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When studying commutator estimates, I have encountered the following problem.

Consider $f\in C^1(\mathbb{R}^d,\mathbb{R})$ with $\nabla f\in L^p$. So $\nabla f(x)\in\mathbb{R}^d$.

My question is what would $\|\nabla f(x)\|_p$ indicate?

I am not sure if the question has been clearly stated, please leave comments if you feel the question is not well formed.

newbie
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    Usually, $\lVert \nabla f(x)\rVert_p^p = \sum \int \lvert \partial f/\partial x_i \rvert^p$. – Daniel Fischer Aug 01 '13 at 11:25
  • The notation suggests that you are integrating the Euclidean norm of the gradient to the $p$-power (the usual notion of the $L^p$-norm) over a suitable domain, and then extracting the $p$-th root. To get a finite integral it is not enough to have a $C^1$ function, so you would need additional hypotheses. – Mikhail Katz Aug 01 '13 at 11:26
  • @user72694 Yes I do have further assumptions, my concern here is more related to what the norm would look like. – newbie Aug 01 '13 at 11:30
  • Well @Daniel Fischer proposed a slightly different definition of the $L^p$ norm from mine. Did't your instructor define his notation? – Mikhail Katz Aug 01 '13 at 11:37
  • @user72694 Actually it is not from a certain course. It's from this book, P111. – newbie Aug 01 '13 at 11:44

1 Answers1

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When you have a function $v \colon X \to E$ on a measure space $X$ with values in a Banach space $E$, saying that $v\in L^p(X,\,E)$ usually means that $v$ is a measurable function such that the real-valued function $n \colon x \mapsto \lVert v(x)\rVert$ is in $L^p(X,\,\mathbb{R})$, and $\lVert v\rVert_p = \left(\int \lVert v(x)\rVert_E^p\right)^{1/p}$.

In the case that $E = \mathbb{R}^d$, firstly, that is equivalent to each component function itself being in $L^p$, and secondly, since all norms on $\mathbb{R}^d$ are equivalent, it doesn't matter which norm we choose on $\mathbb{R}^d$, they all produce equivalent norms on $L^p(X,\,\mathbb{R}^d)$. You then get the most convenient computations if you choose the corresponding $\ell^p$-norm on $\mathbb{R}^d$, which becomes

$$\lVert v\rVert_p = \left(\int_X \sum_{i=1}^d \lvert v_i(x)\rvert^p \,d\mu\right)^{1/p},$$

the $p$-th power of the norm of $v$ is the sum of the $p$-th powers of the norms of the components.

Daniel Fischer
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  • Your answer is very helpful for me. Could you please clarify this point: What is $\mu$ in the last expression of your answer? Thanks in advance. – Janak Jun 02 '15 at 14:20
  • $\mu$ is the measure on $X$ that we consider, @janak. If $X$ is e.g. an open subset of some $\mathbb{R}^n$, it may be (the restriction of) the Lebesgue measure; it may be a probability measure, or whatever measure. The argument is measure-agnostic, it works the same for all measures. – Daniel Fischer Jun 02 '15 at 14:44
  • Thanks for the reply. I still have one more points: the norm of the vector valued function that you have defined, is not satisfying the triangle inequality of the definition of the norm. Here I am assuming the addition of two multivalued function is the component wise addition. – Janak Jun 02 '15 at 15:08
  • Why do you think it doesn't satisfy the triangle inequality? If we denote the $p$-norm on $\mathbb{R}^n$ with $\lvert,\cdot,\rvert_p$, then we have $\lVert v\rVert_p = \lvert (\lVert v_1\rVert_{L^p},\dotsc,\lVert v_n\rVert_{L^p})\rvert_p$. And by the monotonicity of and triangle inequality for $\lvert,\cdot,\rvert_p$, we have – Daniel Fischer Jun 02 '15 at 15:49
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    $$\begin{aligned}\lVert u+v\rVert_p &= \lvert (\lVert u_1+v_1\rVert_{L^p},\dotsc,\lVert u_n+v_n\rVert_{L^p})\rvert_p\ &\leqslant \lvert (\lVert u_1\rVert_{L^p} + \lVert v_1\rVert_{L^p},\dotsc,\lVert u_n\rVert_{L^p} + \lVert v_n\rVert_{L^p})\rvert_p\ &\leqslant \lvert (\lVert u_1\rVert_{L^p},\dotsc, \lVert u_n\rVert_{L^p})\rvert_p + \lvert (\lVert v_1\rVert_{L^p},\dotsc, \lVert v_n\rVert_{L^p})\rvert_p\ &= \lVert u\rVert_p + \lVert v\rVert_p.\end{aligned}$$ – Daniel Fischer Jun 02 '15 at 15:49
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    I suppose that is a typo, @janak, in the "usual definition"? The usual definition is $\lVert v_1\rVert_{L^p} = \Bigl(\int_X \lvert v_1(x)\rvert^p,d\mu\Bigr)^{\frac{1}{p}}$, and thus $$\biggl( \int_X \sum_{i=1}^d \lvert v_i(x)\rvert^p,d\mu\biggr)^{1/p} = \biggl(\sum_{i=1}^d \int_X \lvert v_i(x)\rvert^p,d\mu\biggr)^{1/p} = \biggl(\sum_{i=1}^d \lVert v_i\rVert_{L^p}^p\biggr)^{1/p} = \lvert (\lVert v_1\rVert_{L^p},\dotsc,\lVert v_d\rVert_{L^p})\rvert_p.$$ – Daniel Fischer Jun 03 '15 at 08:36
  • I got the point. Thank you very much for the explanation. – Janak Jun 03 '15 at 08:45
  • You're welcome. – Daniel Fischer Jun 03 '15 at 08:46
  • +1 thanks for the wonderful explanation. If we instead take the "usual" norm on $\mathbb R^d$ we would get $$\lVert v\rVert_p = \left(\int_X \left(\sum_{i=1}^d \lvert v_i(x)\rvert^2\right)^\frac p2 ,d\mu\right)^{1/p}$$ right? But that would be equivalent to the simpler one you wrote. – Ant May 17 '16 at 09:58