why can we use this $x^theta y^{1-theta} le theta x+(1-theta)y$ to prove the $prod limits_{i=1}^{n}x_i ge 1$...
Show that ${x in R^n_+|prod limits_{i=1}^{n}x_i ge 1}$ is convex set
Hint : if $x ,y ge 0$ and $0le theta le 1$,then $x^theta y^{1-theta} le theta x+(1-theta)y$
I don't understand the relation between the problem and the hint,the question wants me to prove the $prod limits_{i=1}^{n}x_i ge 1$ is a convex,why shouldn't we use the definition of convex: $f(theta x+(1-theta )y) le theta f(x)+(1-theta)f(y)$,but use this $x^theta y^{1-theta} le theta x+(1-theta)y$ to prove the $prod limits_{i=1}^{n}x_i ge 1$ is convex set?
convex-analysis convex-optimization
asked Nov 30 at 13:50
shineele
377
add a comment |
Show that ${x in R^n_+|prod limits_{i=1}^{n}x_i ge 1}$ is convex set
Hint : if $x ,y ge 0$ and $0le theta le 1$,then $x^theta y^{1-theta} le theta x+(1-theta)y$
I don't understand the relation between the problem and the hint,the question wants me to prove the $prod limits_{i=1}^{n}x_i ge 1$ is a convex,why shouldn't we use the definition of convex: $f(theta x+(1-theta )y) le theta f(x)+(1-theta)f(y)$,but use this $x^theta y^{1-theta} le theta x+(1-theta)y$ to prove the $prod limits_{i=1}^{n}x_i ge 1$ is convex set?
convex-analysis convex-optimization
asked Nov 30 at 13:50
shineele
377
2
Start with the definition of a convex set (not function) and carry out the rest of your plan. Then you will see when to use the hint.
– Michal Adamaszek
Nov 30 at 13:54
I think that carrying on the usual definition is pretty straightforward. I also can't see what that hint is useful for...not even if it simplifies something the proof...
– DonAntonio
Nov 30 at 14:03
add a comment |
Show that ${x in R^n_+|prod limits_{i=1}^{n}x_i ge 1}$ is convex set
Hint : if $x ,y ge 0$ and $0le theta le 1$,then $x^theta y^{1-theta} le theta x+(1-theta)y$
I don't understand the relation between the problem and the hint,the question wants me to prove the $prod limits_{i=1}^{n}x_i ge 1$ is a convex,why shouldn't we use the definition of convex: $f(theta x+(1-theta )y) le theta f(x)+(1-theta)f(y)$,but use this $x^theta y^{1-theta} le theta x+(1-theta)y$ to prove the $prod limits_{i=1}^{n}x_i ge 1$ is convex set?
convex-analysis convex-optimization
asked Nov 30 at 13:50
shineele
377
Show that ${x in R^n_+|prod limits_{i=1}^{n}x_i ge 1}$ is convex set
Hint : if $x ,y ge 0$ and $0le theta le 1$,then $x^theta y^{1-theta} le theta x+(1-theta)y$
I don't understand the relation between the problem and the hint,the question wants me to prove the $prod limits_{i=1}^{n}x_i ge 1$ is a convex,why shouldn't we use the definition of convex: $f(theta x+(1-theta )y) le theta f(x)+(1-theta)f(y)$,but use this $x^theta y^{1-theta} le theta x+(1-theta)y$ to prove the $prod limits_{i=1}^{n}x_i ge 1$ is convex set?
convex-analysis convex-optimization
convex-analysis convex-optimization
asked Nov 30 at 13:50
shineele
377
asked Nov 30 at 13:50
shineele
377
asked Nov 30 at 13:50
shineele
377
asked Nov 30 at 13:50
shineele
377
asked Nov 30 at 13:50
shineele
377
377
2
Start with the definition of a convex set (not function) and carry out the rest of your plan. Then you will see when to use the hint.
– Michal Adamaszek
Nov 30 at 13:54
I think that carrying on the usual definition is pretty straightforward. I also can't see what that hint is useful for...not even if it simplifies something the proof...
– DonAntonio
Nov 30 at 14:03
add a comment |
2
Start with the definition of a convex set (not function) and carry out the rest of your plan. Then you will see when to use the hint.
– Michal Adamaszek
Nov 30 at 13:54
I think that carrying on the usual definition is pretty straightforward. I also can't see what that hint is useful for...not even if it simplifies something the proof...
– DonAntonio
Nov 30 at 14:03
2
2
Start with the definition of a convex set (not function) and carry out the rest of your plan. Then you will see when to use the hint.
– Michal Adamaszek
Nov 30 at 13:54
Start with the definition of a convex set (not function) and carry out the rest of your plan. Then you will see when to use the hint.
– Michal Adamaszek
Nov 30 at 13:54
I think that carrying on the usual definition is pretty straightforward. I also can't see what that hint is useful for...not even if it simplifies something the proof...
– DonAntonio
Nov 30 at 14:03
I think that carrying on the usual definition is pretty straightforward. I also can't see what that hint is useful for...not even if it simplifies something the proof...
– DonAntonio
Nov 30 at 14:03
add a comment |
1 Answer
1
active
oldest
votes
you may want to use this hint the other way: from right to left...
let's call $mathbb{A}$ your ensemble, let's take $(x,y) in mathbb{A}^2$. to prove that $mathbb{A}$ is convex, we need to prove that $forall t in [0,1], z = tx+(1-t)y ;$ is in $mathbb{A}$
$Pi_{i=1}^n z_i = Pi_{i=1}^n (tx_i+(1-t)y_i)$
now you use the inequality from the right side to the left.
$Pi_{i=1}^n (tx_i+(1-t)y_i) > Pi_{i=1}^n x_i^ty_i^{1-t}$
you are now one line away from the result
answered Nov 30 at 14:05
Alexis
2199
how does $Pi_{i=1}^n x_i^ty_i^{1-t}$ come ?
– shineele
Nov 30 at 14:22
how i came to this formula
– shineele
Nov 30 at 14:26
it's explained in the answer...
– Alexis
Nov 30 at 14:27
i mean why isn't $Pi_{i=1}^n (tx_i+(1-t)y_i) > 0 $ or the other formula or number,why must we $> Pi_{i=1}^n x_i^ty_i^{1-t}$?
– shineele
Nov 30 at 14:29
for two reasons: - one it's the hint, you were looking for a way to use the hint. here it is. if you change this you don't use the hint. - it brings you very very close to the results and it's after that straight forward to prove that z is in $mathbb{A}$
– Alexis
Nov 30 at 14:31
|
show 2 more comments
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you may want to use this hint the other way: from right to left...
let's call $mathbb{A}$ your ensemble, let's take $(x,y) in mathbb{A}^2$. to prove that $mathbb{A}$ is convex, we need to prove that $forall t in [0,1], z = tx+(1-t)y ;$ is in $mathbb{A}$
$Pi_{i=1}^n z_i = Pi_{i=1}^n (tx_i+(1-t)y_i)$
now you use the inequality from the right side to the left.
$Pi_{i=1}^n (tx_i+(1-t)y_i) > Pi_{i=1}^n x_i^ty_i^{1-t}$
you are now one line away from the result
answered Nov 30 at 14:05
Alexis
2199
how does $Pi_{i=1}^n x_i^ty_i^{1-t}$ come ?
– shineele
Nov 30 at 14:22
how i came to this formula
– shineele
Nov 30 at 14:26
it's explained in the answer...
– Alexis
Nov 30 at 14:27
i mean why isn't $Pi_{i=1}^n (tx_i+(1-t)y_i) > 0 $ or the other formula or number,why must we $> Pi_{i=1}^n x_i^ty_i^{1-t}$?
– shineele
Nov 30 at 14:29
for two reasons: - one it's the hint, you were looking for a way to use the hint. here it is. if you change this you don't use the hint. - it brings you very very close to the results and it's after that straight forward to prove that z is in $mathbb{A}$
– Alexis
Nov 30 at 14:31
|
show 2 more comments
you may want to use this hint the other way: from right to left...
let's call $mathbb{A}$ your ensemble, let's take $(x,y) in mathbb{A}^2$. to prove that $mathbb{A}$ is convex, we need to prove that $forall t in [0,1], z = tx+(1-t)y ;$ is in $mathbb{A}$
$Pi_{i=1}^n z_i = Pi_{i=1}^n (tx_i+(1-t)y_i)$
now you use the inequality from the right side to the left.
$Pi_{i=1}^n (tx_i+(1-t)y_i) > Pi_{i=1}^n x_i^ty_i^{1-t}$
you are now one line away from the result
answered Nov 30 at 14:05
Alexis
2199
how does $Pi_{i=1}^n x_i^ty_i^{1-t}$ come ?
– shineele
Nov 30 at 14:22
how i came to this formula
– shineele
Nov 30 at 14:26
it's explained in the answer...
– Alexis
Nov 30 at 14:27
i mean why isn't $Pi_{i=1}^n (tx_i+(1-t)y_i) > 0 $ or the other formula or number,why must we $> Pi_{i=1}^n x_i^ty_i^{1-t}$?
– shineele
Nov 30 at 14:29
for two reasons: - one it's the hint, you were looking for a way to use the hint. here it is. if you change this you don't use the hint. - it brings you very very close to the results and it's after that straight forward to prove that z is in $mathbb{A}$
– Alexis
Nov 30 at 14:31
|
show 2 more comments
you may want to use this hint the other way: from right to left...
let's call $mathbb{A}$ your ensemble, let's take $(x,y) in mathbb{A}^2$. to prove that $mathbb{A}$ is convex, we need to prove that $forall t in [0,1], z = tx+(1-t)y ;$ is in $mathbb{A}$
$Pi_{i=1}^n z_i = Pi_{i=1}^n (tx_i+(1-t)y_i)$
now you use the inequality from the right side to the left.
$Pi_{i=1}^n (tx_i+(1-t)y_i) > Pi_{i=1}^n x_i^ty_i^{1-t}$
you are now one line away from the result
answered Nov 30 at 14:05
Alexis
2199
you may want to use this hint the other way: from right to left...
let's call $mathbb{A}$ your ensemble, let's take $(x,y) in mathbb{A}^2$. to prove that $mathbb{A}$ is convex, we need to prove that $forall t in [0,1], z = tx+(1-t)y ;$ is in $mathbb{A}$
$Pi_{i=1}^n z_i = Pi_{i=1}^n (tx_i+(1-t)y_i)$
now you use the inequality from the right side to the left.
$Pi_{i=1}^n (tx_i+(1-t)y_i) > Pi_{i=1}^n x_i^ty_i^{1-t}$
you are now one line away from the result
answered Nov 30 at 14:05
Alexis
2199
answered Nov 30 at 14:05
Alexis
2199
answered Nov 30 at 14:05
Alexis
2199
answered Nov 30 at 14:05
Alexis
2199
2199
how does $Pi_{i=1}^n x_i^ty_i^{1-t}$ come ?
– shineele
Nov 30 at 14:22
how i came to this formula
– shineele
Nov 30 at 14:26
it's explained in the answer...
– Alexis
Nov 30 at 14:27
i mean why isn't $Pi_{i=1}^n (tx_i+(1-t)y_i) > 0 $ or the other formula or number,why must we $> Pi_{i=1}^n x_i^ty_i^{1-t}$?
– shineele
Nov 30 at 14:29
for two reasons: - one it's the hint, you were looking for a way to use the hint. here it is. if you change this you don't use the hint. - it brings you very very close to the results and it's after that straight forward to prove that z is in $mathbb{A}$
– Alexis
Nov 30 at 14:31
|
show 2 more comments
how does $Pi_{i=1}^n x_i^ty_i^{1-t}$ come ?
– shineele
Nov 30 at 14:22
how i came to this formula
– shineele
Nov 30 at 14:26
it's explained in the answer...
– Alexis
Nov 30 at 14:27
i mean why isn't $Pi_{i=1}^n (tx_i+(1-t)y_i) > 0 $ or the other formula or number,why must we $> Pi_{i=1}^n x_i^ty_i^{1-t}$?
– shineele
Nov 30 at 14:29
for two reasons: - one it's the hint, you were looking for a way to use the hint. here it is. if you change this you don't use the hint. - it brings you very very close to the results and it's after that straight forward to prove that z is in $mathbb{A}$
– Alexis
Nov 30 at 14:31
how does $Pi_{i=1}^n x_i^ty_i^{1-t}$ come ?
– shineele
Nov 30 at 14:22
how does $Pi_{i=1}^n x_i^ty_i^{1-t}$ come ?
– shineele
Nov 30 at 14:22
how i came to this formula
– shineele
Nov 30 at 14:26
how i came to this formula
– shineele
Nov 30 at 14:26
it's explained in the answer...
– Alexis
Nov 30 at 14:27
it's explained in the answer...
– Alexis
Nov 30 at 14:27
i mean why isn't $Pi_{i=1}^n (tx_i+(1-t)y_i) > 0 $ or the other formula or number,why must we $> Pi_{i=1}^n x_i^ty_i^{1-t}$?
– shineele
Nov 30 at 14:29
i mean why isn't $Pi_{i=1}^n (tx_i+(1-t)y_i) > 0 $ or the other formula or number,why must we $> Pi_{i=1}^n x_i^ty_i^{1-t}$?
– shineele
Nov 30 at 14:29
for two reasons: - one it's the hint, you were looking for a way to use the hint. here it is. if you change this you don't use the hint. - it brings you very very close to the results and it's after that straight forward to prove that z is in $mathbb{A}$
– Alexis
Nov 30 at 14:31
for two reasons: - one it's the hint, you were looking for a way to use the hint. here it is. if you change this you don't use the hint. - it brings you very very close to the results and it's after that straight forward to prove that z is in $mathbb{A}$
– Alexis
Nov 30 at 14:31
|
show 2 more comments
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2
Start with the definition of a convex set (not function) and carry out the rest of your plan. Then you will see when to use the hint.
– Michal Adamaszek
Nov 30 at 13:54
I think that carrying on the usual definition is pretty straightforward. I also can't see what that hint is useful for...not even if it simplifies something the proof...
– DonAntonio
Nov 30 at 14:03