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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 $\alpha < \beta$.

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  • $\begingroup$ @Tomás: Yes you are right. I should have assume that $\alpha < \beta$. With this assumtion. I edit immidiately. $\endgroup$
    – passenger
    Nov 4, 2013 at 17:06

1 Answer 1

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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 $\gamma>\alpha\beta$.

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  • $\begingroup$ Dear Tomas, thank you very much for your reply! I would like to ask you if you could give some more details on how to show that $ u \nu \not \in C^{0,\gamma}$, for $\gamma > \alpha$. I am not sure that I have understand how to work to prove that the quantity will be unbounded. Thank's in advance! $\endgroup$
    – passenger
    Nov 4, 2013 at 19:10
  • $\begingroup$ Dear @passenger, did you worked out the first line of my answer? $\endgroup$
    – Tomás
    Nov 4, 2013 at 19:15
  • $\begingroup$ Yes, I did, and I proved the other in a similar way. Am right? I just asked for verification... $\endgroup$
    – passenger
    Nov 4, 2013 at 19:17
  • $\begingroup$ Yes it's right @passenger . Now the idea is: note that $uv=x^{\alpha+\beta}+x^\alpha+x^\beta+1$. As you have proved, $x^\alpha\in C^{0,\alpha}$ but $x^\alpha\notin C^{0,\gamma}$ for all $\gamma>\alpha$, hence the sum $x^{\alpha+\beta}+x^\alpha+x^\beta+1$ belong to $C^{0,\alpha}$, but it cannot belong to $C^{0,\gamma}$ for all $\gamma>\alpha$, because the term $x^\alpha$ is contained in this sum. $\endgroup$
    – Tomás
    Nov 4, 2013 at 19:19
  • $\begingroup$ O.K thank's! Better way this! I worked again as I did for the function $x^{\alpha}$.... $\endgroup$
    – passenger
    Nov 4, 2013 at 19:23

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