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Suppose that we have $$\Omega =\{x\in R^2 | \ \|x\|_2 \le 1 \}$$ I need to show that for any $\alpha>0$, $f(x)=\|x\|^{\alpha}_2$ belongs to the Sobolev space $H^1(\Omega)$ where $$f(x)=\sqrt{(x^2+y^2)^{\alpha}}.$$

Determine the range of $\alpha$ such that $f$ also belongs to $H^2(\Omega)$.

My attempt to show this by induction:

I need to show that $f(x)$, $\frac{\partial f}{\partial x}$, and $\frac{\partial f}{\partial y}$ belongs to $L^2(\Omega)$

So I need to show that every integral $\int_{\Omega}f(x)^{2}\,\mathrm{d}\Omega \lt \infty $.

This is pretty easy for $\alpha=1$, but I don't know how to prove this for $\alpha=n$.

Chickenmancer
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Jack Havis
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1 Answers1

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In general, $\int_0^1\int_0^{2\pi} (r^\beta) rdrd\theta=\frac{2\pi}{\beta+2}$ is finite iff $\beta>-2$.

Let $f(x)=\|x\|^\alpha$. Then, $f\in L^2(\Omega)$ when $\|f\|^2=\|x\|^{2\alpha}=r^{2\alpha}$ has power $2\alpha>-2$, i.e. $\alpha>-1$.

$\nabla f=\alpha\|x\|^{\alpha-2}x$ is in $L^2(\Omega)$ when $\|\nabla f\|^2=\alpha^2\|x\|^{2(\alpha-2)+2}$ has power $2(\alpha-1)>-2$, so $\alpha>0$.

$\nabla^2f=\alpha(\alpha-2)\|x\|^{\alpha-4}x\otimes x+\alpha\|x\|^{\alpha-2}\nabla x$ is in $L^2(\Omega)$ when $\|\nabla^2f\|^2=\alpha\|x\|^{2(\alpha-2)}+\cdots$ has power $2(\alpha-2)>-2$, so $\alpha>1$.

Chrystomath
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