(y_{i} − y_{i−1})(y − y_{i})| ≤ ∑_{i=j+1}^{∞} \frac{d_{m}}{2^{i}} d_{m+1} ≤ d_{m} · d_{m+1}.

Therefore

\frac{|f(y) − f(x) − f_{(1)}(x)(y − x) − \frac{f_{(2)}(x)}{2!}(y − x)^{2}|}{|y − x|^{2}} ≤ \frac{d_{m} · d_{m+1}}{2^{−2} · d_{m}^{2}} .

Using that \frac{d_{m+1}}{d_{m}} → 0 as m → ∞, and y → x implies m → ∞ we obtain that in this case f is also twice Peano differentiable at x.

Suppose that there exists a g ∈ D^{2}(R) such that g|_{F} = f. From this it follows that for every x ∈ R we can find a δ(x) > 0 such that |y − x| < δ(x) implies

|g(y) − g(x) − g_{(1)}(x)(y − x) − \frac{g_{(2)}(x)}{2!}(y − x)^{2}| ≤ (y − x)^{2}.

Since F is a non-empty perfect set, by the Baire Category Theorem we can find a portion I ∩ F of F such that denoting its length by h the set {x : δ(x) > 2 · h} is dense in I ∩ F. Choose an x ∈ I ∩ F such that S(x) contains finitely many, say j, ones and the last one in the sequence S(x) is the m_{0}’th term.

Since g is, by assumption, twice Peano differentiable at x, we can choose a 0 < η < h such that |y − x| < η implies that

|g(y) − g(x) − g_{(1)}(x)(y − x) − \frac{g_{(2)}(x)}{2!}(y − x)^{2}| < 2^{−3(j+4)}(y − x)^{2}.

Choose a y ∈ I ∩ F such that δ(y) > 2h, 0 < y − x < η/4, and S(x)|m_{0} = S(y)|m_{0}. Denote by m_{1}(≥ m_{0}) the number chosen so that S(x)|m_{1} = S(y)|m_{1} and S(x)|(m_{1} + 1) ≠ S(y)|(m_{1} + 1). Since y − x > 0, it is obvious that the (m_{1} + 1)’th terms of S(y) and S(x) are one and zero respectively. We also have

d_{m_{1}}/2 < y − x < d_{m_{1}} < 2(y − x) < η/2,

x_{j} = y_{j} = x, and x_{i} = y_{i} for i = 1, ..., j − 1,
x_{i} = x < y_{i} for i ≥ j + 1,
y_{j+1} − x ≥ d_{m_{1}}/2.
Using also that z_{k} > z_{k−1} for every z ∈ F, and k = 1, 2, ... we obtain

|g_{(1)}(y) − g_{(1)}(x)| = | ∑_{i=1}^{∞} 2^{−i}(y_{i} − y_{i−1}) − ∑_{i=1}^{j−1} 2^{−i}(x_{i} − x_{i−1})| =

2^{−j−1}(y_{j+1} − y_{j}) + ∑_{i=j+2}^{∞} 2^{−i}(y_{i} − y_{i−1} − 0) ≥ 2^{−j−1}(y_{j+1} − y_{j}) =

2^{−j−1}(y_{j+1} − x) ≥ 2^{−j−2} · d_{m_{1}}.

Since g_{(2)}(z) = f_{(2)}(z) = 0 for z ∈ F and |y − x| < η implies that

|g(y) − g(x) − g_{(1)}(x)(y − x) − \frac{g_{(2)}(x)}{2!}(y − x)^{2}| =

|g(y) − g(x) − g_{(1)}(x)(y − x)| < 2^{−3(j+4)}(y − x)^{2}

we infer that for c ∈ R we have

|g(x) + g_{(1)}(x)(y + c · d_{m_{1}} − x) + \frac{g_{(2)}(x)}{2!}(y + c · d_{m_{1}} − x)^{2}−

g(y) − g_{(1)}(y)(c · d_{m_{1}}) − \frac{g_{(2)}(y)}{2!}(c · d_{m_{1}})^{2}| ≥

| |g_{(1)}(x) − g_{(1)}(y)|c · d_{m_{1}} | − |g(y) − g(x) − g_{(1)}(x)(y − x)| > |c| · 2^{−j−2} · d_{m_{1}}^{2} − 2^{−3(j+4)}d_{m_{1}}^{2}.

From the choice of η, |y − x| < η/4, and d_{m_{1}} < η/2. It follows that for |c| < 1 we have |y + c · d_{m_{1}} − x| < η and hence

|g(y + c · d_{m_{1}}) − g(x) − g_{(1)}(x)(y + c · d_{m_{1}} − x)| < 2^{−3(j+4)}(2d_{m_{1}})^{2}.

Furthermore from δ(y) > 2h > η > d_{m_{1}} it follows that

|g(y + c · d_{m_{1}}) − g(y) − g_{(1)}(y) · c · d_{m_{1}} | < c^{2} · d_{m_{1}}^{2}.

The last three displayed inequalities imply that

(4 · 2^{−3(j+4)} + c^{2})d_{m_{1}}^{2} > (|c| · 2^{−j−2} − 2^{−3(j+4)})d_{m_{1}}^{2}.

Choosing c = 2^{−(j+4)} we obtain

2^{−3(j+4)+2} + 2^{−2(j+4)} > 2^{−(2j+6)} − 2^{−3j−12} =
2^{−(j+4)}(2^{−(j+2)} − 2^{−2j−8}) > 2^{−(j+4)} · 2^{−(j+3)} = 2^{−2j−7}

and hence dividing by 2^{−2j−7} we obtain 2^{−j−3} + 2^{−1} > 1 a contradiction proving our theorem.

References

[J] V.Jarník, Sur l’extension du domaine de définition des fonctions d’une variable, qui laisse intacte la dérivabilité de la fonction, Bull. international de l’Acad. Sci. de Bohême, 1923.

[PL] Gy.Petruska and M.Laczkovich, Baire 1 functions, approximately continuous functions and derivatives, Acta Math. Acad. Sci. Hung. (25) 1974, 189-212.