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What does this notation

$||\nabla u|| _p$ mean? I would like an exact definition.

Also I have seen $|\nabla u|_1$. Does this mean the $\sum_i|\partial_i u|$?

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  • $\begingroup$ Maybe I should make it clear the latter means a Euclidean norm of some sort and the former is an L^p norm. I haven't figured it out. As it stands, it might mean $||\sum_i |D_i u| ||_p$, but I do not know. $\endgroup$ – Lost1 Dec 27 '12 at 19:52
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Usually $$\lVert \nabla u \rVert_p^p = \sum_{i=1}^n\lVert D_iu \rVert_p^p$$ where $D_i$ means the $i$th partial derivative. The $\lVert \cdot \rVert_p$ refers to the $L^p$ norm, i.e., $\lVert u \rVert_p = (\int |u|^p)^{\frac{1}{p}}.$

For your second question: set $p =1$, so we have the $L^1$ norm.

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  • $\begingroup$ Do you mean p rather than 2? the | | means Euclidean norm, I think. From the context, I am reading from. but the double line does mean the L^p norm. $\endgroup$ – Lost1 Dec 27 '12 at 19:47
  • $\begingroup$ @Lost Yes you're right, thanks. I guess it depends on the context (maybe you should post that), but I've never seen the $|.|$ as you indicate used in the field of PDEs/analysis (not that I've seen that much..) $\endgroup$ – soup Dec 27 '12 at 19:55
  • $\begingroup$ Yeah, I have not seen this notation either, but from the context, it seems correct. people.bath.ac.uk/masgrb/Sobolev/notes.pdf see page 18, the 3rd derivation, what we said now fits, by using the fact $a_1^p + ... + a_n^p \leq (a_1+...a_n)^p$ for $a_i$ all positive. $\endgroup$ – Lost1 Dec 27 '12 at 20:01
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If $u:\mathbb{R}^n\to\mathbb{R}$ is a function, then $\nabla u=(\partial_1 u,\ldots,\partial_n u)\in\mathbb{R}^n$ is the vector of partial derivatives. For any vector $x$, the $p$-norm $\left\|\cdot\right\|_p$ is defined as $$ \left\|x\right\|_p = \left(\sum_{i}\left|x_i\right|^p\right)^{1/p}. $$ In particular, $$ \left\|\nabla u\right\|_p = \left(\sum_{i}\left|\partial_i u\right|^p\right)^{1/p} $$ and $$ \left\|\nabla u\right\|_1= \sum_{i}\left|\partial_i u\right|. $$

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  • $\begingroup$ see the answer above if you are interested in the answer. I think the lecture who wrote the notes I am reading used || || for L_p norm and | | as the norm you said. $\endgroup$ – Lost1 Dec 27 '12 at 20:02

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