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Consider the open unit disk $\mathbb{D}$ in $\mathbb{R}^2$. In my analysis course, we defined the Sobolev space $H^1(\mathbb{D})$ in a somewhat unusual way. More precisely, $H^1(\mathbb{D})$ was defined to be the completion of the space $$ \left\{ f \in C^1(\mathbb{D}) : \left\Vert{f}\right\Vert_{H^1(\mathbb{D})} < \infty \right\} $$ where $$ \left\Vert f \right\Vert_{H^1(\mathbb{D})} = \left( \int_\mathbb{D} \left\vert f \right\vert^2 + \left\vert \nabla f \right\vert^2 \right)^{1/2}. $$ Now consider the function $$ f(x) = \ln\left( \ln\left(1 + \frac{1}{\left\vert x \right\vert}\right)\right) $$ on $\mathbb{D}$. According to the usual ``weak derivative'' definition of Sobolev spaces (i.e. that used in Evans), I can prove that $f \in H^1(\mathbb{D})$. However, I am unsure of to establish this with the definition we are using.

My attempts so far have involved seeking out functions in $C^1(\mathbb{D})$ converging pointwise a.e. to $f(x)$ and checking whether this sequence also converges to $f$ with respect to the $H^1$-norm. For instance, I considered the sequence in $C^1(\mathbb{D})$ given by $$ f_n(x)= \ln\left( \ln\left(1 + \frac{1}{\sqrt{x_1^2 + x_2^2 + \frac{1}{n}}}\right)\right). $$ Here, I am writing $x = (x_1,x_2) \in \mathbb{D}$.

Unfortunately, I was unable to show that the sequence $(f_n)$ was even Cauchy in $H^1(\mathbb{D})$. Is this the right approach? Or would a different sequence work better? I think mollifiers may work but I was hoping to avoid this in favour of an explicit sequence.

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  • $\begingroup$ Are you defining $|x|^2:=x_1^2 +x_2^2$ ? $\endgroup$ – irchans Oct 7 '18 at 2:50
  • $\begingroup$ Yes, I am using the notation $x = (x_1,x_2) \in \mathbb{D}$. Thank you for pointing this out, I will add this to the question. $\endgroup$ – rolandcyp Oct 7 '18 at 2:52
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Hint: One way to solve the problem is to patch the singularity at the origin.

$f(x)=\log\left(\log\left(1+1/|x|\right)\right).$

$\mathrm{grad}f(x)= \frac{-x}{|x|^3\left(1+1/|x|\right)\log\left(1+1/|x|\right)}.$

Let $z_n = 1/(e^n - 1)$, then $f(z_n, 0) =\log(n)$ and $$\mathrm{grad}f(z_n, 0)=(-\frac{(\exp(n/2)-\exp(-n/2))^2}n,0). $$

Define $f_n(x) := f(x)$ when $|x|\geq z_n$. When $|x|<z_n$ let $f_n$ be the paraboloid which matches the gradient of $f$ and value of $f$ on the circle where $|x|=z_n$ with vertex at the origin. The $f_n$ are $C^1$ and they converge to $f(x)$ in the $H^1$norm.

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