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Some Sobolev spaces are closed under multiplication, making them Banach algebras.

My question is whether $W^{k ,\infty}$ is a Banach algebra?

Since $L^\infty$ is closed under multiplication, I assume $W^{k ,\infty}$ inherits this closure property.

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It indeed is an algebra. To get a Banach algebra, you have to check the norm inequality for the products, i.e., that $\|fg\|\leq \|f\|\cdot\|g\|$ holds. –  abatkai May 31 '12 at 18:48
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A reference: Lipschitz Algebras by Weaver, books.google.com/books/about/… -- this covers only $k=1$, but in a quite general setting. –  user31373 May 31 '12 at 19:35

2 Answers 2

up vote 4 down vote accepted

It may be not a Banach algebra in the usual sense for the usual norm (namely $\sum_{|\alpha|\leq k}\lVert\partial_{\alpha}\cdot\rVert_{\infty}$). However, since Leibniz formula holds (take a test funcion $\varphi$ and a ball which contains the support of $\varphi$, and use the density of test function for the $L^p$ norm over a bounded smooth open set). Hence we have \begin{align} \lVert uv\rVert_{W^{k,\infty}}&=\sum_{|\alpha|\leq k}\lVert\sum_{\beta\leq \alpha}\binom\alpha\beta\partial_{\beta}u\partial_{\alpha-\beta}v\rVert_{\infty}\\ &\leq \sum_{|\alpha|\leq k}\sum_{\beta\leq \alpha}\binom\alpha\beta\lVert\partial_{\beta}u\rVert_{\infty}\lVert\partial_{\alpha-\beta} v\rVert_{\infty}\\ &\leq \lVert u\rVert_{W^{k,\infty}}\cdot \lVert v\rVert_{W^{k,\infty}}\sum_{|\alpha|\leq k}2^{|\alpha|}. \end{align} So we can replace the usual norm on $W^{k,\infty}$ by an equivalent norm $N$ such that $N(uv)\leq N(u)N(v)$ for all $u,v\in W^{k,\infty}$.


So an interesting question:

what is $$\sup\left\{\frac{\lVert uv\rVert_{W^{k,\infty}}}{\lVert u\rVert_{W^{k,\infty}}\lVert v\rVert_{W^{k,\infty}}}, u,v\in W^{k,\infty},uv\neq 0\right\}?$$

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It is in fact sufficient to show that $$\|fg\|\leq C\,\|f\|\cdot\|g\|\tag{1}$$ for some number $C$ independent of the choice of $f$ and $g$ - to be precise, this is sufficient in order to have a Gelfand theory, but it is not the standard definition of Banach algebra.

Now, consider the norm $$\|f\|=\|f\|_{W^{k,\infty}}=\max_{0\leq j\leq k}\left\|\frac{d^jf}{dx^j}\right\|_\infty$$ This norm is equivalent to the standard Sobolev norm, and by the product rule (of the derivative) together with the fact that $L^\infty$ is a Banach algebra (1) follows.

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