find a linear differentiation transformation

[yanyin]

let C^2 be the set of functions with domain R with have first and second derivatives at all points. Defind a linear transformation T: C^2 > C^2 by T(f) = f'', in other words, each input function is mapped to its second derivative function.
what is the kernel of this transformation?

[matt grime]

that isn't the usual definition of C^2, and moreover the map you define isn't a map from C^2 to C^2, but apart from that....

[phoenixthoth]

the kernel is the set of vectors, in this case functions, that are mapped to the zero vector, in this case the zero function.

so your question is the same as asking you to list all functions whose second derivatives are zero. one direction is pretty easy: it's a certain type of polynomial. the other direction, proving that those are the only ones with vanishing second derivatives, is a bit harder and the easiest way i can think of to prove that is to apply the mean value theorem twice (or rolle's theorem).

[matt grime]

it genuinely is more delicate than that - the map given is not well defined. Let f be a function that does not have a third derivative, but has the first two, then T(f) is not in C^2 (there ought to be a constraint on the second derivative being continuous too) T is a map from C^{n+2} to C^{n}. Example: integrate |x| twice the resulting function is twice continuously differentiable, but its image under T is not in C^2, hell it's not even in C^1

[phoenixthoth]

hmm... i guess i overestimated the power of underestimation.

[matt grime]

if the question were rephrased to sidestep these issues then what you did is valid