3
$\begingroup$

Problem: Let $\Omega\subset\mathbb R^2$ denote the open unit ball in $\mathbb R^2$. Prove that the unbounded function $$f(x)=\log\log\left(1+\frac{1}{\vert x\vert}\right)$$ belongs to $H^1(\Omega).$

My Attempt: Let $\{\varepsilon_n\}_{n=1}^\infty\subset[0,1]$ such that $\varepsilon_n\searrow0$ as $n\to\infty$. Put $\Omega_n=B(0,\varepsilon_n)$. Define the sequence of functions $$f_n(x)=\begin{cases}f(x)&\text{if }x\in\Omega\setminus\Omega_n\\0&\text{otherwise.}\end{cases}$$ Note that $\vert f_n\vert^2\nearrow\vert f\vert^2$ as $n\to\infty$, so by the monotone convergence theorem we have $\|f_n\|_{L^2(\Omega)}^2\to\|f\|_{L^2(\Omega)}^2$ as $n\to\infty$. Using integration in polar coordinates, as shown in Folland's Real Analysis text, we have that \begin{align*} \|f_n\|_{L^2(\Omega)}^2 &=\int_{\Omega\setminus\Omega_n} \vert f(x)\vert^2\,dx=\int_{\Omega\setminus\Omega_n}\left\vert\log\log\left(1+\frac{1}{\vert x\vert}\right)\right\vert^2\,dx\\ &=2\pi\int_{\varepsilon_n}^1 r\left\vert\log\log\left(1+\frac1r\right)\right\vert^2\,dr\\ &\leq2\pi\int_{\varepsilon_n}^1 e^r\,dr\\ &\leq2\pi\int_0^1e^r\,dr\\ &=2\pi e\\ &<\infty. \end{align*} Since the bound above does not depend on $n$, letting $n\to\infty$ shows that $f\in L^2(\Omega)$, by the monotone convergence theorem.
Next, observe that $$\nabla f(x)=\left(-\frac{x_1}{\log\left(1+\frac{1}{\vert x\vert}\right)(1+\vert x\vert)\vert x\vert^2},-\frac{x_2}{\log\left(1+\frac{1}{\vert x\vert}\right)(1+\vert x\vert)\vert x\vert^2}\right),$$ so that $$\vert\nabla f(x)\vert^2=\frac{1}{\log\left(1+\frac1{\vert x\vert}\right)^2(1+\vert x\vert)^2\vert x\vert^2}.$$ Using the same method as above we have that $\|\nabla f_n\|_{L^2(\Omega)}^2\to\|\nabla f\|_{L^2(\Omega)}^2$ by the monotone convergence theorem. Then, integrating in polar coordinates once again, we have \begin{align*} \|\nabla f_n\|_{L^2(\Omega)}^2 &=\int_{\Omega\setminus\Omega_n}\vert\nabla f(x)\vert^2\,dx=\int_{\Omega\setminus\Omega_n}\frac{1}{\log\left(1+\frac1{\vert x\vert}\right)^2(1+\vert x\vert)^2\vert x\vert^2}\,dx\\ &=2\pi\int_{\varepsilon_n}^1\frac{1}{\log\left(1+\frac1{r}\right)^2(1+r)^2r^2}\,dr\\ &\to\infty\quad\text{as }n\to\infty. \end{align*} It follows that $f\notin H^1(\Omega)$ since $\vert\nabla f\vert\notin L^2(\Omega)$.


Do you agree with my proof above? I am not sure that I fully understood and applied the definition of the Sobolev Space $H^1(\Omega)$, especially in the second part of the proof. Any clarification if I am in the wrong would be much appreciated.
Thank you for your time and valuable feedback.

$\endgroup$
0
2
$\begingroup$

In the computation of $\int_{\Omega}|\nabla f|^2 \,dx$ you forgot $r\,dr$.

Hence the integral converges since $\frac{1}{\log(1+1/r)^2 r(1+r)^2}$ is integrable close to $r=0$ and we get $f\in H^1(\Omega)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.