Product of Sobolev functions in 1D Let $\sigma \in {W}^{1,\gamma}(0,\infty)$ $(\gamma\geq 1$) and $\phi \in {W}^{1,p}(0,\infty)$ $(p\geq 1$).
I am wondering if $\sigma\phi \in {W}^{1,p}(0,\infty)$ $(1 \leq p <\gamma$)?
Thanks for helps.
 A: You need a Theorem to claim that the function $\sigma\phi$ is at least integrable.
The Hölder's inequality is very useful for this general functions. If $\gamma,p\in[1,\infty]$ and $1/\gamma+1/p=1$, then $\sigma\phi\in L^1(0,\infty)$.
With general functions $\sigma$ and $\phi$ you can't claim that $\sigma\phi\in L^p(0,\infty)$ and this is neccessary for $\sigma\phi\in W^{1,p}(0,\infty)$.
So if $\gamma$ and $p$ verify $1/\gamma+1/p=1$, you can affirm that $\sigma\phi\in W^{1,1}(0,\infty)$.
We assume $1/\gamma+1/p=1$, let's see that the derivate $(\sigma\phi)'$ is integrable.
$$
(\sigma\phi)'=\sigma '\phi+\sigma\phi'
$$
Because $\sigma\in W^{1,\gamma}$ and $\phi\in W^{1,p}$, then $\sigma,\sigma'\in L^\gamma$ and $\phi,\phi'\in L^p$. Using the Hölder inequality:
$\sigma '\phi\in L^1$ and $\sigma\phi'\in L^1$. Because the sum of two functions in $L^1$ is in $L^1$ then $(\sigma\phi)'\in L^1$.
So you can assure that $\sigma\phi\in W^{1,p}(0,\infty)$ when $1/\gamma+1/p=1$. In a general situation you can say nothing.
