The title says it all. Suppose $f \in H^{1}(\mathbb{R}^{n})$ is real-valued and positive almost everywhere. How do I prove that $f^{-1} = 1/f \in H^{1}(\mathbb{R}^{n})$ and $\nabla f^{-1} = -\nabla f/f^{2}$? I honestly don't even understand why $f^{-1} \in L^{2}(\mathbb{R}^{n})$. It does not seem to follow from Hölder inequality. I also tried the reverse Hölder inequality, but got nowhere.
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I wonder if there even is a function $f$ such that both $f$ and $\frac1f$ are in $L^2(\Bbb R^n)$... – Sassatelli Giulio Dec 16 '24 at 14:34
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@SassatelliGiulio wouldn't that mean that $1\in L^1(\Bbb R^n)$? – Sine of the Time Dec 16 '24 at 14:39
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@SineoftheTime That $1\in L^1(\Bbb R^n)$, but point taken, thank you. – Sassatelli Giulio Dec 16 '24 at 14:41
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Corretto :) @SassatelliGiulio – Sine of the Time Dec 16 '24 at 14:42
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It is simply not true: We surely have $$ \exp(-|x|^2) \in H^1(\mathbb{R}^n), $$ but $$ \exp(|x|^2) \notin H^1(\mathbb{R}^n). $$
F. Conrad
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Thanks! I think this can only be true if we use mollifiers: $\varphi/f \in H^{1}(\mathbb{R}^{n})$ for $\varphi$ smooth and compactly supported. – Idontgetit Dec 16 '24 at 14:36