3

Is this double integral $$(f,g)=\int_{x=0}^1\int_{y=0}^1\frac{f(x)g(y)}{|y-x|^{\frac14}}dydx$$ an inner product on continuous functions on $[0,1]$? Namely, is $(f,f)$ always positive for all nonzero continuous functions $f$?

I don't know if this is true, but I conjecture it being correct.

Stephen
  • 33
  • At $(f,f)$ you just get $\int_{x=0}^1\int_{y=0}^1\frac{f^2(x)}{|y-x|^{\frac14}}dydx$ which is strictly positive, being the integral of a strictly positive function. Is this what you are asking about? – user10444 Apr 12 '14 at 00:15
  • @user10444 No, you get $\int_0^1 \int_0^1 \frac{f(x) f(y)}{|y-x|^{1/4}} dy dx $. – Stephen Montgomery-Smith Apr 12 '14 at 01:30

1 Answers1

3

$$ \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{f(x) \overline{g(y)}}{|x-y|^{1/4}} \, dy dx = \int_{-\infty}^\infty \hat f(\xi) \overline{\hat g(\xi)} k(\xi) \, d\xi ,$$ where $k(\xi)$ is the Fourier transform of $x\mapsto |x|^{-1/4}$, that is, $k(\xi) = C |\xi|^{-3/4}$ for some constant $C > 0$. So the answer is "yes."

Stephen Montgomery-Smith
  • 26,430
  • 2
  • 35
  • 64