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.


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 $f\in H^1(\Omega)$.


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