# Is the Sobolev Space $H^k(0,1)$ a banach algebra?

In Adams'book:Sobolev Spaces, I know that if $kp>n,\Omega\subset R^n$ is boundary domain and has cone property, then $W^{k,p}(\Omega)$ could see as a banach algebra. My question is that does it hold for the one-dimensional and $p=2$ case, namely, for $H^k(0,1)$ ? If it is not true, provide a counterexample please.

-

You have $n=1$, $k\ge 1$, and $p=2$. The condition $kp>n$ holds. The domain $(0,1)$ has the cone property in appropriate interpretation, since it is convex. The reason to introduce the cone property is to ensure that the boundary can be easily reached from inside; there is no difficulty in reaching the boundary of $(0,1)$.
Thus, even though the case $n=1$ is not necessarily mentioned by Adams, the result applies.
@ 5PM: Your answer is that $H^k(0,1),k\geqslant 1$ is a Banach algebra, but there is no need to use Adams' results. Am I right? ^-^ –  Darry Jan 24 '13 at 1:42
@Darry Yes, you are right: the good old Leibniz rule is enough. $$(fg)^{(k)} = \sum_j \binom{k}{j}f^{(j)} g^{(k-j)}$$ where all terms are square summable, because $f^{(j)}$ and $g^{(j)}$ are bounded for $j<k$. –  user53153 Jan 24 '13 at 1:55