Mixing
The difference between $(partial f)^{-1}$ and $partial f^*$ on non-reflexive space.
$begingroup$
Let $X$ be a Banach space, $f:Xto Bbb Rcup{infty}$ be a proper function. The multifunction $(partial f)^{-1}$ is defined by
$$
(partial f)^{-1}(x^*) = { xin X : x^* in partial f(x) }.
$$
It can be shown that $(partial f)^{-1} = partial f^*$ if $X$ is reflexive. Of course, $f^*$ is the Fenchel conjugate of $f$.
For a non-reflexive $X$, has anyone seen a concrete example of when
$$
(partial f)^{-1}(x^*) subsetneq partial f^*(x^*)?
$$
More specifically, I am interested in the case that $(partial f)^{-1}(x^*)=emptyset$ but $partial f^*(x^*)$ is non-empty.
real-analysis functional-analysis convex-analysis convex-optimization
asked Feb 1 at 17:41
BigbearZzzBigbearZzz
9,04721653
$endgroup$
add a comment |
$begingroup$
Let $X$ be a Banach space, $f:Xto Bbb Rcup{infty}$ be a proper function. The multifunction $(partial f)^{-1}$ is defined by
$$
(partial f)^{-1}(x^*) = { xin X : x^* in partial f(x) }.
$$
It can be shown that $(partial f)^{-1} = partial f^*$ if $X$ is reflexive. Of course, $f^*$ is the Fenchel conjugate of $f$.
For a non-reflexive $X$, has anyone seen a concrete example of when
$$
(partial f)^{-1}(x^*) subsetneq partial f^*(x^*)?
$$
More specifically, I am interested in the case that $(partial f)^{-1}(x^*)=emptyset$ but $partial f^*(x^*)$ is non-empty.
real-analysis functional-analysis convex-analysis convex-optimization
asked Feb 1 at 17:41
BigbearZzzBigbearZzz
9,04721653
$endgroup$
add a comment |
$begingroup$
Let $X$ be a Banach space, $f:Xto Bbb Rcup{infty}$ be a proper function. The multifunction $(partial f)^{-1}$ is defined by
$$
(partial f)^{-1}(x^*) = { xin X : x^* in partial f(x) }.
$$
It can be shown that $(partial f)^{-1} = partial f^*$ if $X$ is reflexive. Of course, $f^*$ is the Fenchel conjugate of $f$.
For a non-reflexive $X$, has anyone seen a concrete example of when
$$
(partial f)^{-1}(x^*) subsetneq partial f^*(x^*)?
$$
More specifically, I am interested in the case that $(partial f)^{-1}(x^*)=emptyset$ but $partial f^*(x^*)$ is non-empty.
real-analysis functional-analysis convex-analysis convex-optimization
asked Feb 1 at 17:41
BigbearZzzBigbearZzz
9,04721653
$endgroup$
Let $X$ be a Banach space, $f:Xto Bbb Rcup{infty}$ be a proper function. The multifunction $(partial f)^{-1}$ is defined by
$$
(partial f)^{-1}(x^*) = { xin X : x^* in partial f(x) }.
$$
It can be shown that $(partial f)^{-1} = partial f^*$ if $X$ is reflexive. Of course, $f^*$ is the Fenchel conjugate of $f$.
For a non-reflexive $X$, has anyone seen a concrete example of when
$$
(partial f)^{-1}(x^*) subsetneq partial f^*(x^*)?
$$
More specifically, I am interested in the case that $(partial f)^{-1}(x^*)=emptyset$ but $partial f^*(x^*)$ is non-empty.
real-analysis functional-analysis convex-analysis convex-optimization
real-analysis functional-analysis convex-analysis convex-optimization
asked Feb 1 at 17:41
BigbearZzzBigbearZzz
9,04721653
asked Feb 1 at 17:41
BigbearZzzBigbearZzz
9,04721653
asked Feb 1 at 17:41
BigbearZzzBigbearZzz
9,04721653
asked Feb 1 at 17:41
BigbearZzzBigbearZzz
9,04721653
asked Feb 1 at 17:41
BigbearZzzBigbearZzz
9,04721653
9,04721653
add a comment |
add a comment |
1 Answer
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$begingroup$
The following should work:
Let $X=c_0$ and $f=tfrac{1}{2}|cdot|^2$ on $c_0$.
Then $f^*$ looks likewise, but uses the norm on $c_0^*=ell_1$.
Now take a sequence in $ell_1$ such as $x^* =(1/2^n)_{ngeq 1}$.
Then $partial f^*(x^*)={(1,1,ldots)_{ngeq 1}}$
but $(partial f)^{-1}=varnothing$ (because $(1,1,ldots)notin c_0$).
answered Feb 1 at 18:41
max_zornmax_zorn
3,44061429
$endgroup$
$begingroup$
Interesting example. I have one question, if you don't mind. I know the part about getting $f^*$ from $f$ but how did you calculate $partial f^*(x^*)={(1,1,ldots)_{ngeq 1}}$? I've never done calculation like this before.
$endgroup$
– BigbearZzz
Feb 4 at 16:38
$begingroup$
Write $J=partial tfrac{1}{2}|cdot|^2$. Then $(Jz)_n = |z|mathrm{Sign}(z_n)$ by the chain rule, where $mathrm{Sign}$ is the sign function, but with $mathrm{Sign}(0)=[-1,1]$. Unfortunately, I don't have a good reference for this - but websearch duality map or normalized duality map. (If anybody has one, please post.) I recommend you try this for the $ell_1$ norm on $mathbb{R}^2$ - this example is nothing but that generalization.
$endgroup$
– max_zorn
Feb 4 at 17:36
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
The following should work:
Let $X=c_0$ and $f=tfrac{1}{2}|cdot|^2$ on $c_0$.
Then $f^*$ looks likewise, but uses the norm on $c_0^*=ell_1$.
Now take a sequence in $ell_1$ such as $x^* =(1/2^n)_{ngeq 1}$.
Then $partial f^*(x^*)={(1,1,ldots)_{ngeq 1}}$
but $(partial f)^{-1}=varnothing$ (because $(1,1,ldots)notin c_0$).
answered Feb 1 at 18:41
max_zornmax_zorn
3,44061429
$endgroup$
$begingroup$
Interesting example. I have one question, if you don't mind. I know the part about getting $f^*$ from $f$ but how did you calculate $partial f^*(x^*)={(1,1,ldots)_{ngeq 1}}$? I've never done calculation like this before.
$endgroup$
– BigbearZzz
Feb 4 at 16:38
$begingroup$
Write $J=partial tfrac{1}{2}|cdot|^2$. Then $(Jz)_n = |z|mathrm{Sign}(z_n)$ by the chain rule, where $mathrm{Sign}$ is the sign function, but with $mathrm{Sign}(0)=[-1,1]$. Unfortunately, I don't have a good reference for this - but websearch duality map or normalized duality map. (If anybody has one, please post.) I recommend you try this for the $ell_1$ norm on $mathbb{R}^2$ - this example is nothing but that generalization.
$endgroup$
– max_zorn
Feb 4 at 17:36
add a comment |
$begingroup$
The following should work:
Let $X=c_0$ and $f=tfrac{1}{2}|cdot|^2$ on $c_0$.
Then $f^*$ looks likewise, but uses the norm on $c_0^*=ell_1$.
Now take a sequence in $ell_1$ such as $x^* =(1/2^n)_{ngeq 1}$.
Then $partial f^*(x^*)={(1,1,ldots)_{ngeq 1}}$
but $(partial f)^{-1}=varnothing$ (because $(1,1,ldots)notin c_0$).
answered Feb 1 at 18:41
max_zornmax_zorn
3,44061429
$endgroup$
$begingroup$
Interesting example. I have one question, if you don't mind. I know the part about getting $f^*$ from $f$ but how did you calculate $partial f^*(x^*)={(1,1,ldots)_{ngeq 1}}$? I've never done calculation like this before.
$endgroup$
– BigbearZzz
Feb 4 at 16:38
$begingroup$
Write $J=partial tfrac{1}{2}|cdot|^2$. Then $(Jz)_n = |z|mathrm{Sign}(z_n)$ by the chain rule, where $mathrm{Sign}$ is the sign function, but with $mathrm{Sign}(0)=[-1,1]$. Unfortunately, I don't have a good reference for this - but websearch duality map or normalized duality map. (If anybody has one, please post.) I recommend you try this for the $ell_1$ norm on $mathbb{R}^2$ - this example is nothing but that generalization.
$endgroup$
– max_zorn
Feb 4 at 17:36
add a comment |
$begingroup$
The following should work:
Let $X=c_0$ and $f=tfrac{1}{2}|cdot|^2$ on $c_0$.
Then $f^*$ looks likewise, but uses the norm on $c_0^*=ell_1$.
Now take a sequence in $ell_1$ such as $x^* =(1/2^n)_{ngeq 1}$.
Then $partial f^*(x^*)={(1,1,ldots)_{ngeq 1}}$
but $(partial f)^{-1}=varnothing$ (because $(1,1,ldots)notin c_0$).
answered Feb 1 at 18:41
max_zornmax_zorn
3,44061429
$endgroup$
The following should work:
Let $X=c_0$ and $f=tfrac{1}{2}|cdot|^2$ on $c_0$.
Then $f^*$ looks likewise, but uses the norm on $c_0^*=ell_1$.
Now take a sequence in $ell_1$ such as $x^* =(1/2^n)_{ngeq 1}$.
Then $partial f^*(x^*)={(1,1,ldots)_{ngeq 1}}$
but $(partial f)^{-1}=varnothing$ (because $(1,1,ldots)notin c_0$).
answered Feb 1 at 18:41
max_zornmax_zorn
3,44061429
answered Feb 1 at 18:41
max_zornmax_zorn
3,44061429
answered Feb 1 at 18:41
max_zornmax_zorn
3,44061429
answered Feb 1 at 18:41
max_zornmax_zorn
3,44061429
3,44061429
$begingroup$
Interesting example. I have one question, if you don't mind. I know the part about getting $f^*$ from $f$ but how did you calculate $partial f^*(x^*)={(1,1,ldots)_{ngeq 1}}$? I've never done calculation like this before.
$endgroup$
– BigbearZzz
Feb 4 at 16:38
$begingroup$
Write $J=partial tfrac{1}{2}|cdot|^2$. Then $(Jz)_n = |z|mathrm{Sign}(z_n)$ by the chain rule, where $mathrm{Sign}$ is the sign function, but with $mathrm{Sign}(0)=[-1,1]$. Unfortunately, I don't have a good reference for this - but websearch duality map or normalized duality map. (If anybody has one, please post.) I recommend you try this for the $ell_1$ norm on $mathbb{R}^2$ - this example is nothing but that generalization.
$endgroup$
– max_zorn
Feb 4 at 17:36
add a comment |
$begingroup$
Interesting example. I have one question, if you don't mind. I know the part about getting $f^*$ from $f$ but how did you calculate $partial f^*(x^*)={(1,1,ldots)_{ngeq 1}}$? I've never done calculation like this before.
$endgroup$
– BigbearZzz
Feb 4 at 16:38
$begingroup$
Write $J=partial tfrac{1}{2}|cdot|^2$. Then $(Jz)_n = |z|mathrm{Sign}(z_n)$ by the chain rule, where $mathrm{Sign}$ is the sign function, but with $mathrm{Sign}(0)=[-1,1]$. Unfortunately, I don't have a good reference for this - but websearch duality map or normalized duality map. (If anybody has one, please post.) I recommend you try this for the $ell_1$ norm on $mathbb{R}^2$ - this example is nothing but that generalization.
$endgroup$
– max_zorn
Feb 4 at 17:36
$begingroup$
Interesting example. I have one question, if you don't mind. I know the part about getting $f^*$ from $f$ but how did you calculate $partial f^*(x^*)={(1,1,ldots)_{ngeq 1}}$? I've never done calculation like this before.
$endgroup$
– BigbearZzz
Feb 4 at 16:38
$begingroup$
Interesting example. I have one question, if you don't mind. I know the part about getting $f^*$ from $f$ but how did you calculate $partial f^*(x^*)={(1,1,ldots)_{ngeq 1}}$? I've never done calculation like this before.
$endgroup$
– BigbearZzz
Feb 4 at 16:38
$begingroup$
Write $J=partial tfrac{1}{2}|cdot|^2$. Then $(Jz)_n = |z|mathrm{Sign}(z_n)$ by the chain rule, where $mathrm{Sign}$ is the sign function, but with $mathrm{Sign}(0)=[-1,1]$. Unfortunately, I don't have a good reference for this - but websearch duality map or normalized duality map. (If anybody has one, please post.) I recommend you try this for the $ell_1$ norm on $mathbb{R}^2$ - this example is nothing but that generalization.
$endgroup$
– max_zorn
Feb 4 at 17:36
$begingroup$
Write $J=partial tfrac{1}{2}|cdot|^2$. Then $(Jz)_n = |z|mathrm{Sign}(z_n)$ by the chain rule, where $mathrm{Sign}$ is the sign function, but with $mathrm{Sign}(0)=[-1,1]$. Unfortunately, I don't have a good reference for this - but websearch duality map or normalized duality map. (If anybody has one, please post.) I recommend you try this for the $ell_1$ norm on $mathbb{R}^2$ - this example is nothing but that generalization.
$endgroup$
– max_zorn
Feb 4 at 17:36
add a comment |
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