Let $U \subset \mathbb R^n$ be open, bounded and connected, and let $
\alpha, \beta \in (0,1]$ with $\alpha < \beta$.If $f \in C^{0,\alpha} (U)$ and $g \in C^{0,\beta}(U)$ then what is
the maximum value of $\gamma \in (0,1]$ such that $ fg\in C^{0,
\gamma} (U)$ ?
I came up with this while I was reading this http://terrytao.wordpress.com/2009/04/30/245c-notes-4-sobolev-spaces/
It is clear that for $\gamma=\alpha$ the conclusion holds, but can how can we prove that this is indeed the maximum such value of $\gamma$. Am I missing something obvious?
Moreover, can we prove something about the composition $f \circ g$? I suspect that in this case the desired maximum value is the product of the exponents $\alpha \beta$, but I can't prove it. Am I right?
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Just a recall from the theory: We denote by $C^{0,\alpha}(U)$ the space of Holder continuous functions with exponent $\alpha\in (0,1]$ i.e.,
$ f: U \to \mathbb R^m$ such that
$$ [f]_{\alpha,U} := \sup_{x\neq y\, \in U} \frac{ |f(x)-f(y)|}{|x-y|^{\alpha}}< \infty$$
The space $C^{0,\alpha}(U)$ eqquiped with the norm
$$ \| f\|_{C^{0,\alpha}} := \|f\|_{\infty} + [f]_{\alpha,U} $$
is a Banach space.
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edit: I submit the assumption $\alpha < \beta$.
Best Answer
To begin, consider $U=(0,1)$ and $\alpha\in (0,1)$. Prove that $x^\alpha\in C^{0,\alpha}(U)$, but $x^\alpha\notin C^{0,\beta}(U)$ for $\beta>\alpha$.
Now consider the functions $u(x)=x^\alpha+1$ and $v(x)=x^\beta +1$ with $\alpha<\beta$. As you alread know $uv\in C^{0,\alpha}(U)$. Moreover, you can verify that if $\gamma>\alpha$ then $uv \notin C^{0,\gamma}(U)$, hence, $\alpha$ is optimal.
It is worth to note that $x^\alpha x^\beta=x^{\alpha+\beta}\in C^{0,\alpha+\beta}(U)$.
With respect to the second question note that: $$\frac{|f(g(x))-f(g(y))|}{|x-y|^{\alpha\beta}}\leq \frac{|f(g(x))-f(g(y))|}{|g(x)-g(y)|^\alpha}\frac{|g(x)-g(y)|}{|x-y|^{\beta}},\ g(x)\neq g(y)$$
Can you conclude?
Concerning the optimality, consider the example $u(x)=x^\alpha$ and $v(x)=x^\beta$ for $x\in (0,1)$ and $\alpha<\beta$. We have that $u(v(x))=x^{\alpha\beta}\in C^{0,\alpha\beta}(U)$ and because of the first line of this answer, we have that $uv\notin C^{0,\gamma}(U)$ for all $\gamma>\alpha\beta$.