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$$ \lim_{p\rightarrow \infty} \int_{\mathbb R^N} \left( \frac{\left\lvert \nabla u \right\rvert}{\left\| \nabla u \right\|_p} \right)^{p-2} \frac{\nabla u}{\left\| \nabla u \right\|_p}\cdot \nabla v \,dx \label{\star}\tag{$\star$}$$

where $\,\left\lvert\,\cdot\,\right\rvert\,$ is the Euclidean $2$-norm and $\;\left\| \nabla u \right\|_p = \left\|\, \left\lvert \nabla u \right\rvert \, \right\|_p$.

Could anyone give me some hint for finding an explicit form for the expression involving $u$? I feel like my approach is not quite correct...

First I know the term in the limit is bounded $$ \begin{aligned} \eqref{\star} & \leq \int_{\mathbb R^N} \left( \frac{\left\lvert \nabla u \right\rvert}{\left\| \nabla u \right\|_p}\right)^{p-1} \left\lvert \nabla v \right\rvert \,dx \\ & = \left\| \nabla u \right\|_p^{1-p} \int_{\mathbb R^N} \left\lvert \nabla u \right\rvert^{p-1} \left\lvert \nabla v \right\rvert \,dx \\ &\leq \left\| \nabla u \right\|_p^{1-p} \Bigg(\int_{\mathbb R^N} \Big(\left\lvert \nabla u \right\rvert^{p-1}\Big)^{\frac{p}{p-1}}\,dx\Bigg)^{\frac{p-1}{p}} \Bigg(\int_{\mathbb R^N} \left\lvert \nabla v \right\rvert ^ p \,dx\Bigg)^{\frac{1}{p}} \\ & = \left\| \nabla u \right\|_p^{1-p}\left\| \nabla u \right\|_p^{p-1} \left\| \nabla v\right\|_p \\ & = \left\| \nabla v\right\|_p. \end{aligned} $$ with the additional assumption that $ \nabla v \in L^\infty\!\left(\mathbb R^N, \mathbb R^N\right)$.

Define the set $$ D:= \big\{\left\lvert \nabla u \right\rvert \leq \|\nabla u \|_\infty -\delta\big\},$$ the limit $\eqref{\star}$ is zero on the set $D$. To see this, we know there exists $N\in \mathbb N$ such that $$\left|\, \left\|\nabla u \right\|_\infty - \left\|\nabla u \right\|_p \,\right| \leq \frac{\delta}{2} \quad \forall \; p\geq N.$$ On the set $D$ we have $$ \left\|\nabla u \right\|_p - \left\lvert \nabla u \right\rvert \geq \frac{\delta}{2} \quad \forall \; p\geq N$$ which means $$\frac{\left\lvert \nabla u \right\rvert}{\left\| \nabla u \right\|_p} \leq 1 - \frac{\delta}{2} \quad \forall \; p\geq N$$ and the term in the limit $\left(\dfrac{\left\lvert \nabla u \right\rvert}{\left\| \nabla u \right\|_p}\right)^{p-2}$ goes to $0$ uniformly.

Next I looked at the set $K: = \big\{\left\lvert \nabla u \right\rvert = \left\|\nabla u \right\|_\infty\big\}$, the term $$ \left(\frac{\left\lvert \nabla u \right\rvert}{\left\| \nabla u \right\|_p}\right)^{p-2} = \left(\frac{\left\| \nabla u \right\|_\infty}{\left\| \nabla u \right\|_p}\right)^{p-2} $$ is in the form $``1^\infty"$, so I tried to use L'Hopital's rule. And we can calculate the limit (assuming the integrals are defined and finite, I just want to see what the limit might look like). $$ \begin{aligned} \lim_{p\rightarrow \infty} \left(\frac{\left\| \nabla u \right\|_\infty}{\left\| \nabla u \right\|_p}\right)^{p-2} &= \lim_{p\rightarrow \infty} \exp\left( \left(p-2\right)\log\left(\frac{\left\| \nabla u \right\|_\infty}{\left\| \nabla u \right\|_p}\right)\right) \\ &=\lim_{p\rightarrow \infty} \exp\left( \frac{\log\left(\frac{\left\| \nabla u \right\|_\infty}{\left\| \nabla u \right\|_p}\right)}{\frac{1}{p-2}}\right) \\ &=\lim_{p\rightarrow \infty} \exp\left( \frac{\frac{d}{dp} \left[-\log\left(\frac{\left\| \nabla u \right\|_p}{\left\| \nabla u \right\|_\infty}\right)\right]}{\frac{-1}{(p-2)^2}}\right) \\ &=\lim_{p\rightarrow \infty} \exp \left( \dfrac{ \left( \dfrac{\left\| \nabla u \right\|_\infty}{\left\| \nabla u \right\|_p} \right) \dfrac{ \frac{d}{dp} \big( \left\| \nabla u \right\|_p \big) }{ \left\|\nabla u \right\|_\infty} } { \dfrac{1}{\left(p-2\right)^2 } } \right) \end{aligned} $$ where $$ \begin{aligned} \frac{d}{dp}\Big[\left\| \nabla u \right\|_p \Big] &= \frac{d}{dp}\left[\left(\int_{\mathbb R^N} \left\lvert\nabla u \right\rvert^p \,dx \right)^{1/p} \right] \\ &=\frac{d}{dp}\left[\exp\left(\frac{1}{p} \log\left(\int_{\mathbb R^N} \left\lvert\nabla u \right\rvert^p \,dx \right)\right)\right] \\ &=\|\nabla u \|_p \left\{\frac{-1}{p^2}\log\left(\int_{\mathbb R^N} \left\lvert\nabla u \right\rvert^p \,dx \right) + \dfrac{1}{p} \dfrac{1}{\int_{\mathbb R^N} \left\lvert\nabla u \right\rvert^p \,dx } \int_{\mathbb R^N} \left\lvert \nabla u \right\rvert^p \log\big(\left\lvert\nabla u \right\rvert\big) \,dx \right\} \end{aligned} $$

And I am a little bit stuck. Thank you very much!

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  • $\begingroup$ @whacka Yes $\|\nabla u\|_p = \int |\nabla u |^p dx$ where $|\cdot|$ is the 2-norm for vectors in $\mathbb{R}^N$. $\endgroup$ – Xiao Aug 17 '15 at 2:58
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Thanks for the help from user zhw.

I claim that $$\lim_{p\rightarrow \infty} \int_{\mathbb R^N}\Bigg\{\left(\frac{|\nabla u|}{\|\nabla u\|_p}\right)^{p-2}\frac{\nabla u}{\|\nabla u\|_p}\Bigg\} \cdot \nabla v dx = \int_K \frac{1}{m(K)}\frac{\nabla u}{\|\nabla u\|_\infty} \cdot \nabla v dx$$ where $K:= \{|\nabla u| = \|\nabla u \|_\infty\}$. We assume $m(K)>0$ or else the limit is $0$.

On subset $\{\mathbb R^N - K\}$, The term $\left(\frac{|\nabla u |}{\|\nabla u\|_p}\right)^{p-2}$ goes to $0$ pointwise a.e. on $\mathbb R^N - K$, by Lebesgue dominated convergence theorem, we have $$\lim_{p\to\infty} \int_{\mathbb R^N - K} \left(\frac{|\nabla u |}{\|\nabla u\|_p}\right)^{p-2} \frac{\nabla u}{\|\nabla u\|_p} \cdot \nabla v dx = 0$$

Next on the set $K: = \{|\nabla u| = \|\nabla u \|_\infty\}$ I claim that the term
$$\lim_{p\to\infty} \left(\frac{|\nabla u |}{\|\nabla u\|_p}\right)^{p-2} = \lim_{p\to\infty} \left(\frac{\|\nabla u \|_\infty}{\|\nabla u\|_p}\right)^{p-2} = \frac{1}{m(K)}.$$

We can rewrite $\left(\frac{\|\nabla u \|_\infty}{\|\nabla u\|_p}\right)^{p-2}$ as \begin{align*} &\left(\frac{\|\nabla u \|_\infty}{\|\nabla u\|_p}\right)^{p-2} \\ =&\left(\frac{\|\nabla u \|_\infty}{\|\nabla u\|_p}\right)^{p}\frac{\|\nabla u\|_p^2}{\|\nabla u\|_\infty^2}\\ =& \frac{\|\nabla u\|_\infty^p}{\|\nabla u\|_\infty^p m(K) + \|\nabla u\|_\infty^p \int_{\mathbb R^N - K} \left(\frac{|\nabla u|}{\|\nabla u\|_\infty}\right)^p dx} \frac{\|\nabla u\|_p^2}{\|\nabla u\|_\infty^2}\\ =& \frac{1}{ m(K) + \int_{\mathbb R^N - K} \left(\frac{|\nabla u|}{\|\nabla u\|_\infty}\right)^p dx} \frac{\|\nabla u\|_p^2}{\|\nabla u\|_\infty^2} \end{align*} Take the limit $p\to \infty$ and by Lebesgue dominated convergence theorem we get $$\lim_{p\to\infty} \left(\frac{|\nabla u |}{\|\nabla u\|_p}\right)^{p-2} = \frac{1}{m(K)}$$ on the set $K$.

Thus the integral $$\lim_{p\to\infty} \int_K \left(\frac{|\nabla u |}{\|\nabla u\|_p}\right)^{p-2} \frac{\nabla u}{\|\nabla u\|_p} \cdot \nabla v dx =\int_K \frac{1}{m(K)}\frac{\nabla u}{\|\nabla u\|_\infty} \cdot \nabla v dx$$

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