Abstract :
Consider an operator T :C1(R)→C(R) satisfying the Leibniz rule functional equation
T (f · g) = (Tf ) · g +f · (T g), f, g ∈ C1(R).
We prove that all solution operators T have the form
Tf (x) = c(x)f (x) +d(x)f (x) ln f (x) , f∈ C1(R), x ∈ R
where c, d ∈ C(R) are suitable continuous functions. If T acts on the smaller space Ck(R) for some k 2,
there are no further solutions. If T maps all of C(R) into C(R), c = 0 and we only have the entropy
function cf ln |f | solution. We also consider the case of C1-functions f :Rn → R. More generally, if
T :C1(R)→C(R) and A1,A2 :C(R)→C(R) are operators satisfying the generalized Leibniz rule equation
T (f · g) = (Tf ) · (A1g) + (A2f ) · (T g), f, g ∈ C1(R),
and some weak additional assumptions, the operators A1 and A2 are of a very restricted type and anyTf (x) = c(x)f (x) f (x)
p(x) sgn f (x) +d(x) ln f (x)
f (x)
p(x)+1 sgnf (x) .
Here c, d,p ∈ C1(R) are continuous functions with Im(p) ⊂ [0,∞) and the factor {sgnf (x)} may be
present or not, yielding two different solutions. If c = 0, A1 and A2 must be equal and are uniquely determined
by T ,
A1f (x) = A2f (x) = f (x)
p(x)+1 sgnf (x) .
In the case that c(x) = 0, we show that there are two further types of solutions of the functional equation
depending only on x and f (x).
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