Fourier Transforms: Jump Discontinuities & Continuous Functions in G(R)

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If a function belongs to G(R) but has points that are jump discontinuities, it's Fourier transform will not belong to G(R).
But would it be correct to claim that if a function in G(R) is continuous than its Fourier transform also belongs to G(R)? I guess it's not true, but can someone suggest a counterexample?
Thanks.
 
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Frst, what do YOU mean by "G(R)"?
 
Sorry, I wasn't sure whether this sign is well known.
G(R) is the space of functions that might have points of discontinuity only of first kind, and which are absolutely integrable.
 
ideas? anyone?...:frown:
 

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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