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How can I show that $f: (-1,1) \to \mathbb{R}, \ x \mapsto \ln(|x|)$ has no weak derivative but in $B_1(0)\in \mathbb{R}^2$ it has? I know that every classical solution is also a weak solution in this case.

Any help is greatly appreciated.

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    $\begingroup$ are you considering $\ln|x|$ as a function $\mathbb{R} \to\mathbb{R}$ and then $\frac{1}{2}\ln(x_1^2+x_2^2)$ as a function $\mathbb{R}^2 \to\mathbb{R}$ ? $\endgroup$ – reuns Apr 27 '16 at 20:31
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    $\begingroup$ and from what I understand, if $D^\alpha f$ is in $L^1_{loc}$ then it is clearly the weak derivative of $f$, hence it should reduce to proving that $\frac{x_1}{x_1^2+x_2^2}$ is in $L^1_{loc}(\mathbb{R}^2)$ $\endgroup$ – reuns Apr 27 '16 at 20:37
  • $\begingroup$ yes $ln|x|$ with R-->R! But then in the exercise they only tell, that in $B_1(0)$ I should find a weak derivative on $x_1$. Nothing more. i would guess that your solution also works with that, right? $\endgroup$ – Thesinus Apr 27 '16 at 20:55
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Let be help you by showing that $\ln|\cdot|: (-1,1) \to \mathbb{R}_{<0}$ is not weakly differentiable.

Lemma: Let $u'$ be the weak derivative of $u$ on $(a,b)$. Then for all intervals $(\alpha, \beta) \subset (a,b)$ it holds that $u'|_{(\alpha, \beta)}$ is also the weak derivative of $u|_{(\alpha, \beta)}$ on $(\alpha, \beta)$.

Proof. Let $(\alpha, \beta) \subset (a,b)$ and $\phi \in \mathcal{C}_{\text{c}}^{\infty}(\alpha, \beta)$ and define the trivial extension of $\phi$ by $\tilde{\phi} \in \mathcal{C}_{\text{c}}^{\infty}(a,b)$. Then, we conclude \begin{equation*} \int_{\alpha}^{\beta} u(x) \phi'(x) dx = \int_{a}^{b} u(x) \tilde{\phi}'(x) dx = - \int_{a}^{b} u'(x) \tilde{\phi}(x) dx = - \int_{\alpha}^{\beta} u'(x) \phi(x) dx, \end{equation*} which implies the proposition.$\ \square$

Back to your question: This implies that the only candidate for the weak derivative of your function has to be \begin{equation*} \omega(x) = \begin{cases} -\frac{1}{x}, & \text{if } x \in (-1,0), \\ \frac{1}{x}, & \text{if } x \in (0,1) \end{cases}. \end{equation*} Now, for all $\phi \in \mathcal{C}_{\text{c}}((-1,1))$ (with $\phi(-1) = \phi(1) = 0$ ($\star$)) we have by integration by parts (IBP) \begin{align*} \int_{-1}^{1} \ln(|x|) \phi'(x) \, dx & = \int_{-1}^{0} \ln(-x) \phi'(x) \, dx + \int_{0}^{1} \ln(x) \phi'(x) \, dx \\ & \overset{\textrm{(IBP)}}{=} \big[\ln(-x) \phi(x) \big]_{x = - 1}^{0} + \int_{-1}^{0} \frac{\phi(x)}{x} \, dx + \big[\ln(x) \phi(x) \big]_{x = 0}^{1} - \int_{0}^{1} \frac{\phi(x)}{x} \, dx \\ & = - \int_{-1}^{1} \omega(x) \phi(x) \, dx + \big[\ln(-x) \phi(x) \big]_{x = - 1}^{0} + \big[\ln(x) \phi(x) \big]_{x = 0}^{1} \\ & \overset{(\star)}{=} \int_{-1}^{1} \omega(x) \phi(x) \, dx + \lim_{x \nearrow 0} \ln(-x) + \lim_{x \searrow 0} \ln(x) \\ & \neq - \int_{-1}^{1} \omega(x) \phi(x) \, dx. \end{align*} and therefore, your function is not weakly differentiable on $(-1,1)$.

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