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.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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