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If $f$ is proper, lsc, and $fracf(x) + f(y)2 = f^**left(fracx + y2right) implies x = y$, is $f$ necessarily convex?
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Suppose $X$ is a real Hilbert Space and $f : X to (-infty, infty]$ is a lower semicontinuous, proper function. Further, suppose $f$ satisfies the following, for all $x, y in operatornamedom f$:
$$fracf(x) + f(y)2 = f^**left(fracx + y2right) implies x = y.$$
Is $f$ necessarily a convex function?
Here $^*$ refers to the Fenchel conjugate, and $operatornamedom f$ is the set of points $x in X$ such that $f(x) neq infty$.
I know that:
- $f^**(x) le f(x)$ for all $x$ and $f^**(x) = f(x)$ for all $x$ if and only if $f$ is convex (and lsc).
- In fact, $f^**$ is the greatest convex minorant of $f$.
- This means that
$$fracf(x) + f(y)2 ge fracf^**(x) + f^**(y)2 ge f^**left(fracx + y2right)$$
for all $x, y in operatornamedom f$. - Therefore,
$$fracf(x) + f(y)2 = f^**left(fracx + y2right)$$
implies that $f(x) = f^**(x)$ and $f(y) = f^**(y)$. - Another consequence is that $f^**(lambda x + (1 - lambda y)) = lambda f^**(x) + (1 - lambda)f^**(y)$ for all $lambda in [0, 1]$.
My thoughts:
- Really, I just need to establish that $f^**(x) = f(x)$ for all $x$.
- Despite biduals showing up both in the premises and the above desired conclusion, there doesn't seem to be a direct path to manipulate one to the other, especially since not every point in $overlineoperatornameconv operatornamedom f$ can be expressed as $fracx+y2$ where $x, y in operatornamedom f$.
- The function $g(x, y) = fracf(x)+f(y)2 - f^**left(fracx + y2right)$ is not a metric in general, even if $g(x, y) = 0 implies x = y$.
- I get a feeling that Stegall's variational principle might help, for a variety of reasons, but one handy reason is that we may add any linear functional to $f$, without changing $g$.
Any thoughts are welcome!
functional-analysis convex-analysis variational-analysis
asked Aug 1 at 5:21
Theo Bendit
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This question has an open bounty worth +400
reputation from Theo Bendit ending ending at 2018-08-14 04:59:49Z">in 5 days.
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Suppose $X$ is a real Hilbert Space and $f : X to (-infty, infty]$ is a lower semicontinuous, proper function. Further, suppose $f$ satisfies the following, for all $x, y in operatornamedom f$:
$$fracf(x) + f(y)2 = f^**left(fracx + y2right) implies x = y.$$
Is $f$ necessarily a convex function?
Here $^*$ refers to the Fenchel conjugate, and $operatornamedom f$ is the set of points $x in X$ such that $f(x) neq infty$.
I know that:
- $f^**(x) le f(x)$ for all $x$ and $f^**(x) = f(x)$ for all $x$ if and only if $f$ is convex (and lsc).
- In fact, $f^**$ is the greatest convex minorant of $f$.
- This means that
$$fracf(x) + f(y)2 ge fracf^**(x) + f^**(y)2 ge f^**left(fracx + y2right)$$
for all $x, y in operatornamedom f$. - Therefore,
$$fracf(x) + f(y)2 = f^**left(fracx + y2right)$$
implies that $f(x) = f^**(x)$ and $f(y) = f^**(y)$. - Another consequence is that $f^**(lambda x + (1 - lambda y)) = lambda f^**(x) + (1 - lambda)f^**(y)$ for all $lambda in [0, 1]$.
My thoughts:
- Really, I just need to establish that $f^**(x) = f(x)$ for all $x$.
- Despite biduals showing up both in the premises and the above desired conclusion, there doesn't seem to be a direct path to manipulate one to the other, especially since not every point in $overlineoperatornameconv operatornamedom f$ can be expressed as $fracx+y2$ where $x, y in operatornamedom f$.
- The function $g(x, y) = fracf(x)+f(y)2 - f^**left(fracx + y2right)$ is not a metric in general, even if $g(x, y) = 0 implies x = y$.
- I get a feeling that Stegall's variational principle might help, for a variety of reasons, but one handy reason is that we may add any linear functional to $f$, without changing $g$.
Any thoughts are welcome!
functional-analysis convex-analysis variational-analysis
asked Aug 1 at 5:21
Theo Bendit
11.7k1841
This question has an open bounty worth +400
reputation from Theo Bendit ending ending at 2018-08-14 04:59:49Z">in 5 days.
This question has not received enough attention.
What does "proper" mean?
– mathworker21
yesterday
Proper means $operatornamedom f neq emptyset$.
– Theo Bendit
13 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose $X$ is a real Hilbert Space and $f : X to (-infty, infty]$ is a lower semicontinuous, proper function. Further, suppose $f$ satisfies the following, for all $x, y in operatornamedom f$:
$$fracf(x) + f(y)2 = f^**left(fracx + y2right) implies x = y.$$
Is $f$ necessarily a convex function?
Here $^*$ refers to the Fenchel conjugate, and $operatornamedom f$ is the set of points $x in X$ such that $f(x) neq infty$.
I know that:
- $f^**(x) le f(x)$ for all $x$ and $f^**(x) = f(x)$ for all $x$ if and only if $f$ is convex (and lsc).
- In fact, $f^**$ is the greatest convex minorant of $f$.
- This means that
$$fracf(x) + f(y)2 ge fracf^**(x) + f^**(y)2 ge f^**left(fracx + y2right)$$
for all $x, y in operatornamedom f$. - Therefore,
$$fracf(x) + f(y)2 = f^**left(fracx + y2right)$$
implies that $f(x) = f^**(x)$ and $f(y) = f^**(y)$. - Another consequence is that $f^**(lambda x + (1 - lambda y)) = lambda f^**(x) + (1 - lambda)f^**(y)$ for all $lambda in [0, 1]$.
My thoughts:
- Really, I just need to establish that $f^**(x) = f(x)$ for all $x$.
- Despite biduals showing up both in the premises and the above desired conclusion, there doesn't seem to be a direct path to manipulate one to the other, especially since not every point in $overlineoperatornameconv operatornamedom f$ can be expressed as $fracx+y2$ where $x, y in operatornamedom f$.
- The function $g(x, y) = fracf(x)+f(y)2 - f^**left(fracx + y2right)$ is not a metric in general, even if $g(x, y) = 0 implies x = y$.
- I get a feeling that Stegall's variational principle might help, for a variety of reasons, but one handy reason is that we may add any linear functional to $f$, without changing $g$.
Any thoughts are welcome!
functional-analysis convex-analysis variational-analysis
asked Aug 1 at 5:21
Theo Bendit
11.7k1841
Suppose $X$ is a real Hilbert Space and $f : X to (-infty, infty]$ is a lower semicontinuous, proper function. Further, suppose $f$ satisfies the following, for all $x, y in operatornamedom f$:
$$fracf(x) + f(y)2 = f^**left(fracx + y2right) implies x = y.$$
Is $f$ necessarily a convex function?
Here $^*$ refers to the Fenchel conjugate, and $operatornamedom f$ is the set of points $x in X$ such that $f(x) neq infty$.
I know that:
- $f^**(x) le f(x)$ for all $x$ and $f^**(x) = f(x)$ for all $x$ if and only if $f$ is convex (and lsc).
- In fact, $f^**$ is the greatest convex minorant of $f$.
- This means that
$$fracf(x) + f(y)2 ge fracf^**(x) + f^**(y)2 ge f^**left(fracx + y2right)$$
for all $x, y in operatornamedom f$. - Therefore,
$$fracf(x) + f(y)2 = f^**left(fracx + y2right)$$
implies that $f(x) = f^**(x)$ and $f(y) = f^**(y)$. - Another consequence is that $f^**(lambda x + (1 - lambda y)) = lambda f^**(x) + (1 - lambda)f^**(y)$ for all $lambda in [0, 1]$.
My thoughts:
- Really, I just need to establish that $f^**(x) = f(x)$ for all $x$.
- Despite biduals showing up both in the premises and the above desired conclusion, there doesn't seem to be a direct path to manipulate one to the other, especially since not every point in $overlineoperatornameconv operatornamedom f$ can be expressed as $fracx+y2$ where $x, y in operatornamedom f$.
- The function $g(x, y) = fracf(x)+f(y)2 - f^**left(fracx + y2right)$ is not a metric in general, even if $g(x, y) = 0 implies x = y$.
- I get a feeling that Stegall's variational principle might help, for a variety of reasons, but one handy reason is that we may add any linear functional to $f$, without changing $g$.
Any thoughts are welcome!
functional-analysis convex-analysis variational-analysis
asked Aug 1 at 5:21
Theo Bendit
11.7k1841
asked Aug 1 at 5:21
Theo Bendit
11.7k1841
asked Aug 1 at 5:21
Theo Bendit
11.7k1841
asked Aug 1 at 5:21
Theo Bendit
11.7k1841
11.7k1841
This question has an open bounty worth +400
reputation from Theo Bendit ending ending at 2018-08-14 04:59:49Z">in 5 days.
This question has not received enough attention.
This question has an open bounty worth +400
reputation from Theo Bendit ending ending at 2018-08-14 04:59:49Z">in 5 days.
This question has not received enough attention.
What does "proper" mean?
– mathworker21
yesterday
Proper means $operatornamedom f neq emptyset$.
– Theo Bendit
13 hours ago
add a comment |Â
What does "proper" mean?
– mathworker21
yesterday
Proper means $operatornamedom f neq emptyset$.
– Theo Bendit
13 hours ago
What does "proper" mean?
– mathworker21
yesterday
What does "proper" mean?
– mathworker21
yesterday
Proper means $operatornamedom f neq emptyset$.
– Theo Bendit
13 hours ago
Proper means $operatornamedom f neq emptyset$.
– Theo Bendit
13 hours ago
add a comment |Â
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What does "proper" mean?
– mathworker21
yesterday
Proper means $operatornamedom f neq emptyset$.
– Theo Bendit
13 hours ago