an upper bound of a $d$ dimensional vector under a length constraint.
Let $k = (k_1, cdots, k_d) in mathbb{N}^d$ and $s = (s_1, cdots, s_d) in mathbb{R}_+^d$. define
$$k^s = (k_1^{s_1},cdots,k_d^{s_d})in mathbb{R}_+^d.$$
Let $|cdot|_p$ denote the $p$-norm on Euclidean space $mathbb{R}^d$.
Under the constraint $|k|_2 ge n>0$ ($p = 2$ is somehow conventional only, as all norms are equivalent), what is the upper bound (possibly with a dimension related constant) for
$$sup_{|k|_2 ge n} |k^{-s}|_2,$$
where $k^{-s} = (k_1^{-s_1},cdots,k_d^{-s_d})$ and $s_1, cdots, s_d >0$.
A bit of background. The question rises in an attempt to prove some embedding theorem for (anisotropic) Sobolev space. When $s_1 = cdots = s_d$, one can see that the upper bound actually gives the error bound on the approximation in the span of the first $n$ bases.
inequality sobolev-spaces
edited Nov 29 at 14:46
Alex Ravsky
38.8k32079
asked Nov 26 at 22:51
newbie
1,49611936
add a comment |
Let $k = (k_1, cdots, k_d) in mathbb{N}^d$ and $s = (s_1, cdots, s_d) in mathbb{R}_+^d$. define
$$k^s = (k_1^{s_1},cdots,k_d^{s_d})in mathbb{R}_+^d.$$
Let $|cdot|_p$ denote the $p$-norm on Euclidean space $mathbb{R}^d$.
Under the constraint $|k|_2 ge n>0$ ($p = 2$ is somehow conventional only, as all norms are equivalent), what is the upper bound (possibly with a dimension related constant) for
$$sup_{|k|_2 ge n} |k^{-s}|_2,$$
where $k^{-s} = (k_1^{-s_1},cdots,k_d^{-s_d})$ and $s_1, cdots, s_d >0$.
A bit of background. The question rises in an attempt to prove some embedding theorem for (anisotropic) Sobolev space. When $s_1 = cdots = s_d$, one can see that the upper bound actually gives the error bound on the approximation in the span of the first $n$ bases.
inequality sobolev-spaces
edited Nov 29 at 14:46
Alex Ravsky
38.8k32079
asked Nov 26 at 22:51
newbie
1,49611936
add a comment |
Let $k = (k_1, cdots, k_d) in mathbb{N}^d$ and $s = (s_1, cdots, s_d) in mathbb{R}_+^d$. define
$$k^s = (k_1^{s_1},cdots,k_d^{s_d})in mathbb{R}_+^d.$$
Let $|cdot|_p$ denote the $p$-norm on Euclidean space $mathbb{R}^d$.
Under the constraint $|k|_2 ge n>0$ ($p = 2$ is somehow conventional only, as all norms are equivalent), what is the upper bound (possibly with a dimension related constant) for
$$sup_{|k|_2 ge n} |k^{-s}|_2,$$
where $k^{-s} = (k_1^{-s_1},cdots,k_d^{-s_d})$ and $s_1, cdots, s_d >0$.
A bit of background. The question rises in an attempt to prove some embedding theorem for (anisotropic) Sobolev space. When $s_1 = cdots = s_d$, one can see that the upper bound actually gives the error bound on the approximation in the span of the first $n$ bases.
inequality sobolev-spaces
edited Nov 29 at 14:46
Alex Ravsky
38.8k32079
asked Nov 26 at 22:51
newbie
1,49611936
Let $k = (k_1, cdots, k_d) in mathbb{N}^d$ and $s = (s_1, cdots, s_d) in mathbb{R}_+^d$. define
$$k^s = (k_1^{s_1},cdots,k_d^{s_d})in mathbb{R}_+^d.$$
Let $|cdot|_p$ denote the $p$-norm on Euclidean space $mathbb{R}^d$.
Under the constraint $|k|_2 ge n>0$ ($p = 2$ is somehow conventional only, as all norms are equivalent), what is the upper bound (possibly with a dimension related constant) for
$$sup_{|k|_2 ge n} |k^{-s}|_2,$$
where $k^{-s} = (k_1^{-s_1},cdots,k_d^{-s_d})$ and $s_1, cdots, s_d >0$.
A bit of background. The question rises in an attempt to prove some embedding theorem for (anisotropic) Sobolev space. When $s_1 = cdots = s_d$, one can see that the upper bound actually gives the error bound on the approximation in the span of the first $n$ bases.
inequality sobolev-spaces
inequality sobolev-spaces
edited Nov 29 at 14:46
Alex Ravsky
38.8k32079
asked Nov 26 at 22:51
newbie
1,49611936
edited Nov 29 at 14:46
Alex Ravsky
38.8k32079
asked Nov 26 at 22:51
newbie
1,49611936
edited Nov 29 at 14:46
Alex Ravsky
38.8k32079
edited Nov 29 at 14:46
Alex Ravsky
38.8k32079
edited Nov 29 at 14:46
Alex Ravsky
38.8k32079
38.8k32079
asked Nov 26 at 22:51
newbie
1,49611936
asked Nov 26 at 22:51
newbie
1,49611936
asked Nov 26 at 22:51
newbie
1,49611936
1,49611936
add a comment |
add a comment |
1 Answer
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For each $s$ we have
$$|k^{-s}|_2=left|sum k_i^{-2s_i}right|_2le left|sum k_i^{0}right|_2=sqrt{d}.$$
On the other hand, put $k=(1,1,dots,1, lceilsqrt{n}rceil)$. Then $|k|_2ge n$, but
$$|k^{-s}|_2>left|sum_{i=1}^{d-1} 1^{-2s_i}right|_2=sqrt{d-1}.$$
answered Nov 29 at 14:43
Alex Ravsky
38.8k32079
add a comment |
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1 Answer
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1 Answer
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For each $s$ we have
$$|k^{-s}|_2=left|sum k_i^{-2s_i}right|_2le left|sum k_i^{0}right|_2=sqrt{d}.$$
On the other hand, put $k=(1,1,dots,1, lceilsqrt{n}rceil)$. Then $|k|_2ge n$, but
$$|k^{-s}|_2>left|sum_{i=1}^{d-1} 1^{-2s_i}right|_2=sqrt{d-1}.$$
answered Nov 29 at 14:43
Alex Ravsky
38.8k32079
add a comment |
For each $s$ we have
$$|k^{-s}|_2=left|sum k_i^{-2s_i}right|_2le left|sum k_i^{0}right|_2=sqrt{d}.$$
On the other hand, put $k=(1,1,dots,1, lceilsqrt{n}rceil)$. Then $|k|_2ge n$, but
$$|k^{-s}|_2>left|sum_{i=1}^{d-1} 1^{-2s_i}right|_2=sqrt{d-1}.$$
answered Nov 29 at 14:43
Alex Ravsky
38.8k32079
add a comment |
For each $s$ we have
$$|k^{-s}|_2=left|sum k_i^{-2s_i}right|_2le left|sum k_i^{0}right|_2=sqrt{d}.$$
On the other hand, put $k=(1,1,dots,1, lceilsqrt{n}rceil)$. Then $|k|_2ge n$, but
$$|k^{-s}|_2>left|sum_{i=1}^{d-1} 1^{-2s_i}right|_2=sqrt{d-1}.$$
answered Nov 29 at 14:43
Alex Ravsky
38.8k32079
For each $s$ we have
$$|k^{-s}|_2=left|sum k_i^{-2s_i}right|_2le left|sum k_i^{0}right|_2=sqrt{d}.$$
On the other hand, put $k=(1,1,dots,1, lceilsqrt{n}rceil)$. Then $|k|_2ge n$, but
$$|k^{-s}|_2>left|sum_{i=1}^{d-1} 1^{-2s_i}right|_2=sqrt{d-1}.$$
answered Nov 29 at 14:43
Alex Ravsky
38.8k32079
answered Nov 29 at 14:43
Alex Ravsky
38.8k32079
answered Nov 29 at 14:43
Alex Ravsky
38.8k32079
answered Nov 29 at 14:43
Alex Ravsky
38.8k32079
38.8k32079
add a comment |
add a comment |
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