Unicity of the projection on a closed convex subset of $L^p$
$begingroup$
The original problem (not full) statement is the following :
Let $(Omega, mathfrak{B},mu)$ a measured space and $p>1$ and let $C$ be a closed convex subset of $L^p$.
For $uin L^p$ let $d_u:=inf{|phi-u |, phi in C}$
Through some questions, the problem shows the existence of a $phi in C$ such that $|phi-u |=d_u$.
The problem I have is with unicity. I read somewhere on MSE that the $L^p$ norm is strictly convex, which implies unicity.
As far as I know, no norm is strictly convex, so the previous affirmation might be wrong so how could we deal with unicity.
functional-analysis measure-theory lebesgue-measure lp-spaces projection
asked Dec 5 '18 at 11:21
Oussama SarihOussama Sarih
47827
$endgroup$
add a comment |
$begingroup$
The original problem (not full) statement is the following :
Let $(Omega, mathfrak{B},mu)$ a measured space and $p>1$ and let $C$ be a closed convex subset of $L^p$.
For $uin L^p$ let $d_u:=inf{|phi-u |, phi in C}$
Through some questions, the problem shows the existence of a $phi in C$ such that $|phi-u |=d_u$.
The problem I have is with unicity. I read somewhere on MSE that the $L^p$ norm is strictly convex, which implies unicity.
As far as I know, no norm is strictly convex, so the previous affirmation might be wrong so how could we deal with unicity.
functional-analysis measure-theory lebesgue-measure lp-spaces projection
asked Dec 5 '18 at 11:21
Oussama SarihOussama Sarih
47827
$endgroup$
1
$begingroup$
See en.wikipedia.org/wiki/Strictly_convex_space
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 11:52
2
$begingroup$
is $p=infty$ possible? Because i think then the solution is not unique
$endgroup$
– supinf
Dec 5 '18 at 11:59
1
$begingroup$
Small remark on the jargon: One usually speaks about "uniqueness" and not "unicity".
$endgroup$
– MaoWao
Dec 5 '18 at 13:28
add a comment |
$begingroup$
The original problem (not full) statement is the following :
Let $(Omega, mathfrak{B},mu)$ a measured space and $p>1$ and let $C$ be a closed convex subset of $L^p$.
For $uin L^p$ let $d_u:=inf{|phi-u |, phi in C}$
Through some questions, the problem shows the existence of a $phi in C$ such that $|phi-u |=d_u$.
The problem I have is with unicity. I read somewhere on MSE that the $L^p$ norm is strictly convex, which implies unicity.
As far as I know, no norm is strictly convex, so the previous affirmation might be wrong so how could we deal with unicity.
functional-analysis measure-theory lebesgue-measure lp-spaces projection
asked Dec 5 '18 at 11:21
Oussama SarihOussama Sarih
47827
$endgroup$
The original problem (not full) statement is the following :
Let $(Omega, mathfrak{B},mu)$ a measured space and $p>1$ and let $C$ be a closed convex subset of $L^p$.
For $uin L^p$ let $d_u:=inf{|phi-u |, phi in C}$
Through some questions, the problem shows the existence of a $phi in C$ such that $|phi-u |=d_u$.
The problem I have is with unicity. I read somewhere on MSE that the $L^p$ norm is strictly convex, which implies unicity.
As far as I know, no norm is strictly convex, so the previous affirmation might be wrong so how could we deal with unicity.
functional-analysis measure-theory lebesgue-measure lp-spaces projection
functional-analysis measure-theory lebesgue-measure lp-spaces projection
asked Dec 5 '18 at 11:21
Oussama SarihOussama Sarih
47827
asked Dec 5 '18 at 11:21
Oussama SarihOussama Sarih
47827
asked Dec 5 '18 at 11:21
Oussama SarihOussama Sarih
47827
asked Dec 5 '18 at 11:21
Oussama SarihOussama Sarih
47827
asked Dec 5 '18 at 11:21
Oussama SarihOussama Sarih
47827
47827
1
$begingroup$
See en.wikipedia.org/wiki/Strictly_convex_space
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 11:52
2
$begingroup$
is $p=infty$ possible? Because i think then the solution is not unique
$endgroup$
– supinf
Dec 5 '18 at 11:59
1
$begingroup$
Small remark on the jargon: One usually speaks about "uniqueness" and not "unicity".
$endgroup$
– MaoWao
Dec 5 '18 at 13:28
add a comment |
1
$begingroup$
See en.wikipedia.org/wiki/Strictly_convex_space
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 11:52
2
$begingroup$
is $p=infty$ possible? Because i think then the solution is not unique
$endgroup$
– supinf
Dec 5 '18 at 11:59
1
$begingroup$
Small remark on the jargon: One usually speaks about "uniqueness" and not "unicity".
$endgroup$
– MaoWao
Dec 5 '18 at 13:28
1
1
$begingroup$
See en.wikipedia.org/wiki/Strictly_convex_space
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 11:52
$begingroup$
See en.wikipedia.org/wiki/Strictly_convex_space
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 11:52
2
2
$begingroup$
is $p=infty$ possible? Because i think then the solution is not unique
$endgroup$
– supinf
Dec 5 '18 at 11:59
$begingroup$
is $p=infty$ possible? Because i think then the solution is not unique
$endgroup$
– supinf
Dec 5 '18 at 11:59
1
1
$begingroup$
Small remark on the jargon: One usually speaks about "uniqueness" and not "unicity".
$endgroup$
– MaoWao
Dec 5 '18 at 13:28
$begingroup$
Small remark on the jargon: One usually speaks about "uniqueness" and not "unicity".
$endgroup$
– MaoWao
Dec 5 '18 at 13:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
For $pin(1,infty)$, assume there are two different solutions $phi,psiin C$.
Then, by convexity of the norm, one can show that $(phi+psi)/2$ is closer to $u$ than $phi$ or $psi$.
In order to show the last part, you have to use the fact that for real numbers $a,b$ it holds that
$$
| (a+b)/2 | ^p = (|a|^p+|b|^p)/2
quadLeftrightarrowquad
a = b.
$$
answered Dec 5 '18 at 12:11
supinfsupinf
6,3111028
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
For $pin(1,infty)$, assume there are two different solutions $phi,psiin C$.
Then, by convexity of the norm, one can show that $(phi+psi)/2$ is closer to $u$ than $phi$ or $psi$.
In order to show the last part, you have to use the fact that for real numbers $a,b$ it holds that
$$
| (a+b)/2 | ^p = (|a|^p+|b|^p)/2
quadLeftrightarrowquad
a = b.
$$
answered Dec 5 '18 at 12:11
supinfsupinf
6,3111028
$endgroup$
add a comment |
$begingroup$
Hint:
For $pin(1,infty)$, assume there are two different solutions $phi,psiin C$.
Then, by convexity of the norm, one can show that $(phi+psi)/2$ is closer to $u$ than $phi$ or $psi$.
In order to show the last part, you have to use the fact that for real numbers $a,b$ it holds that
$$
| (a+b)/2 | ^p = (|a|^p+|b|^p)/2
quadLeftrightarrowquad
a = b.
$$
answered Dec 5 '18 at 12:11
supinfsupinf
6,3111028
$endgroup$
add a comment |
$begingroup$
Hint:
For $pin(1,infty)$, assume there are two different solutions $phi,psiin C$.
Then, by convexity of the norm, one can show that $(phi+psi)/2$ is closer to $u$ than $phi$ or $psi$.
In order to show the last part, you have to use the fact that for real numbers $a,b$ it holds that
$$
| (a+b)/2 | ^p = (|a|^p+|b|^p)/2
quadLeftrightarrowquad
a = b.
$$
answered Dec 5 '18 at 12:11
supinfsupinf
6,3111028
$endgroup$
Hint:
For $pin(1,infty)$, assume there are two different solutions $phi,psiin C$.
Then, by convexity of the norm, one can show that $(phi+psi)/2$ is closer to $u$ than $phi$ or $psi$.
In order to show the last part, you have to use the fact that for real numbers $a,b$ it holds that
$$
| (a+b)/2 | ^p = (|a|^p+|b|^p)/2
quadLeftrightarrowquad
a = b.
$$
answered Dec 5 '18 at 12:11
supinfsupinf
6,3111028
answered Dec 5 '18 at 12:11
supinfsupinf
6,3111028
answered Dec 5 '18 at 12:11
supinfsupinf
6,3111028
answered Dec 5 '18 at 12:11
supinfsupinf
6,3111028
6,3111028
add a comment |
add a comment |
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1
$begingroup$
See en.wikipedia.org/wiki/Strictly_convex_space
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 11:52
2
$begingroup$
is $p=infty$ possible? Because i think then the solution is not unique
$endgroup$
– supinf
Dec 5 '18 at 11:59
1
$begingroup$
Small remark on the jargon: One usually speaks about "uniqueness" and not "unicity".
$endgroup$
– MaoWao
Dec 5 '18 at 13:28