Mixing
Inclusion of Holder-Besov space
I'm sorry if this is a stupid question. But I am currently reading a paper which makes heavy use of the Holder-Besov space
$$mathcal{C}^{s}(mathbb{R}^{n}; mathbb{R}) = B^{s}_{infty,infty}(mathbb{R}^{n};mathbb{R})$$
with norms defined using the Littlewood-Paley decomposition in the sense that
$$ | u |_{mathcal{C}^{s}} = Big| ( 2^{js} | Delta_{j}u|_{L^{infty}} )_{j geq -1} Big|_{ell^{infty}(mathbb{Z})}. $$
Now there is the fact
$$ s leq r Longrightarrow | u |_{mathcal{C}^{s}} leq C | u |_{mathcal{C}^{r}} $$
where $C$ can be chosen independent of $s$ and $r$. The best I can do is to choose $C$ such that $C$ depends on $ s $ and $r$. For example, if $ s，epsilon > 0 $ are given. Then I will have to estimate
$$ 2^{-s} | Delta_{-1} u |_{L^{infty}} leq 2^{-s} C| u |_{L^{infty}} leq 2^{s} C sum_{j geq -1} | Delta_{j}u |_{L^{infty}} leq 2^{s} | u|_{mathcal{C}^{s+epsilon}}C sum_{j geq -1} 2^{-j(s+epsilon)} = C_{s,epsilon} | u |_{mathcal{C}^{s+epsilon}}.$$
It's not clear to me how to get a better bound for the above case $j=-1$. So my question is. How do we get a sharper uniform bound?
Thanks in advance!
functional-analysis besov-space
asked Nov 21 '18 at 15:20
Meagain
16910
add a comment |
I'm sorry if this is a stupid question. But I am currently reading a paper which makes heavy use of the Holder-Besov space
$$mathcal{C}^{s}(mathbb{R}^{n}; mathbb{R}) = B^{s}_{infty,infty}(mathbb{R}^{n};mathbb{R})$$
with norms defined using the Littlewood-Paley decomposition in the sense that
$$ | u |_{mathcal{C}^{s}} = Big| ( 2^{js} | Delta_{j}u|_{L^{infty}} )_{j geq -1} Big|_{ell^{infty}(mathbb{Z})}. $$
Now there is the fact
$$ s leq r Longrightarrow | u |_{mathcal{C}^{s}} leq C | u |_{mathcal{C}^{r}} $$
where $C$ can be chosen independent of $s$ and $r$. The best I can do is to choose $C$ such that $C$ depends on $ s $ and $r$. For example, if $ s，epsilon > 0 $ are given. Then I will have to estimate
$$ 2^{-s} | Delta_{-1} u |_{L^{infty}} leq 2^{-s} C| u |_{L^{infty}} leq 2^{s} C sum_{j geq -1} | Delta_{j}u |_{L^{infty}} leq 2^{s} | u|_{mathcal{C}^{s+epsilon}}C sum_{j geq -1} 2^{-j(s+epsilon)} = C_{s,epsilon} | u |_{mathcal{C}^{s+epsilon}}.$$
It's not clear to me how to get a better bound for the above case $j=-1$. So my question is. How do we get a sharper uniform bound?
Thanks in advance!
functional-analysis besov-space
asked Nov 21 '18 at 15:20
Meagain
16910
add a comment |
I'm sorry if this is a stupid question. But I am currently reading a paper which makes heavy use of the Holder-Besov space
$$mathcal{C}^{s}(mathbb{R}^{n}; mathbb{R}) = B^{s}_{infty,infty}(mathbb{R}^{n};mathbb{R})$$
with norms defined using the Littlewood-Paley decomposition in the sense that
$$ | u |_{mathcal{C}^{s}} = Big| ( 2^{js} | Delta_{j}u|_{L^{infty}} )_{j geq -1} Big|_{ell^{infty}(mathbb{Z})}. $$
Now there is the fact
$$ s leq r Longrightarrow | u |_{mathcal{C}^{s}} leq C | u |_{mathcal{C}^{r}} $$
where $C$ can be chosen independent of $s$ and $r$. The best I can do is to choose $C$ such that $C$ depends on $ s $ and $r$. For example, if $ s，epsilon > 0 $ are given. Then I will have to estimate
$$ 2^{-s} | Delta_{-1} u |_{L^{infty}} leq 2^{-s} C| u |_{L^{infty}} leq 2^{s} C sum_{j geq -1} | Delta_{j}u |_{L^{infty}} leq 2^{s} | u|_{mathcal{C}^{s+epsilon}}C sum_{j geq -1} 2^{-j(s+epsilon)} = C_{s,epsilon} | u |_{mathcal{C}^{s+epsilon}}.$$
It's not clear to me how to get a better bound for the above case $j=-1$. So my question is. How do we get a sharper uniform bound?
Thanks in advance!
functional-analysis besov-space
asked Nov 21 '18 at 15:20
Meagain
16910
I'm sorry if this is a stupid question. But I am currently reading a paper which makes heavy use of the Holder-Besov space
$$mathcal{C}^{s}(mathbb{R}^{n}; mathbb{R}) = B^{s}_{infty,infty}(mathbb{R}^{n};mathbb{R})$$
with norms defined using the Littlewood-Paley decomposition in the sense that
$$ | u |_{mathcal{C}^{s}} = Big| ( 2^{js} | Delta_{j}u|_{L^{infty}} )_{j geq -1} Big|_{ell^{infty}(mathbb{Z})}. $$
Now there is the fact
$$ s leq r Longrightarrow | u |_{mathcal{C}^{s}} leq C | u |_{mathcal{C}^{r}} $$
where $C$ can be chosen independent of $s$ and $r$. The best I can do is to choose $C$ such that $C$ depends on $ s $ and $r$. For example, if $ s，epsilon > 0 $ are given. Then I will have to estimate
$$ 2^{-s} | Delta_{-1} u |_{L^{infty}} leq 2^{-s} C| u |_{L^{infty}} leq 2^{s} C sum_{j geq -1} | Delta_{j}u |_{L^{infty}} leq 2^{s} | u|_{mathcal{C}^{s+epsilon}}C sum_{j geq -1} 2^{-j(s+epsilon)} = C_{s,epsilon} | u |_{mathcal{C}^{s+epsilon}}.$$
It's not clear to me how to get a better bound for the above case $j=-1$. So my question is. How do we get a sharper uniform bound?
Thanks in advance!
functional-analysis besov-space
functional-analysis besov-space
asked Nov 21 '18 at 15:20
Meagain
16910
asked Nov 21 '18 at 15:20
Meagain
16910
asked Nov 21 '18 at 15:20
Meagain
16910
asked Nov 21 '18 at 15:20
Meagain
16910
asked Nov 21 '18 at 15:20
Meagain
16910
16910
add a comment |
add a comment |
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