Let $$f : (0,1) \to \mathbb{R}$$ and $$g(x) = |f(x)|^{r-1} f(x)$$$r \in \mathbb{N}$. It is known that $g\in \mathcal{L}^2(0,1)$ and the $r^{th}$ weak derivative, $ g^{(r)} \in \mathcal{L}^2(0,1)$. I need help to show that the first weak derivative $\require{enclose}\enclose{horizontalstrike}f^{(1)} \in \mathcal{L}^{2r}(0,1) $,

edit : need to show that the first weak derivative $$g^{(1)} \in \mathcal{L}^{2r}(0,1) $$

PS : 1. : If I differentiate $g$ $r$ times, apart from other terms (in sum) I get a term $f'(x)^r$, so I am trying to get the result, if I could prove that all other terms also are integrable.

  1. Also since it is a $L^2$ norm, can I leverage the asymptotics of Fourier series coefficients to solve it?
  • Where did this problem arise? What do you know already? – David Roberts Aug 10 at 2:37
  • @DavidRoberts : If I differentiate $g$ $r$ times, apart from other terms (in sum) I get a term $f'(x)^r$, so I am trying to get the result, if I could prove that all other terms also are integrable. – Rajesh Dachiraju Aug 10 at 2:41
  • @DavidRoberts : Also since it is a $L^2$ norm, can I leverage the asymptotics of Fourier series coefficients to solve it? – Rajesh Dachiraju Aug 10 at 2:45
  • You haven't said where this problem arose. The context may help potential answerers. – David Roberts Aug 10 at 5:40
  • 1
    This does not seem to be true: if $f(x) = x^{1/r}$, then $g(x) = x$ is infinitely smooth, but $f'(x) = r^{-1} x^{1/r - 1}$ is not square integrable if $r \geqslant 2$. – Mateusz Kwaśnicki Aug 10 at 6:31

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