Maximizing a strictly concave function over a compact convex set
$begingroup$
Let $f: S to mathbb{R}$ be a (strictly) concave function, where
$S := {y in mathbb{R}^m: ygeq 0,, sum_{i=1}^m y_i=1 }$.
I want to show that there is a $y^*in S$, which maximizes $f$. $S$ is a convex set and I found some results from convex optimization that strictly convex functions on convex sets have unique minimizers if local minima exist. Here, the set $S$ is also compact; how can we use this to prove the existence of a maximizing $y^*in S$?
Thanks in advance!
optimization convex-analysis convex-optimization
asked Dec 2 '18 at 22:55
Max93Max93
31029
$endgroup$
add a comment |
$begingroup$
Let $f: S to mathbb{R}$ be a (strictly) concave function, where
$S := {y in mathbb{R}^m: ygeq 0,, sum_{i=1}^m y_i=1 }$.
I want to show that there is a $y^*in S$, which maximizes $f$. $S$ is a convex set and I found some results from convex optimization that strictly convex functions on convex sets have unique minimizers if local minima exist. Here, the set $S$ is also compact; how can we use this to prove the existence of a maximizing $y^*in S$?
Thanks in advance!
optimization convex-analysis convex-optimization
asked Dec 2 '18 at 22:55
Max93Max93
31029
$endgroup$
$begingroup$
Apply the result yo have quoted to $-f$.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:15
$begingroup$
This result needs the existence of local minima of -$f$. How does this follow from the compactness of $S$?
$endgroup$
– Max93
Dec 2 '18 at 23:28
1
$begingroup$
@Max93: All convex functions are continuous (in fact, locally Lipschitz continuous), so you can appeal to the Extreme Value Theorem which says that continuous functions attain their maxima / minima over compact sets.
$endgroup$
– VHarisop
Dec 4 '18 at 19:17
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@VHarisop : Thank you! For the Extreme Value Theorem one needs that the function is continuous on the compact set, but doesn't it only follow from the convexity that the function is continuous on the interior? Why isn't this a problem?
$endgroup$
– Max93
Dec 4 '18 at 20:21
$begingroup$
@Max93: My apologies, I read the statement hastily and thought $f$ was defined everywhere.
$endgroup$
– VHarisop
Dec 4 '18 at 20:57
add a comment |
$begingroup$
Let $f: S to mathbb{R}$ be a (strictly) concave function, where
$S := {y in mathbb{R}^m: ygeq 0,, sum_{i=1}^m y_i=1 }$.
I want to show that there is a $y^*in S$, which maximizes $f$. $S$ is a convex set and I found some results from convex optimization that strictly convex functions on convex sets have unique minimizers if local minima exist. Here, the set $S$ is also compact; how can we use this to prove the existence of a maximizing $y^*in S$?
Thanks in advance!
optimization convex-analysis convex-optimization
asked Dec 2 '18 at 22:55
Max93Max93
31029
$endgroup$
Let $f: S to mathbb{R}$ be a (strictly) concave function, where
$S := {y in mathbb{R}^m: ygeq 0,, sum_{i=1}^m y_i=1 }$.
I want to show that there is a $y^*in S$, which maximizes $f$. $S$ is a convex set and I found some results from convex optimization that strictly convex functions on convex sets have unique minimizers if local minima exist. Here, the set $S$ is also compact; how can we use this to prove the existence of a maximizing $y^*in S$?
Thanks in advance!
optimization convex-analysis convex-optimization
optimization convex-analysis convex-optimization
asked Dec 2 '18 at 22:55
Max93Max93
31029
asked Dec 2 '18 at 22:55
Max93Max93
31029
asked Dec 2 '18 at 22:55
Max93Max93
31029
asked Dec 2 '18 at 22:55
Max93Max93
31029
asked Dec 2 '18 at 22:55
Max93Max93
31029
31029
$begingroup$
Apply the result yo have quoted to $-f$.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:15
$begingroup$
This result needs the existence of local minima of -$f$. How does this follow from the compactness of $S$?
$endgroup$
– Max93
Dec 2 '18 at 23:28
1
$begingroup$
@Max93: All convex functions are continuous (in fact, locally Lipschitz continuous), so you can appeal to the Extreme Value Theorem which says that continuous functions attain their maxima / minima over compact sets.
$endgroup$
– VHarisop
Dec 4 '18 at 19:17
$begingroup$
@VHarisop : Thank you! For the Extreme Value Theorem one needs that the function is continuous on the compact set, but doesn't it only follow from the convexity that the function is continuous on the interior? Why isn't this a problem?
$endgroup$
– Max93
Dec 4 '18 at 20:21
$begingroup$
@Max93: My apologies, I read the statement hastily and thought $f$ was defined everywhere.
$endgroup$
– VHarisop
Dec 4 '18 at 20:57
add a comment |
$begingroup$
Apply the result yo have quoted to $-f$.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:15
$begingroup$
This result needs the existence of local minima of -$f$. How does this follow from the compactness of $S$?
$endgroup$
– Max93
Dec 2 '18 at 23:28
1
$begingroup$
@Max93: All convex functions are continuous (in fact, locally Lipschitz continuous), so you can appeal to the Extreme Value Theorem which says that continuous functions attain their maxima / minima over compact sets.
$endgroup$
– VHarisop
Dec 4 '18 at 19:17
$begingroup$
@VHarisop : Thank you! For the Extreme Value Theorem one needs that the function is continuous on the compact set, but doesn't it only follow from the convexity that the function is continuous on the interior? Why isn't this a problem?
$endgroup$
– Max93
Dec 4 '18 at 20:21
$begingroup$
@Max93: My apologies, I read the statement hastily and thought $f$ was defined everywhere.
$endgroup$
– VHarisop
Dec 4 '18 at 20:57
$begingroup$
Apply the result yo have quoted to $-f$.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:15
$begingroup$
Apply the result yo have quoted to $-f$.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:15
$begingroup$
This result needs the existence of local minima of -$f$. How does this follow from the compactness of $S$?
$endgroup$
– Max93
Dec 2 '18 at 23:28
$begingroup$
This result needs the existence of local minima of -$f$. How does this follow from the compactness of $S$?
$endgroup$
– Max93
Dec 2 '18 at 23:28
1
1
$begingroup$
@Max93: All convex functions are continuous (in fact, locally Lipschitz continuous), so you can appeal to the Extreme Value Theorem which says that continuous functions attain their maxima / minima over compact sets.
$endgroup$
– VHarisop
Dec 4 '18 at 19:17
$begingroup$
@Max93: All convex functions are continuous (in fact, locally Lipschitz continuous), so you can appeal to the Extreme Value Theorem which says that continuous functions attain their maxima / minima over compact sets.
$endgroup$
– VHarisop
Dec 4 '18 at 19:17
$begingroup$
@VHarisop : Thank you! For the Extreme Value Theorem one needs that the function is continuous on the compact set, but doesn't it only follow from the convexity that the function is continuous on the interior? Why isn't this a problem?
$endgroup$
– Max93
Dec 4 '18 at 20:21
$begingroup$
@VHarisop : Thank you! For the Extreme Value Theorem one needs that the function is continuous on the compact set, but doesn't it only follow from the convexity that the function is continuous on the interior? Why isn't this a problem?
$endgroup$
– Max93
Dec 4 '18 at 20:21
$begingroup$
@Max93: My apologies, I read the statement hastily and thought $f$ was defined everywhere.
$endgroup$
– VHarisop
Dec 4 '18 at 20:57
$begingroup$
@Max93: My apologies, I read the statement hastily and thought $f$ was defined everywhere.
$endgroup$
– VHarisop
Dec 4 '18 at 20:57
add a comment |
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$begingroup$
Apply the result yo have quoted to $-f$.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:15
$begingroup$
This result needs the existence of local minima of -$f$. How does this follow from the compactness of $S$?
$endgroup$
– Max93
Dec 2 '18 at 23:28
1
$begingroup$
@Max93: All convex functions are continuous (in fact, locally Lipschitz continuous), so you can appeal to the Extreme Value Theorem which says that continuous functions attain their maxima / minima over compact sets.
$endgroup$
– VHarisop
Dec 4 '18 at 19:17
$begingroup$
@VHarisop : Thank you! For the Extreme Value Theorem one needs that the function is continuous on the compact set, but doesn't it only follow from the convexity that the function is continuous on the interior? Why isn't this a problem?
$endgroup$
– Max93
Dec 4 '18 at 20:21
$begingroup$
@Max93: My apologies, I read the statement hastily and thought $f$ was defined everywhere.
$endgroup$
– VHarisop
Dec 4 '18 at 20:57