Operator equations and domain dependence Hermann Knig Kiel, Germany - - PowerPoint PPT Presentation

Operator equations and domain dependence

Hermann König

Kiel, Germany

Bedlewo, July 2014

Hermann König (Kiel) Operator equations Bedlewo, July 2014 1 / 37

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Hermann König (Kiel) Operator equations Bedlewo, July 2014 2 / 37

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Hermann König (Kiel) Operator equations Bedlewo, July 2014 3 / 37

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Hermann König (Kiel) Operator equations Bedlewo, July 2014 4 / 37

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Hermann König (Kiel) Operator equations Bedlewo, July 2014 5 / 37

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Hermann König (Kiel) Operator equations Bedlewo, July 2014 6 / 37

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Hermann König (Kiel) Operator equations Bedlewo, July 2014 7 / 37

Operator equations

Basic questions Olek’s interest included: Isomorphic classification of C k(Ω)-spaces and of Sobolev spaces. These are natural domains for differential operators. We consider operator equations in C k-spaces modeling derivatives. Joint work with Vitali Milman.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 8 / 37

Operator equations

Basic questions Olek’s interest included: Isomorphic classification of C k(Ω)-spaces and of Sobolev spaces. These are natural domains for differential operators. We consider operator equations in C k-spaces modeling derivatives. Joint work with Vitali Milman. Aim: Characterize derivatives by simple properties, like the Chain rule: D(f ◦ g) = (Df ) ◦ g · Dg ; f , g ∈ C 1(R), D2(f ◦ g) = (D2f ) ◦ g · (Dg)2 + (Df ) ◦ g · D2g ; f , g ∈ C 2(R).

Hermann König (Kiel) Operator equations Bedlewo, July 2014 8 / 37

Operator equations

Basic questions Olek’s interest included: Isomorphic classification of C k(Ω)-spaces and of Sobolev spaces. These are natural domains for differential operators. We consider operator equations in C k-spaces modeling derivatives. Joint work with Vitali Milman. Aim: Characterize derivatives by simple properties, like the Chain rule: D(f ◦ g) = (Df ) ◦ g · Dg ; f , g ∈ C 1(R), D2(f ◦ g) = (D2f ) ◦ g · (Dg)2 + (Df ) ◦ g · D2g ; f , g ∈ C 2(R). Replace D or D2 by a general operator T : C k(R) → C(R) and study T(f ◦ g) = (Tf ) ◦ g · Tg ; f , g ∈ C k(R), (1) T(f ◦ g) = (Tf ) ◦ g · A1g + (A2f ) ◦ g · Tg ; f , g ∈ C k(R), (2) with A1, A2 : C l(R) → C(R) , l < k. Similar equations for the Leibniz rule.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 8 / 37

Operator equations

Basic questions Olek’s interest included: Isomorphic classification of C k(Ω)-spaces and of Sobolev spaces. These are natural domains for differential operators. We consider operator equations in C k-spaces modeling derivatives. Joint work with Vitali Milman. Aim: Characterize derivatives by simple properties, like the Chain rule: D(f ◦ g) = (Df ) ◦ g · Dg ; f , g ∈ C 1(R), D2(f ◦ g) = (D2f ) ◦ g · (Dg)2 + (Df ) ◦ g · D2g ; f , g ∈ C 2(R). Replace D or D2 by a general operator T : C k(R) → C(R) and study T(f ◦ g) = (Tf ) ◦ g · Tg ; f , g ∈ C k(R), (1) T(f ◦ g) = (Tf ) ◦ g · A1g + (A2f ) ◦ g · Tg ; f , g ∈ C k(R), (2) with A1, A2 : C l(R) → C(R) , l < k. Similar equations for the Leibniz rule.

• General solutions of (1) and (2) ?
• Dependence of the solutions on the domain and range spaces?

Hermann König (Kiel) Operator equations Bedlewo, July 2014 8 / 37

Chain rule operator equation The chain rule functional equation

The chain rule functional equation Chain rule for f , g ∈ C 1(R) : D(f ◦ g) = (Df ) ◦ g · Dg. Let T : C 1(R) → C(R) be an operator satisfying the functional equation T(f ◦ g)(x) = (Tf )(g(x)) · (Tg)(x); f , g ∈ C 1(R), x ∈ R. (1) Which operators T satisfy (1)?

Hermann König (Kiel) Operator equations Bedlewo, July 2014 9 / 37

Chain rule operator equation The chain rule functional equation

The chain rule functional equation Chain rule for f , g ∈ C 1(R) : D(f ◦ g) = (Df ) ◦ g · Dg. Let T : C 1(R) → C(R) be an operator satisfying the functional equation T(f ◦ g)(x) = (Tf )(g(x)) · (Tg)(x); f , g ∈ C 1(R), x ∈ R. (1) Which operators T satisfy (1)? Examples: a) p > 0, (Tf )(x) = |f ′(x)|p and (Tf )(x) = sgn f ′(x) |f ′(x)|p both satisfy (1). b) Let H ∈ C(R), H > 0. Define (Tf )(x) = H(f (x))/H(x). Then T satisfies (1). c) Consider T : C 1(R) → C(R), (Tf )(x) =

• f ′(x)

f ∈ C 1(R) bijective else

• .

Then T satisfies (1).

Hermann König (Kiel) Operator equations Bedlewo, July 2014 9 / 37

Chain rule operator equation Solutions of the chain rule operator equation

Solutions of the chain rule operator equation T : C k(R) → C(R) is C k-non-degenerate if T|C k

b (R) = 0 where C k

b (R) are

the (half-) bounded functions in C k(R). Here k ∈ N0. Multiplying two solutions of the chain rule yields again a solution.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 10 / 37

Chain rule operator equation Solutions of the chain rule operator equation

Solutions of the chain rule operator equation T : C k(R) → C(R) is C k-non-degenerate if T|C k

b (R) = 0 where C k

b (R) are

the (half-) bounded functions in C k(R). Here k ∈ N0. Multiplying two solutions of the chain rule yields again a solution.

Theorem

Assume T : C k(R) → C(R) satisfies the chain rule T(f ◦ g) = (Tf ) ◦ g · Tg; f , g ∈ C k(R) for k ∈ N0 and that T is C k-non-degenerate. Then there is p ≥ 0 and H ∈ C>0(R) such that for any f ∈ C k(R) (Tf )(x) = H(f (x)) H(x) |f ′(x)|p{sgn f ′(x)}, and this also holds for k = ∞, i.e. f ∈ C ∞(R).

Hermann König (Kiel) Operator equations Bedlewo, July 2014 10 / 37

Chain rule operator equation Chain rule operator equation

Steps in the Proof a) Show localization: There is F : Rk+2 → R such that Tf (x) = F(x, f (x), ..., f (k)(x)) for all f ∈ C k(R) and x ∈ R. b) Analyze the structure of the representing function F : F(x, α0, ..., αk) = H(α0)

H(x) K(α1), K multiplicative,

F independent of α2, ..., αk if k ≥ 2. c) Show the measurability and then the continuity of the coefficient functions occurring in F.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 11 / 37

Chain rule operator equation Localization

Localization in the case k = 1

Proposition

If T satisfies the chain rule (1) in C 1(R) and is C 1-non-degenerate, there is F : R3 → R such that (Tf )(x) = F(x, f (x), f ′(x)); x ∈ R, f , g ∈ C 1(R).

Hermann König (Kiel) Operator equations Bedlewo, July 2014 12 / 37

Chain rule operator equation Localization

Localization in the case k = 1

Proposition

If T satisfies the chain rule (1) in C 1(R) and is C 1-non-degenerate, there is F : R3 → R such that (Tf )(x) = F(x, f (x), f ′(x)); x ∈ R, f , g ∈ C 1(R). Step 1. First show localization on open intervals J ⊂ R: a) If g|J = Id, Tg|J = 1: For x ∈ J, there are h ∈ C 1(R), y ∈ R with Imh ⊂ J, h(y) = x and Th(y) = 0. Then g ◦ h = h and 0 = Th(y) = Tg(x) Th(y), i.e. Tg(x) = 1, Tg|J = 1.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 12 / 37

Chain rule operator equation Localization

Localization in the case k = 1

Proposition

If T satisfies the chain rule (1) in C 1(R) and is C 1-non-degenerate, there is F : R3 → R such that (Tf )(x) = F(x, f (x), f ′(x)); x ∈ R, f , g ∈ C 1(R). Step 1. First show localization on open intervals J ⊂ R: a) If g|J = Id, Tg|J = 1: For x ∈ J, there are h ∈ C 1(R), y ∈ R with Imh ⊂ J, h(y) = x and Th(y) = 0. Then g ◦ h = h and 0 = Th(y) = Tg(x) Th(y), i.e. Tg(x) = 1, Tg|J = 1. b) If f1|J = f2|J for f1, f2 ∈ C 1(R), then (Tf1)|J = (Tf2)|J: For x ∈ J, there are J1 ⊂ J, g ∈ C 1(R) with x ∈ J1, g|J1 = Id, Img ⊂ J. Hence Tg(x) = 1, f1 ◦ g = f2 ◦ g. Thus Tf1(x) = Tf1(g(x)Tg(x) = T(f1 ◦ g)(x) = T(f2 ◦ g)(x) = Tf2(x), i.e. Tf1|J = Tf2|J.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 12 / 37

Chain rule operator equation Localization

Step 2. For f ∈ C 1(R) and x0 ∈ R, define g to be the tangent line at x0, g(x) := f (x0) + f ′(x0)(x − x0) ; x ∈ R. Let h(x) := f (x) x < x0 g(x) x ≥ x0

• . By construction h ∈ C 1(R).

Let I1 = (−∞, x0), I2 = (x0, ∞). Then f |I1 = h |I1, h |I2 = g |I2. By Step 1, (Tf ) |I1 = (Th) |I1, (Th) |I2 = (Tg) |I2.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 13 / 37

Chain rule operator equation Localization

Step 2. For f ∈ C 1(R) and x0 ∈ R, define g to be the tangent line at x0, g(x) := f (x0) + f ′(x0)(x − x0) ; x ∈ R. Let h(x) := f (x) x < x0 g(x) x ≥ x0

• . By construction h ∈ C 1(R).

Let I1 = (−∞, x0), I2 = (x0, ∞). Then f |I1 = h |I1, h |I2 = g |I2. By Step 1, (Tf ) |I1 = (Th) |I1, (Th) |I2 = (Tg) |I2. Since Tf , Th, Tg are continuous functions, this also holds on {x0} = ¯ I1 ∩ ¯ I2, i.e. (Tf )(x0) = (Th)(x0) = (Tg)(x0). But the tangent g only depends on x0, f (x0), f ′(x0); (Tf )(x0) = F(x0, f (x0), f ′(x0)).

• Hermann König (Kiel)

Operator equations Bedlewo, July 2014 13 / 37

Chain rule operator equation Analysis of the function F

Analysis of the function F We now analyze the function F with (Tf )(x) = F(x, f (x), f ′(x)). For x0, y0, z0 ∈ R and u, v ∈ R, choose f , g ∈ C 1(R) with g(x0) = y0, f (y0) = z0, g′(x0) = u, f ′(y0) = v

Hermann König (Kiel) Operator equations Bedlewo, July 2014 14 / 37

Chain rule operator equation Analysis of the function F

Analysis of the function F We now analyze the function F with (Tf )(x) = F(x, f (x), f ′(x)). For x0, y0, z0 ∈ R and u, v ∈ R, choose f , g ∈ C 1(R) with g(x0) = y0, f (y0) = z0, g′(x0) = u, f ′(y0) = v Since T(f ◦ g)(x0) = (Tf )(y0) · (Tg)(x0), we have F(x0, z0, vu) = F(y0, z0, v)F(x0, y0, u). F(x0, x0, vu) = F(y0, x0, v)F(x0, y0, u) = F(x0, y0, u)F(y0, x0, v) = F(y0, y0, vu) Hence K(u) := F(x0, x0, u) is independent of x0 ∈ R and for y0 = x0 K(uv) = K(u)K(v), K(1) = 1, K measurable , K(u) = |u|p or K(u) = |u|p sgn(u). F(x0, y0, u) = G(x0, y0)K(u), G(x0, y0) := 1/F(y0, x0, 1).

Hermann König (Kiel) Operator equations Bedlewo, July 2014 14 / 37

Chain rule operator equation Analysis of the function F

Analysis of F for k ≥ 2 Since (f ◦ g)(k) is complicated, the chain rule formula for F is complicated. To show that F does not depend on f ′′(x), · · · , f (k)(x), one uses inductively that (f ◦ g)(k) has the form (f ◦g)(k) = f (k)◦g·g′k+ϕk(f ′◦g, · · · , f (k−1)◦g, g′, · · · , g(k−1))+f ′◦g·g(k) and that for k ≥ 2 the power g′k is different from the one of f ′ ◦ g in F(x,(f ◦ g)(x), ..., (f ◦ g)(k)(x)) = F(g(x), f (g(x)), ..., f (k)(g(x))) F(x, g(x), ..., g(k)(x)), yielding equality of terms F(x, z, α1β1, α2β2

1 + β2α1, · · · ) = F(x, z, α1β1, α2β1 + β2α2 1, · · · ),

which is used to show independence of the last variables with indices ≥ 2.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 15 / 37

Second order chain rule Second order chain rule formula

Second order chain rule formulas For f , g ∈ C 2(R) one has D2(f ◦ g) = D2f ◦ g · g′2 + f ′ ◦ g · D2g.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 16 / 37

Second order chain rule Second order chain rule formula

Second order chain rule formulas For f , g ∈ C 2(R) one has D2(f ◦ g) = D2f ◦ g · g′2 + f ′ ◦ g · D2g. We study the solutions of a generalized operator functional equation: Let k ≥ 2 and T : C k(R) → C(R), A1, A2 : C k−1(R) → C(R) be such that T(f ◦ g) = Tf ◦ g · A1g + A2f ◦ g · Tg; f , g ∈ C k(R). (2)

Hermann König (Kiel) Operator equations Bedlewo, July 2014 16 / 37

Second order chain rule Second order chain rule formula

Second order chain rule formulas For f , g ∈ C 2(R) one has D2(f ◦ g) = D2f ◦ g · g′2 + f ′ ◦ g · D2g. We study the solutions of a generalized operator functional equation: Let k ≥ 2 and T : C k(R) → C(R), A1, A2 : C k−1(R) → C(R) be such that T(f ◦ g) = Tf ◦ g · A1g + A2f ◦ g · Tg; f , g ∈ C k(R). (2) A1 is isotropic if it commutes with all shift operators. (T, A1) is C k-non-degenerate if for all open sets J ⊂ R and x ∈ J there are y1, y2 ∈ R and g1, g2 ∈ C k(R) such that g1(y1) = x = g2(y2) and zi = (Tgi(yi), A1gi(yi)) ∈ R2 are linearly independent for i = 1, 2.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 16 / 37

Second order chain rule Second order chain rule formula

Second order chain rule formulas For f , g ∈ C 2(R) one has D2(f ◦ g) = D2f ◦ g · g′2 + f ′ ◦ g · D2g. We study the solutions of a generalized operator functional equation: Let k ≥ 2 and T : C k(R) → C(R), A1, A2 : C k−1(R) → C(R) be such that T(f ◦ g) = Tf ◦ g · A1g + A2f ◦ g · Tg; f , g ∈ C k(R). (2) A1 is isotropic if it commutes with all shift operators. (T, A1) is C k-non-degenerate if for all open sets J ⊂ R and x ∈ J there are y1, y2 ∈ R and g1, g2 ∈ C k(R) such that g1(y1) = x = g2(y2) and zi = (Tgi(yi), A1gi(yi)) ∈ R2 are linearly independent for i = 1, 2.

Note: For A1 = A2 = 1

2T, (2) is just the chain rule operator equation (1).

Non-degeneration excludes this. Equation (2) resembles the addition formula for the sin-function. Let f1, f2 be smooth functions with f1(0) = 0, f ′(0) = 1, f ′′

1 (0) = 0 and

f2(0) = f ′′

2 (0) = 0, f ′ 2(0) = 1, f ′′′ 2 (0) = 0, e.g. f1(x) = x + x2/2, f2(x) = x + x3/6. Hermann König (Kiel) Operator equations Bedlewo, July 2014 16 / 37

Second order chain rule Second order chain rule equation

Second order chain rule equation

Theorem

Let T : C k(R) → C(R) and A1, A2 : C k−1(R) → C(R) be such that T(f ◦ g) = (Tf ) ◦ g · A1g + (A2f ) ◦ g · Tg; f , g ∈ C k(R) (2) is satisfied. Suppose (T, A1) is C k-non-degenerate and that A1 and A2 are isotropic and C k−1-pointwise continuous. Assume first that (Tf2)(0) = 0 and k ≥ 3. Then there are p ≥ 2, c = 0 and a function H ∈ C(R) which is determined by T(2Id) such that A2f = |f ′|p{sgn f ′}, A1f = f ′2 · A2f Tf =

• c · (f ′f ′′′ − 3

2f ′′2) + [H ◦ f · f ′2 − H] · f ′2

• |f ′|p−2 {sgn f ′}.

Note that f ′f ′′′ − 3

2f ′′2 = f ′2 · Sf where Sf = f ′′′ f ′ − 3 2

• f ′′

f ′

2 is the Schwarzian derivative of f when f ′ = 0.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 17 / 37

Second order chain rule Second order chain rule equation

Second order chain rule equation

Theorem

Let T : C k(R) → C(R) and A1, A2 : C k−1(R) → C(R) be such such that T(f ◦ g) = (Tf ) ◦ g · A1g + (A2f ) ◦ g · Tg; f , g ∈ C k(R) (2) is satisfied. Suppose (T, A1) is C k-non-degenerate and that A1 and A2 are isotropic and C k−1-pointwise continuous. Assume that (Tf1)(0) = 0, Tf2(0) = 0 and k ≥ 2. Then there are p ≥ 1, c = 0 and a function H ∈ C(R) which is determined by T(2Id) such that A2f = |f ′|p{sgn f ′}, A1f = f ′ · A2f , Tf =

• c f ′′ + [H ◦ f · f ′ − H] · f ′

|f ′|p−1 {sgn f ′}.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 18 / 37

Second order chain rule Second order chain rule equation

Second order chain rule equation

Theorem

Same assumptions as before, except that now (Tf2)(0) = 0, Tf1(0) = 0 and k ≥ 1. Then there are three types of solutions with p ≥ 0 Tf = (c ln |f ′| + H ◦ f − H)|f ′|p{sgn f ′} A1f = A2f = |f ′|p{sgn f ′}, Tf = H ◦ f |f ′|q[sgn f ′] − H|f ′|p{sgn f ′} A1f = |f ′|q[sgn f ′], A2f = |f ′|p{sgn f ′}, Tf = c|f ′|p sin(d ln |f ′|){sgn f ′} A1f = A2f = |f ′|p cos(d ln |f ′|){sgn f ′}

Hermann König (Kiel) Operator equations Bedlewo, July 2014 19 / 37

Second order chain rule Second order chain rule equation

Second order chain rule equation Remarks. (1) All operators in the previous three Theorems satisfy the operator equation (2) and -taken together- they constitute all solutions. (2) Note that there are no solutions of (2) depending on the fourth or higher derivatives of f . (3) The main solutions for T are the second derivative f ′′, k = 2, f ′2 times the Schwarzian derivative Sf , k = 3, and ln |f ′|, k = 1. So the natural domains of solutions of (2) are C k(R) for k ∈ {1, 2, 3}. Two initial conditions may determine the form of T, e.g. T(x2) = 2, T(x3) = 6x yields Tf = f ′′, A1f = f ′2, A2f = f ′ and T(x2) = −6, T(x3) = −36x2 implies Tf = f ′2Sf , A1f = f ′4, A2f = f ′2. (4) The solutions for A1, A2 only depend on f ′. Therefore the natural domain for A1 and A2 is C 1(R). (5) The general form of (T, A1, A2) may also be determined if (A1, A2) are not isotropic. Then there are further multiplicative terms K(f )

K(Id).

Hermann König (Kiel) Operator equations Bedlewo, July 2014 20 / 37

Second order chain rule Second order chain rule equation

Localization By non-degeneracy, for I ⊂ R, x ∈ I, there are yi ∈ R, gi ∈ C k(R), gi(yi) = x, Imgi ⊂ I, zi := (Tgi(yi), A1gi(yi)) ∈ R2 linearly independent for i ∈ {1, 2}. Let f , f1, f2 ∈ C k(R). (a) If f |I = Id, f ◦ gi = gi. Hence by the functional equation (2) Tgi(yi)(1 − A2f (x)) = A1gi(yi)Tf (x). Therefore A2f (x) = 1 and Tf (x) = 0: Tf |I = 0, A2|I = 1. Also A1f |I = 1.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 21 / 37

Second order chain rule Second order chain rule equation

Localization By non-degeneracy, for I ⊂ R, x ∈ I, there are yi ∈ R, gi ∈ C k(R), gi(yi) = x, Imgi ⊂ I, zi := (Tgi(yi), A1gi(yi)) ∈ R2 linearly independent for i ∈ {1, 2}. Let f , f1, f2 ∈ C k(R). (a) If f |I = Id, f ◦ gi = gi. Hence by the functional equation (2) Tgi(yi)(1 − A2f (x)) = A1gi(yi)Tf (x). Therefore A2f (x) = 1 and Tf (x) = 0: Tf |I = 0, A2|I = 1. Also A1f |I = 1. (b) If f1|I = f2|I, f1 ◦ gi = f2 ◦ gi and by the operator equation (2) 0 = T(f1 ◦ gi) − T(f2 ◦ gi) = A1gi(yi)(Tf1(x) − Tf2(x)) + Tgi(yi)(A2f1(x) − A2f2(x)) which yields Tf1|I = Tf2|I and A2f1|I = A2f2|I. Similarly for A1.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 21 / 37

Second order chain rule Second order chain rule equation

Proposition

Under the assumptions of the Theorem, there are functions F : Rk+2 → R and A1, A2 : Rk → R such that Tf (x) = F(x, f (x), · · · , f (k)(x)) , Aif (x) = Bi(f (x), · · · , f (k−1)(x)) , i = 1, 2. The pointwise continuity of the A′

is yields that Aif (x) is independent of

f (k)(x), the isotropicity of the A′

is implies that Aif (x) is independent of x.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 22 / 37

Second order chain rule Additive and multiplicative functions

Additive functions

Proposition (Banach, Sierpinski)

Let a : R → R be additive, a(x + y) = a(x) + a(y). (a) If a is measurable, a is linear, i.e. a(x) = c · x. (b) If a is not linear, it is unbounded on any small open interval J ⊂ R.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 23 / 37

Second order chain rule Additive and multiplicative functions

Additive functions

Proposition (Banach, Sierpinski)

Let a : R → R be additive, a(x + y) = a(x) + a(y). (a) If a is measurable, a is linear, i.e. a(x) = c · x. (b) If a is not linear, it is unbounded on any small open interval J ⊂ R. Note: a(mx) = ma(x), a x

n

• = 1

na(x) for n, m ∈ N easily gives that

a(rx) = ra(x),r ∈ Q, x ∈ R. Thus continuous additive functions are linear (Cauchy).

Corollary

Let K : R → R be multiplicative, K(uv) = K(u) K(v) , u, v ∈ R, and

• measurable. Then there is p ∈ R such that

K(u) = |u|p

• r

K(u) = |u|p sgnu.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 23 / 37

Second order chain rule Additive functions

Proof of (a). Let a : R → R be measurable, t ∈ R, t = 0 be fixed and φ(x) := a(x) − a(t) t x , ψ(x) := 1 1 + |φ(x)| , x ∈ R . Then φ(x + t) = φ(x), ψ(x + t) = ψ(x), |φ(2x)| = 2|φ(x)| and t ψ(x) dx = 2t

t

ψ(y) dy = 1 2 2t ψ(y) dy = t ψ(2x) dx ,

Hermann König (Kiel) Operator equations Bedlewo, July 2014 24 / 37

Second order chain rule Additive functions

Proof of (a). Let a : R → R be measurable, t ∈ R, t = 0 be fixed and φ(x) := a(x) − a(t) t x , ψ(x) := 1 1 + |φ(x)| , x ∈ R . Then φ(x + t) = φ(x), ψ(x + t) = ψ(x), |φ(2x)| = 2|φ(x)| and t ψ(x) dx = 2t

t

ψ(y) dy = 1 2 2t ψ(y) dy = t ψ(2x) dx , 0 = t (ψ(x) − ψ(2x)) dx = t |φ(x)| (1 + |φ(x)|)(1 + 2|φ(x)|) dx . Hence φ = 0 a. e. in x, a(x) = a(t)

t

x a.e. in x. Thus for any t ∈ R there is x0 ∈ R such that a(x0) = a(t) t x0 , a(x0) = a(1) 1 x0 , i.e. a(t) = a(1) t.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 24 / 37

Second order chain rule Regularity

Continuity of additive functions

Proposition (Faifman)

Let ax,j : R → R be a family of additive functions, for j = 1, · · · , d such that for all g ∈ C ∞(R) ax,1(g(x)) + · · · + ax,d

• g(d−1)(x)
• is continuous in x ∈ R. Then ax,j(α) = aj(x)α for all α ∈ R, j = 1, · · · , d,

and the aj ∈ C(R) are continuous.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 25 / 37

Second order chain rule Analysis of the representing functions

The operator equation in terms of F, A1, A2 T(f ◦ g) = Tf ◦ g · A1g + A2f ◦ g · Tg; f , g ∈ C k(R). (2) Tf (x) = F(x, f (x), · · · , f (k)(x)) , Aif (x) = Bi(f (x), · · · , f (k−1)(x)). Equation (2) means for k = 2 (as illustration) F(x, z,α1β1, α2β2

1 + α1β2) =

F(y, z, α1, α2)B1(y, β1) + F(x, y, β1, β2)B2(z, α1) (3) for all x, y, z, α1, α2, β1, β2 ∈ R. T(Id) = 0, Ai(Id) = 1 means F(x, x, 1, 0) = 0, Bi(y, 1) = 1.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 26 / 37

Second order chain rule Analysis of the representing functions

The operator equation in terms of F, A1, A2 T(f ◦ g) = Tf ◦ g · A1g + A2f ◦ g · Tg; f , g ∈ C k(R). (2) Tf (x) = F(x, f (x), · · · , f (k)(x)) , Aif (x) = Bi(f (x), · · · , f (k−1)(x)). Equation (2) means for k = 2 (as illustration) F(x, z,α1β1, α2β2

1 + α1β2) =

F(y, z, α1, α2)B1(y, β1) + F(x, y, β1, β2)B2(z, α1) (3) for all x, y, z, α1, α2, β1, β2 ∈ R. T(Id) = 0, Ai(Id) = 1 means F(x, x, 1, 0) = 0, Bi(y, 1) = 1. For α1 = β1 = 1, F(x, z, 1, α2 + β2) = F(y, z, 1, α2) + F(x, y, 1, β2) , F(z, z, 1, α2 + β2) = F(y, z, 1, α2) + F(z, y, 1, β2) = F(y, y, α2 + β2) = F(y, y, 1, α2) + F(y, y, 1, β2) is additive and independent of y, z ∈ R. After showing measurability in α2, F(z, z, 1, α2) = c α2, c = T(x + x2/2)(0). With G(z) := −F(z, 0, 1, 0), F(x, z, 1, α2) = c α2 + G(z) − G(x) . (4)

Hermann König (Kiel) Operator equations Bedlewo, July 2014 26 / 37

Second order chain rule Analysis of the representing functions

The operator equation in terms of F, A1, A2 F(x, z,α1β1, α2β2

1 + α1β2) =

F(y, z, α1, α2)B1(y, β1) + F(x, y, β1, β2)B2(z, α1) (3) For β1 = 1, α1 = 1, using (4) F(x, z, α1, α2 + α1β′

2) = F(y, z, α1, α2) + F(x, y, 1, β′ 2)B2(z, α1)

= F(y, z, α1, α2) + (c β′

2 + G(y) − G(x)) B2(z, α1).

In terms of β2 = α1β′

2 this means if α1 = 0

F(x, z,α1, α2 + β2) = F(y, z, α1, α2) + c β2 α1

• B2(z, α1) + (G(y) − G(x))B2(z, α1)

= F(y, z, α1, β2) + c α2 α1

• B2(z, α1) + (G(y) − G(x))B2(z, α1).

Hermann König (Kiel) Operator equations Bedlewo, July 2014 27 / 37

Second order chain rule Analysis of the representing functions

The operator equation in terms of F, A1, A2 : Taking differences F(y, z, α1, α2) + c β2 α1

• B2(z, α1) = F(y, z, α1, β2) + c

α2 α1

• B2(z, α1),

F(y, z, α1, α2) = F(y, z, α1, 0) + c α2 α1

• B2(z, α1).

Insert this into (3), and regroup terms according to occurrence of α2, β2, c β2 β1

• (B2(z, α1β1) − B2(z, α1)B2(y, β1))

+

• c
• β1

α2 α1

• B2(z, α1β1) − c

α2 α1

• B2(z, α1)B1(y, β1)
• (5)

= F(y, z, α1, 0)B1(y, β1) + F(x, y, β1, 0)B2(z, α1) − F(x, z, α1β1, 0). The right side is independent of α2, β2, hence B2(z, α1β1) = B2(z, α1)B2(y, β1) , B1(y, β1) = β1B2(y, β1). This implies B2(z, α1) = |α1|p{sgnα1}.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 28 / 37

Second order chain rule Analysis of the representing functions

The operator equation in terms of F, A1, A2 Since the left side of (5) is now 0, also the right side is 0: F(x, z, α1β1, 0) = F(y, z, α1, 0)B1(y, β1) + F(x, y, β1, 0)B2(z, α1). This quickly yields F(x, z, α1, 0) = (H(z)α1 − H(x)) |α1|p {sgnα1}, F(x, z, α1, α2) = (cα2 + (α1H(z) − H(x))α1) |α1|p−1{sgnα1}. for a suitable continuous function H.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 29 / 37

Second order chain rule Analysis of the representing functions

The operator equation in terms of F, A1, A2 Since the left side of (5) is now 0, also the right side is 0: F(x, z, α1β1, 0) = F(y, z, α1, 0)B1(y, β1) + F(x, y, β1, 0)B2(z, α1). This quickly yields F(x, z, α1, 0) = (H(z)α1 − H(x)) |α1|p {sgnα1}, F(x, z, α1, α2) = (cα2 + (α1H(z) − H(x))α1) |α1|p−1{sgnα1}. for a suitable continuous function H. For T : C k(R) → C(R) with k ≥ 4 there are too many independent variables in (f ◦ g)(k)(x) so that equation (2) -or the functional equation corresponding to (3)- cannot be satisfied even for x = y = z. For k = 3, a similar analysis as above gives the Schwarzian derivative terms.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 29 / 37

Derivations in Spaces Ck (R) Leibniz rule, Derivations in Ck (R)

Leibniz rule operator equation The derivative as a map T : C 1(R) → C(R) satisfies the Leibniz rule T(f · g) = Tf · g + f · Tg ; f , g ∈ C 1(R).

Hermann König (Kiel) Operator equations Bedlewo, July 2014 30 / 37

Derivations in Spaces Ck (R) Leibniz rule, Derivations in Ck (R)

Leibniz rule operator equation The derivative as a map T : C 1(R) → C(R) satisfies the Leibniz rule T(f · g) = Tf · g + f · Tg ; f , g ∈ C 1(R).

Theorem

Let k ∈ N0 and suppose that T : C k(R) → C(R) is a map satisfying T(f · g)(x) = (Tf )(x) · g(x) + f (x) · (Tg)(x) ; f , g ∈ C 1(R), x ∈ R. Then there are continuous functions a, b ∈ C(R) such that T has the form (Tf )(x) = b(x)f ′(x) + a(x)f (x) ln |f (x)|. For k = 0, b = 0 and T extends to C(R). For k ≥ 2, T extends to C 1(R) by the same formula.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 30 / 37

Derivations in Spaces Ck (R) Leibniz rule

Remarks. (1) No continuity of T is assumed, neither linearity. (2) The case k = 0 (entropy function) is due to Goldmann, ˘ Semrl (1994) (3) Alternatively, the solution for T can be written (Tf )(x) = f (x)[b(x)(ln |f |)′(x) + a(x)(ln |f |)(x)]. (4) If T also maps C 2(R) into C 1(R), Tf = bf ′. (5) The Theorem can be extended to functions f : Rn → R. If T : C 1(Rn, R) → C(Rn, Rn) satisfies T(f ·g)(x) = (Tf )(x)·g(x)+f (x)·(Tg)(x) ; f , g ∈ C 1(Rn, R), x ∈ Rn, there are functions b ∈ C(Rn, L(Rn, Rn)) and a ∈ C(Rn, Rn) such that (Tf )(x) = b(x)f ′(x) + a(x)f (x) ln |f (x)|.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 31 / 37

Derivations in Spaces Ck (R) Second order derivations

Second order derivations For f , g ∈ C 2(R), (f · g)′′ = f ′′ · g + f · g′′ + 2 · f ′ · g′. We replace this equation by a general operator equation on C 2(R), T(f · g) = Tf · g + f · Tg + Af · Ag, (6) where T : C 2(R) → C(R) and A : C 2(R) → C(R) are unknown operators.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 32 / 37

Derivations in Spaces Ck (R) Second order derivations

Second order derivations For f , g ∈ C 2(R), (f · g)′′ = f ′′ · g + f · g′′ + 2 · f ′ · g′. We replace this equation by a general operator equation on C 2(R), T(f · g) = Tf · g + f · Tg + Af · Ag, (6) where T : C 2(R) → C(R) and A : C 2(R) → C(R) are unknown operators. Clearly (6) contains additively the solutions S of the homogeneous equation. S(f · g) = Sf · g + f · Sg, Sf (x) = b(x) f ′(x) + a(x) f (x) ln |f (x)|, with the same A. A is assumed to be non-degenerate in the following sense:

• Definition. An operator A : C 2(R) → C(R) is non-degenerate if for any
• pen interval J ⊂ R and any x ∈ J there are functions g1, g2 ∈ C 2(R) with

support in J such that the two vectors (gi(x), Agi(x)) ∈ R2, i ∈ {1, 2} are linearly independent.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 32 / 37

Derivations in Spaces Ck (R) Solution of the second order derivation equation

Solution of the second order derivation equation

Theorem

Let T : C 2(R) → C(R) and A : C 2(R) → C(R) be operators such that T(f · g) = Tf · g + f · Tg + Af · Ag; f , g ∈ C 2(R) (6)

• holds. Assume A is non-degenerate. Then there are a, b, d, e, p ∈ C(R) such that with

(Sf )(x) := b(x)f ′(x) + a(x)f (x) ln |f (x)|, x ∈ R all operators T and A verifying (6) are of one of the following three types (Tf )(x) = (T1f )(x) + (Sf )(x),

Hermann König (Kiel) Operator equations Bedlewo, July 2014 33 / 37

Derivations in Spaces Ck (R) Solution of the second order derivation equation

Solution of the second order derivation equation

Theorem

Let T : C 2(R) → C(R) and A : C 2(R) → C(R) be operators such that T(f · g) = Tf · g + f · Tg + Af · Ag; f , g ∈ C 2(R) (6)

• holds. Assume A is non-degenerate. Then there are a, b, d, e, p ∈ C(R) such that with

(Sf )(x) := b(x)f ′(x) + a(x)f (x) ln |f (x)|, x ∈ R all operators T and A verifying (6) are of one of the following three types (Tf )(x) = (T1f )(x) + (Sf )(x),

• T1f (x) = d(x)2

2

f ′′(x), Af (x) = d(x)f ′(x)

• T1f (x) = e(x)2

2

f (x)(ln |f (x)|)2, Af (x) = e(x)f (x) ln |f (x)|

• T1f (x) = e(x)2f (x)({sgn f (x)} |f (x)|p(x) − 1),

Af (x) = e(x)f (x)({sgn f (x)} |f (x)|p(x) − 1) Conversely, these operators (T, A) satisfy (6).

Hermann König (Kiel) Operator equations Bedlewo, July 2014 33 / 37

Derivations in Spaces Ck (R) Idea of the Proof

Idea of the Proof First Step. Using that A is non-degenerate, one shows localization, first

• n intervals, and then pointwise:

Tf (x) = Fx(f (x), f ′(x), f ′′(x)), Af (x) = Bx(f (x), f ′(x)). (7)

Hermann König (Kiel) Operator equations Bedlewo, July 2014 34 / 37

Derivations in Spaces Ck (R) Idea of the Proof

Idea of the Proof First Step. Using that A is non-degenerate, one shows localization, first

• n intervals, and then pointwise:

Tf (x) = Fx(f (x), f ′(x), f ′′(x)), Af (x) = Bx(f (x), f ′(x)). (7) Analysis of Fx. Separate variables α2 = f ′′(x), α1 = f ′(x), α0 = f (x) as much as possible, get independent functional equations: Insert (7) into (6) for relations between Fx and Bx. Find Fx(α0, α1, α2) = α0

• cx

α2 α0

• + Fx
• 1, α1

α0 , 0

• + Fx(α0, 0, 0)

+ Bx(α0, 0)Bx

• 1, α1

α0

• .

Here Fx(α0, 0, 0) and Fx(1, α1, 0) satisfy two functional equations with three possible solution forms each. They are given in the following two Propositions.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 34 / 37

Derivations in Spaces Ck (R) Functional Equations

Functional Equations F = Fx(1, ·, 0) is a solution of the following equation.

Proposition

Let F, B : R → R be functions such that F(α + β) = F(α) + F(β) + B(α)B(β); α, β ∈ R.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 35 / 37

Derivations in Spaces Ck (R) Functional Equations

Functional Equations F = Fx(1, ·, 0) is a solution of the following equation.

Proposition

Let F, B : R → R be functions such that F(α + β) = F(α) + F(β) + B(α)B(β); α, β ∈ R. Then there are additive functions c, d : R → R and γ ∈ R such that F and B are of one of the following forms (a) F(α) = 1

2(c[α])2 + d[α],

B(α) = c[α] (b) F(α) = γ2(ec[α] − 1) + d[α], B(α) = γ(ec[α] − 1) (c) F(α) = −γ2 + d[α], B(α) = γ.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 35 / 37

Derivations in Spaces Ck (R) Functional Equations

F = Fx(·, 0, 0) is a solution of the following equation.

Proposition

Let F, B : R → R be functions such that F(αβ) = F(α)β + F(β)α + B(α)B(β); α, β ∈ R.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 36 / 37

Derivations in Spaces Ck (R) Functional Equations

F = Fx(·, 0, 0) is a solution of the following equation.

Proposition

Let F, B : R → R be functions such that F(αβ) = F(α)β + F(β)α + B(α)B(β); α, β ∈ R. Then there are additive functions c, d : R → R and γ ∈ R such that F and B are of one of the following forms (a) F(α) = α 1

2(c[ln |α|])2 + d[ln |α|]

• ,

B(α) = αc[ln |α|] (b) F(α) = α

• γ2(ec[ln |α|] − 1) + d[ln |α|]
• ,

B(α) = αγ(ec[ln |α|] − 1) (c) F(α) = α

• −γ2 + d[ln |α|]
• ,

B(α) = αγ.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 36 / 37

Derivations in Spaces Ck (R) A case of non-localization

Example. If T, A : C 2(R) → C(R) satisfy T(f · g) = Tf · g + f · Tg + Af · Ag , f , g ∈ C 2(R) , and A is degenerate, in general there are non-localized solutions:

Hermann König (Kiel) Operator equations Bedlewo, July 2014 37 / 37

Derivations in Spaces Ck (R) A case of non-localization

Example. If T, A : C 2(R) → C(R) satisfy T(f · g) = Tf · g + f · Tg + Af · Ag , f , g ∈ C 2(R) , and A is degenerate, in general there are non-localized solutions: Let φ, p : R → R be continuous functions with φ(x) = x. Define operators T, A : C 2(R) → C(R) by (Tf )(x) = −f (x) + |f (φ(x))|p(x) , (Af )(x) = f (x) − |f (φ(x))|p(x). One quickly checks that (6) is satisfied by T and A. On intervals J of length < |φ(x) − x| around x ∈ J and functions f with support in J, (Af )(x) = f (x). Therefore A is degenerate. T and A are not locally defined in one point x, depending also on φ(x). These operators T and A remind of the third solution of (6) in the case of φ(x) = x.

Hermann König (Kiel) Operator equations Bedlewo, July 2014 37 / 37