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 March 5th, 2013, 12:07 PM #1 Newbie   Joined: Mar 2013 Posts: 6 Thanks: 0 Holder's spces Hello, I want to know, with any norm we can provide $C^{2+\alpha,1+\beta}(\Omega \times (0,T])$ with $T=>0,0=<\alpha,\beta=<=1$ and $\Omega$ bounded ,open domain in $R$. thank's
 March 5th, 2013, 01:41 PM #2 Newbie   Joined: Mar 2013 Posts: 6 Thanks: 0 Re: Holder's spces no one, .
 March 5th, 2013, 02:13 PM #3 Global Moderator   Joined: May 2007 Posts: 6,684 Thanks: 658 Re: Holder's spces You might get a response if you define your notation.
 March 6th, 2013, 01:50 AM #4 Newbie   Joined: Mar 2013 Posts: 6 Thanks: 0 Re: Holder's spces Thank you, We say a function in ${C}^{2+\alpha,1+\beta}(\Omega \times (0,T])$ if and only if is ${C}^{2,\alpha}$ from the first variable, ${C}^{1,\beta}$ from the second variable. But, How I can define an norm? !
March 6th, 2013, 01:28 PM   #5
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Re: Holder's spces

Quote:
 Originally Posted by x-member Thank you, We say a function in ${C}^{2+\alpha,1+\beta}(\Omega \times (0,T])$ if and only if is ${C}^{2,\alpha}$ from the first variable, ${C}^{1,\beta}$ from the second variable. But, How I can define an norm? !
${C}^{2,\alpha}$
${C}^{1,\beta}$
${C}^{2+\alpha,1+\beta}(\Omega \times (0,T])$

All of the above use notation I have never seen before, especially ${C}^{n,\z}$

 March 6th, 2013, 01:48 PM #6 Newbie   Joined: Mar 2013 Posts: 6 Thanks: 0 Re: Holder's spces $C^{2,\alpha}$ means the $C^{2}$ function with this amount $sup {D^{2}u(x)-D^{2}u(y)}/{|x-y|^{\alpha}}$ is finished.
March 7th, 2013, 01:24 PM   #7
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Re: Holder's spces

Quote:
 Originally Posted by x-member $C^{2,\alpha}$ means the $C^{2}$ function with this amount $sup {D^{2}u(x)-D^{2}u(y)}/{|x-y|^{\alpha}}$ is finished.
$sup {D^{2}u(x)-D^{2}u(y)}/{|x-y|^{\alpha}}$, what is D? Are both D terms divided?

finished ?

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