A generalized function whose kth derivative is 0

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Homework Statement


Let f be a distribution on R and suppose that its kth derivative is 0. Prove that f is a polynomial.

2. The attempt at a solution
I honestly haven't a clue how to start. If I could treat f like a "regular" function, this would so easy.
 
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Maybe worth to try integrating k times?=)
 
I believe integration does not exist for distributions so that will not work.
 
Well the first question to ask it "what does the derivative of a distribution mean?", i.e. what's the definition of the derivative of a distribution?
 
The derivative f' is the distribution defined by (f', g) = -(f, g'), where g is any test function.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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