For k Z: T( x ) = T( x + 0 ) = T( x ) + T( 0 ) T( 0 ) = 0 0 = T( 0 ) = T( x + (– x )) = T( x ) + T(– x ) T(– x ) = –T( x ) and similarly for T(–2 x ) etc.
2 1 x +
2 1 x ) = T(
2 1 x )+T(
2 1 x ) = 2T(
2 1 x ) T(
2 1 x ) =
2 1 T( x )
For k Q:
T( x ) = T(
1 x ) etc. then combine for T( 3
2 x ) etc.
and similarly for T( 3
The question now seems to be whether we can extend this to all real numbers.
Condition (2) does not imply condition (1) for all complex values of k
As a quick aside, we can note if we allow complex number then we can find a transformation which
z z
z z
) Re( ) Re(
1
1
satisfies (2) but not (1), e.g. T
=
satisfies (1) but not (1) for complex k .
2
2
i
i
i
i
3 2
1 3 1 2 ) =
1
3 2
3 2
3 2
For example, T( i
) = T(
≠ i T
= i
=
i
i
i
i
1
Returning to real numbers
The key question remains, does condition (2) imply condition (1) for all real k ?
It is perhaps worth noting that if T is known to be continuous then it would follow from the implication being true for rational k that it was true for all real k . The argument can be made more rigorous than what follows, but the essential idea is that this follows from the rational numbers being what is called dense, meaning that in any small interval you can always find a new rational number, e.g. between 3 1 and 4 1 there is 7 2 . If the implication from (2) to (1) failed at an irrational number k but was true for arbitrarily close rational numbers either side of k then the transformation would not be continuous at k . It follows that if a transformation is continuous then (2) implies (1) for all real k . It might be pointed out that if a transformation is linear then it is continuous. Again a non-rigorous argument for this would be to note that if it is discontinuous then linearity fails at points close to the discontinuity. This is not really the point though – we are looking at whether (1) and (2) are independent, in particular whether (2) implies (1). Any function which satisfies (2) but not (1) will be non-linear as it fails (1) and so noting that linearity implies continuity does not help answer the question.
A transformation exists which satisfies (2) but not (1) for some real k
We now come to the most interesting part of the issue, I believe, namely that there are transformations which satisfy (2) but not (1), without the need for complex numbers. The following example comes from chapter 2 section 3 of ‘Real Variables’ by A. Torchinsky, under a section titled ‘Applications of Zorn’s lemma’.
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