Mixing
Prove inequality that is used to prove convexity
$begingroup$
I'm looking for the proof of an inequality such that I can prove the problem below (this is not the problem I need an answer to, it is the inequality later on that I don't get).
Problem
$$text{Let } V = { x in mathbb{R}^2: 1-x_1^2-x_2^2 leq alpha} text{ be a given set. Prove that for any } alpha in mathbb{R} text{, set V is convex.}$$
Solution
$$text{For } lambda in [0,1] text{ and } x,y in V text{, we want to show that } lambda x+(1-lambda)y in V.$$
$$text{Since } x,y in V, text{we know that } 1-x_1^2-x_2^2 leq alpha text{ and } 1-y_1^2-y_2^2 leq alpha$$
$$text{Now we want to show that } 1-[lambda x_1 + (1-lambda)y_1]^2-[lambda x_2 + (1-lambda)y_2]^2 leq alpha$$
$$ text{By rewriting the equation above to }lambda[1-x_1^2-x_2^2] + (1-lambda)[1-y_1^2-y_2^2] leq lambdaalpha + (1-lambda)alpha = alpha$$
$$text{Since } alpha, x text{ and } y text{ were given arbitrarily, this holds for any value of these variables.}$$
Actual question
In order to do this, the answer sheet states that the following inequality holds (similar for $x_2$ and $y_2$)
$$lambda^2 x_1^2 + 2lambda(1-lambda)x_1y_1 + (1-lambda)^2y_1^2 leq lambda x_1^2 + (1-lambda)y_1^2$$
How can I derive this inequality? How can I show that this inequality holds?
EDIT: I know that $2xy leq x^2+y^2$, inserting this gives $lambda^2 x_1^2 +lambda(1-lambda)(x_1^2+y_1^2)+(1-lambda)^2y_1^2$. I don't see how this is less than $lambda x_1^2 + (1-lambda)y_1^2$ since $lambda(1-lambda)(x_1^2+y_1^2) geq 0$?
inequality convex-analysis convexity-inequality
asked Jan 24 at 13:41
user2177152user2177152
11
$endgroup$
add a comment |
$begingroup$
I'm looking for the proof of an inequality such that I can prove the problem below (this is not the problem I need an answer to, it is the inequality later on that I don't get).
Problem
$$text{Let } V = { x in mathbb{R}^2: 1-x_1^2-x_2^2 leq alpha} text{ be a given set. Prove that for any } alpha in mathbb{R} text{, set V is convex.}$$
Solution
$$text{For } lambda in [0,1] text{ and } x,y in V text{, we want to show that } lambda x+(1-lambda)y in V.$$
$$text{Since } x,y in V, text{we know that } 1-x_1^2-x_2^2 leq alpha text{ and } 1-y_1^2-y_2^2 leq alpha$$
$$text{Now we want to show that } 1-[lambda x_1 + (1-lambda)y_1]^2-[lambda x_2 + (1-lambda)y_2]^2 leq alpha$$
$$ text{By rewriting the equation above to }lambda[1-x_1^2-x_2^2] + (1-lambda)[1-y_1^2-y_2^2] leq lambdaalpha + (1-lambda)alpha = alpha$$
$$text{Since } alpha, x text{ and } y text{ were given arbitrarily, this holds for any value of these variables.}$$
Actual question
In order to do this, the answer sheet states that the following inequality holds (similar for $x_2$ and $y_2$)
$$lambda^2 x_1^2 + 2lambda(1-lambda)x_1y_1 + (1-lambda)^2y_1^2 leq lambda x_1^2 + (1-lambda)y_1^2$$
How can I derive this inequality? How can I show that this inequality holds?
EDIT: I know that $2xy leq x^2+y^2$, inserting this gives $lambda^2 x_1^2 +lambda(1-lambda)(x_1^2+y_1^2)+(1-lambda)^2y_1^2$. I don't see how this is less than $lambda x_1^2 + (1-lambda)y_1^2$ since $lambda(1-lambda)(x_1^2+y_1^2) geq 0$?
inequality convex-analysis convexity-inequality
asked Jan 24 at 13:41
user2177152user2177152
11
$endgroup$
$begingroup$
Hint: $2x_1y_1 le x_1^2 + y_1^2$
$endgroup$
– Martin R
Jan 24 at 13:45
$begingroup$
I assume that the last term on the LHS should be $(1-lambda)^2 y_1^2 $ ?
$endgroup$
– Martin R
Jan 24 at 13:48
$begingroup$
Yes, thanks! I just edited it. I already knew your hint, but I just don't see it yet.
$endgroup$
– user2177152
Jan 24 at 13:51
$begingroup$
But then you are (almost) done. It only remains to expand $lambda^2 x_1^2 +lambda(1-lambda)(x_1^2+y_1^2)+(1-lambda)^2y_1^2$ and collect the terms with $x_1^2$ and $y_1^2$, respectively.
$endgroup$
– Martin R
Jan 24 at 13:51
$begingroup$
Thank you so much! I feel like such an idiot right now, but I finally get it, thank you!
$endgroup$
– user2177152
Jan 24 at 13:53
add a comment |
$begingroup$
I'm looking for the proof of an inequality such that I can prove the problem below (this is not the problem I need an answer to, it is the inequality later on that I don't get).
Problem
$$text{Let } V = { x in mathbb{R}^2: 1-x_1^2-x_2^2 leq alpha} text{ be a given set. Prove that for any } alpha in mathbb{R} text{, set V is convex.}$$
Solution
$$text{For } lambda in [0,1] text{ and } x,y in V text{, we want to show that } lambda x+(1-lambda)y in V.$$
$$text{Since } x,y in V, text{we know that } 1-x_1^2-x_2^2 leq alpha text{ and } 1-y_1^2-y_2^2 leq alpha$$
$$text{Now we want to show that } 1-[lambda x_1 + (1-lambda)y_1]^2-[lambda x_2 + (1-lambda)y_2]^2 leq alpha$$
$$ text{By rewriting the equation above to }lambda[1-x_1^2-x_2^2] + (1-lambda)[1-y_1^2-y_2^2] leq lambdaalpha + (1-lambda)alpha = alpha$$
$$text{Since } alpha, x text{ and } y text{ were given arbitrarily, this holds for any value of these variables.}$$
Actual question
In order to do this, the answer sheet states that the following inequality holds (similar for $x_2$ and $y_2$)
$$lambda^2 x_1^2 + 2lambda(1-lambda)x_1y_1 + (1-lambda)^2y_1^2 leq lambda x_1^2 + (1-lambda)y_1^2$$
How can I derive this inequality? How can I show that this inequality holds?
EDIT: I know that $2xy leq x^2+y^2$, inserting this gives $lambda^2 x_1^2 +lambda(1-lambda)(x_1^2+y_1^2)+(1-lambda)^2y_1^2$. I don't see how this is less than $lambda x_1^2 + (1-lambda)y_1^2$ since $lambda(1-lambda)(x_1^2+y_1^2) geq 0$?
inequality convex-analysis convexity-inequality
asked Jan 24 at 13:41
user2177152user2177152
11
$endgroup$
I'm looking for the proof of an inequality such that I can prove the problem below (this is not the problem I need an answer to, it is the inequality later on that I don't get).
Problem
$$text{Let } V = { x in mathbb{R}^2: 1-x_1^2-x_2^2 leq alpha} text{ be a given set. Prove that for any } alpha in mathbb{R} text{, set V is convex.}$$
Solution
$$text{For } lambda in [0,1] text{ and } x,y in V text{, we want to show that } lambda x+(1-lambda)y in V.$$
$$text{Since } x,y in V, text{we know that } 1-x_1^2-x_2^2 leq alpha text{ and } 1-y_1^2-y_2^2 leq alpha$$
$$text{Now we want to show that } 1-[lambda x_1 + (1-lambda)y_1]^2-[lambda x_2 + (1-lambda)y_2]^2 leq alpha$$
$$ text{By rewriting the equation above to }lambda[1-x_1^2-x_2^2] + (1-lambda)[1-y_1^2-y_2^2] leq lambdaalpha + (1-lambda)alpha = alpha$$
$$text{Since } alpha, x text{ and } y text{ were given arbitrarily, this holds for any value of these variables.}$$
Actual question
In order to do this, the answer sheet states that the following inequality holds (similar for $x_2$ and $y_2$)
$$lambda^2 x_1^2 + 2lambda(1-lambda)x_1y_1 + (1-lambda)^2y_1^2 leq lambda x_1^2 + (1-lambda)y_1^2$$
How can I derive this inequality? How can I show that this inequality holds?
EDIT: I know that $2xy leq x^2+y^2$, inserting this gives $lambda^2 x_1^2 +lambda(1-lambda)(x_1^2+y_1^2)+(1-lambda)^2y_1^2$. I don't see how this is less than $lambda x_1^2 + (1-lambda)y_1^2$ since $lambda(1-lambda)(x_1^2+y_1^2) geq 0$?
inequality convex-analysis convexity-inequality
inequality convex-analysis convexity-inequality
asked Jan 24 at 13:41
user2177152user2177152
11
asked Jan 24 at 13:41
user2177152user2177152
11
edited Jan 24 at 13:49
user2177152
asked Jan 24 at 13:41
user2177152user2177152
11
asked Jan 24 at 13:41
user2177152user2177152
11
asked Jan 24 at 13:41
user2177152user2177152
11
11
$begingroup$
Hint: $2x_1y_1 le x_1^2 + y_1^2$
$endgroup$
– Martin R
Jan 24 at 13:45
$begingroup$
I assume that the last term on the LHS should be $(1-lambda)^2 y_1^2 $ ?
$endgroup$
– Martin R
Jan 24 at 13:48
$begingroup$
Yes, thanks! I just edited it. I already knew your hint, but I just don't see it yet.
$endgroup$
– user2177152
Jan 24 at 13:51
$begingroup$
But then you are (almost) done. It only remains to expand $lambda^2 x_1^2 +lambda(1-lambda)(x_1^2+y_1^2)+(1-lambda)^2y_1^2$ and collect the terms with $x_1^2$ and $y_1^2$, respectively.
$endgroup$
– Martin R
Jan 24 at 13:51
$begingroup$
Thank you so much! I feel like such an idiot right now, but I finally get it, thank you!
$endgroup$
– user2177152
Jan 24 at 13:53
add a comment |
$begingroup$
Hint: $2x_1y_1 le x_1^2 + y_1^2$
$endgroup$
– Martin R
Jan 24 at 13:45
$begingroup$
I assume that the last term on the LHS should be $(1-lambda)^2 y_1^2 $ ?
$endgroup$
– Martin R
Jan 24 at 13:48
$begingroup$
Yes, thanks! I just edited it. I already knew your hint, but I just don't see it yet.
$endgroup$
– user2177152
Jan 24 at 13:51
$begingroup$
But then you are (almost) done. It only remains to expand $lambda^2 x_1^2 +lambda(1-lambda)(x_1^2+y_1^2)+(1-lambda)^2y_1^2$ and collect the terms with $x_1^2$ and $y_1^2$, respectively.
$endgroup$
– Martin R
Jan 24 at 13:51
$begingroup$
Thank you so much! I feel like such an idiot right now, but I finally get it, thank you!
$endgroup$
– user2177152
Jan 24 at 13:53
$begingroup$
Hint: $2x_1y_1 le x_1^2 + y_1^2$
$endgroup$
– Martin R
Jan 24 at 13:45
$begingroup$
Hint: $2x_1y_1 le x_1^2 + y_1^2$
$endgroup$
– Martin R
Jan 24 at 13:45
$begingroup$
I assume that the last term on the LHS should be $(1-lambda)^2 y_1^2 $ ?
$endgroup$
– Martin R
Jan 24 at 13:48
$begingroup$
I assume that the last term on the LHS should be $(1-lambda)^2 y_1^2 $ ?
$endgroup$
– Martin R
Jan 24 at 13:48
$begingroup$
Yes, thanks! I just edited it. I already knew your hint, but I just don't see it yet.
$endgroup$
– user2177152
Jan 24 at 13:51
$begingroup$
Yes, thanks! I just edited it. I already knew your hint, but I just don't see it yet.
$endgroup$
– user2177152
Jan 24 at 13:51
$begingroup$
But then you are (almost) done. It only remains to expand $lambda^2 x_1^2 +lambda(1-lambda)(x_1^2+y_1^2)+(1-lambda)^2y_1^2$ and collect the terms with $x_1^2$ and $y_1^2$, respectively.
$endgroup$
– Martin R
Jan 24 at 13:51
$begingroup$
But then you are (almost) done. It only remains to expand $lambda^2 x_1^2 +lambda(1-lambda)(x_1^2+y_1^2)+(1-lambda)^2y_1^2$ and collect the terms with $x_1^2$ and $y_1^2$, respectively.
$endgroup$
– Martin R
Jan 24 at 13:51
$begingroup$
Thank you so much! I feel like such an idiot right now, but I finally get it, thank you!
$endgroup$
– user2177152
Jan 24 at 13:53
$begingroup$
Thank you so much! I feel like such an idiot right now, but I finally get it, thank you!
$endgroup$
– user2177152
Jan 24 at 13:53
add a comment |
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$begingroup$
Hint: $2x_1y_1 le x_1^2 + y_1^2$
$endgroup$
– Martin R
Jan 24 at 13:45
$begingroup$
I assume that the last term on the LHS should be $(1-lambda)^2 y_1^2 $ ?
$endgroup$
– Martin R
Jan 24 at 13:48
$begingroup$
Yes, thanks! I just edited it. I already knew your hint, but I just don't see it yet.
$endgroup$
– user2177152
Jan 24 at 13:51
$begingroup$
But then you are (almost) done. It only remains to expand $lambda^2 x_1^2 +lambda(1-lambda)(x_1^2+y_1^2)+(1-lambda)^2y_1^2$ and collect the terms with $x_1^2$ and $y_1^2$, respectively.
$endgroup$
– Martin R
Jan 24 at 13:51
$begingroup$
Thank you so much! I feel like such an idiot right now, but I finally get it, thank you!
$endgroup$
– user2177152
Jan 24 at 13:53