Find a linear differentiation transformation

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The discussion centers on defining a linear transformation T: C^2 > C^2, where T(f) = f'', mapping functions to their second derivatives. Participants highlight that the kernel of this transformation consists of functions whose second derivatives are zero, primarily certain types of polynomials. However, there is contention regarding the definition of C^2, as the transformation may not be well-defined if the input function lacks a third derivative. An example illustrates that integrating a function like |x| twice results in a function not belonging to C^2, raising concerns about continuity. The conversation suggests that rephrasing the question could clarify these issues and validate the original approach.
yanyin
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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?
 
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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...
 
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).
 
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
 
hmm... i guess i overestimated the power of underestimation.
 
if the question were rephrased to sidestep these issues then what you did is valid
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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