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How to prove the set ${(x,y,z)in mathbb R^3|zgeq 0,x^2+y^2leq z^2}$ is convex?
$begingroup$
How to prove the set $S={(x,y,z)in mathbb R^3|zgeq 0,x^2+y^2leq z^2}$ is convex?
So to prove this I took $(x,y,z),(x_1,y_1,z_1)in S$ and $lambdain (0,1).$
Then $x^2+y^2leq z^2$ and $x_1^2+y_1^2leq z_1^2$ and $z,z_1geq 0.$
Now to prove ${lambda x+(1-lambda)x_1}^2+{lambda y+(1-lambda)y_1}^2leq {lambda z+(1-lambda)z_1}^2$ , it's coming like $lambda^2(x^2+y^2-z^2)+(1-lambda)^2(x_1^2+y_1^2-z_1^2)+2lambda(1-lambda)(xx_1+yy_1-zz_1)leq 0.$
So from here, I was trying to prove $(xx_1+yy_1-zz_1)leq 0$ because the former terms are already negative in the above inequality.
I'm not sure whether $(xx_1+yy_1-zz_1)leq 0$ is true or not?
Can we prove the inequality $xx_1+yy_1-zz_1leq 0$? Or there is another way to prove the convexity of the set $S$?
Any help is appreciated. Thank you.
convex-analysis
asked Jan 20 at 17:26
nurun neshanurun nesha
1,0362623
$endgroup$
add a comment |
$begingroup$
How to prove the set $S={(x,y,z)in mathbb R^3|zgeq 0,x^2+y^2leq z^2}$ is convex?
So to prove this I took $(x,y,z),(x_1,y_1,z_1)in S$ and $lambdain (0,1).$
Then $x^2+y^2leq z^2$ and $x_1^2+y_1^2leq z_1^2$ and $z,z_1geq 0.$
Now to prove ${lambda x+(1-lambda)x_1}^2+{lambda y+(1-lambda)y_1}^2leq {lambda z+(1-lambda)z_1}^2$ , it's coming like $lambda^2(x^2+y^2-z^2)+(1-lambda)^2(x_1^2+y_1^2-z_1^2)+2lambda(1-lambda)(xx_1+yy_1-zz_1)leq 0.$
So from here, I was trying to prove $(xx_1+yy_1-zz_1)leq 0$ because the former terms are already negative in the above inequality.
I'm not sure whether $(xx_1+yy_1-zz_1)leq 0$ is true or not?
Can we prove the inequality $xx_1+yy_1-zz_1leq 0$? Or there is another way to prove the convexity of the set $S$?
Any help is appreciated. Thank you.
convex-analysis
asked Jan 20 at 17:26
nurun neshanurun nesha
1,0362623
$endgroup$
add a comment |
$begingroup$
How to prove the set $S={(x,y,z)in mathbb R^3|zgeq 0,x^2+y^2leq z^2}$ is convex?
So to prove this I took $(x,y,z),(x_1,y_1,z_1)in S$ and $lambdain (0,1).$
Then $x^2+y^2leq z^2$ and $x_1^2+y_1^2leq z_1^2$ and $z,z_1geq 0.$
Now to prove ${lambda x+(1-lambda)x_1}^2+{lambda y+(1-lambda)y_1}^2leq {lambda z+(1-lambda)z_1}^2$ , it's coming like $lambda^2(x^2+y^2-z^2)+(1-lambda)^2(x_1^2+y_1^2-z_1^2)+2lambda(1-lambda)(xx_1+yy_1-zz_1)leq 0.$
So from here, I was trying to prove $(xx_1+yy_1-zz_1)leq 0$ because the former terms are already negative in the above inequality.
I'm not sure whether $(xx_1+yy_1-zz_1)leq 0$ is true or not?
Can we prove the inequality $xx_1+yy_1-zz_1leq 0$? Or there is another way to prove the convexity of the set $S$?
Any help is appreciated. Thank you.
convex-analysis
asked Jan 20 at 17:26
nurun neshanurun nesha
1,0362623
$endgroup$
How to prove the set $S={(x,y,z)in mathbb R^3|zgeq 0,x^2+y^2leq z^2}$ is convex?
So to prove this I took $(x,y,z),(x_1,y_1,z_1)in S$ and $lambdain (0,1).$
Then $x^2+y^2leq z^2$ and $x_1^2+y_1^2leq z_1^2$ and $z,z_1geq 0.$
Now to prove ${lambda x+(1-lambda)x_1}^2+{lambda y+(1-lambda)y_1}^2leq {lambda z+(1-lambda)z_1}^2$ , it's coming like $lambda^2(x^2+y^2-z^2)+(1-lambda)^2(x_1^2+y_1^2-z_1^2)+2lambda(1-lambda)(xx_1+yy_1-zz_1)leq 0.$
So from here, I was trying to prove $(xx_1+yy_1-zz_1)leq 0$ because the former terms are already negative in the above inequality.
I'm not sure whether $(xx_1+yy_1-zz_1)leq 0$ is true or not?
Can we prove the inequality $xx_1+yy_1-zz_1leq 0$? Or there is another way to prove the convexity of the set $S$?
Any help is appreciated. Thank you.
convex-analysis
convex-analysis
asked Jan 20 at 17:26
nurun neshanurun nesha
1,0362623
asked Jan 20 at 17:26
nurun neshanurun nesha
1,0362623
asked Jan 20 at 17:26
nurun neshanurun nesha
1,0362623
asked Jan 20 at 17:26
nurun neshanurun nesha
1,0362623
asked Jan 20 at 17:26
nurun neshanurun nesha
1,0362623
1,0362623
add a comment |
add a comment |
4 Answers
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Yes, you can show the inequality $$x x_1+y y_1 leq z z_1.$$ It is sufficient to show the inequality
$$(x x_1 + y y_1)^2 leq (z z_1)^2.$$ (Note that sufficiency follows from the fact that $z, z_1 geq 0$.)
To show the squared form, multiply the inequalities
$$x^2 + y^2 leq z^2$$
$$x_1^2 + y_1^2 leq z_1^2$$
together. Then use the inequality $a^2 + b^2 geq 2ab$ with
for $a=x y_1$ and
$ b= x_1y$
. This gives
$$z^2 z_1^2 geq (x^2+y^2)(x_1^2+y_1^2)=(xx_1)^2+(xy_1)^2+(x_1 y)^2+(y y_1)^2 geq (x x_1)2+2(x x_1y y_1)+(y y_1)2=(x x_1+y y_1)2.$$
answered Jan 20 at 17:58
Jordan GreenJordan Green
1,133310
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$begingroup$
Note that $(x,y,z)in Simplies lambda (x,y,z)in S$ since $$(lambda x)^2+(lambda y)^2=lambda^2(x^2+y^2)le lambda^2z^2=(lambda z)^2$$
Thus $S$ is a cone.
Prove that a cone is convex if and only if it is closed under addition.
Then you see that your condition $xx_1+yy_1le zz_1$ is the closure under addition of $(x,y,z)$ and $(x_1,y_1,z_1)$.
This one is obtained using $able frac {a^2+b^2}2$ since both $begin{cases}x^2+y^2le z^2\{x_1}^2+{y_1}^2le{z_1}^2end{cases}$ are verified for points of $S$.
answered Jan 20 at 18:11
zwimzwim
12.5k831
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add a comment |
$begingroup$
Here we have
$$
xx_1 + yy_1 le sqrt{(x^2+y^2)(x_1^2+y_1^2)} le sqrt{z^2}sqrt{z_1^2}= zz_1,.
$$
The final equality relies on $z,z_1 ge 0$.
The first inequality is an example of the Cauchy-Schwarz inequality, and can be directly shown in this case. (Note that the inequality is trivial if $xx_1+yy_1 < 0$.)
$$
x^2y_1^2 - 2xyx_1y_1 + y^2x_1^2 = (xy_1-yx_1)^2 ge 0 \
x^2y_1^2 + y^2x_1^2 ge 2xyx_1y_1 \
x^2x_1^2+ x^2y_1^2 + y^2x_1^2 + y^2y_1^2 ge x^2x_1^2+ 2xyx_1y_1 + y^2y_1^2 \
(x^2+y^2)(x_1^2+y_1^2) ge (xx_1+yy_1)^2
$$
edited Jan 21 at 16:00
Martin Sleziak
44.8k10119272
answered Jan 20 at 18:01
Rolf HoyerRolf Hoyer
11.3k31629
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add a comment |
$begingroup$
Let us consider the usual norm $|vec a|=sqrt{a_1^2+a_2^2}$ on $mathbb R^2$
Notice also that the condition describing $S$ is exactly $sqrt{x^2+y^2} le z$, so
$$S={(x,y,z); zge0, |(x,y)|le z}.$$
So now if we have two points $(x_1,y_1,z_1),(x_2,y_2,z_2)in S$, then we get for any $lambdain(0,1)$
$$|(lambda x_1+(1-lambda)x_2,lambda y_1+(1-lambda)y_2)| overset{(*)}le
lambda|(x_1,y_1)| + (1-lambda)|(x_2,y_2)| le lambda z_1+(1-lambda)z_2.$$
The inequality $(*)$ follows from triangle inequality applied to vectors $(lambda x_1,lambda y_1)$ and $((1-lambda)x_2,(1-lambda)y_2)$. But we can view it also as convexity of the function $(x,y)mapsto|(x,y)|$. See, for example Why is every $p$-norm convex? and other posts linked there.
We have shown that
$$|(lambda x_1+(1-lambda)x_2,lambda y_1+(1-lambda)y_2)| le lambda z_1+(1-lambda)z_2,$$
which means that also the convex combination $(lambda x_1+(1-lambda)x_2,lambda y_1+(1-lambda)y_2,lambda z_1+(1-lambda)z_2)$ belongs to $S$.
Notice that similar approach would work to prove convexity of $S={(x_1,dots,x_{n+1}); x_{n+1}ge 0, sum_{k=1}^n x_k^2 le x_{n+1}^2}$ in $mathbb R^{n+1}$. (This condition can be equivalently stated as $|(x_1,dots,x_n)|lesqrt{x_{n+1}}$.)
answered Jan 21 at 16:59
Martin SleziakMartin Sleziak
44.8k10119272
$endgroup$
add a comment |
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4 Answers
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4 Answers
4
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$begingroup$
Yes, you can show the inequality $$x x_1+y y_1 leq z z_1.$$ It is sufficient to show the inequality
$$(x x_1 + y y_1)^2 leq (z z_1)^2.$$ (Note that sufficiency follows from the fact that $z, z_1 geq 0$.)
To show the squared form, multiply the inequalities
$$x^2 + y^2 leq z^2$$
$$x_1^2 + y_1^2 leq z_1^2$$
together. Then use the inequality $a^2 + b^2 geq 2ab$ with
for $a=x y_1$ and
$ b= x_1y$
. This gives
$$z^2 z_1^2 geq (x^2+y^2)(x_1^2+y_1^2)=(xx_1)^2+(xy_1)^2+(x_1 y)^2+(y y_1)^2 geq (x x_1)2+2(x x_1y y_1)+(y y_1)2=(x x_1+y y_1)2.$$
answered Jan 20 at 17:58
Jordan GreenJordan Green
1,133310
$endgroup$
add a comment |
$begingroup$
Yes, you can show the inequality $$x x_1+y y_1 leq z z_1.$$ It is sufficient to show the inequality
$$(x x_1 + y y_1)^2 leq (z z_1)^2.$$ (Note that sufficiency follows from the fact that $z, z_1 geq 0$.)
To show the squared form, multiply the inequalities
$$x^2 + y^2 leq z^2$$
$$x_1^2 + y_1^2 leq z_1^2$$
together. Then use the inequality $a^2 + b^2 geq 2ab$ with
for $a=x y_1$ and
$ b= x_1y$
. This gives
$$z^2 z_1^2 geq (x^2+y^2)(x_1^2+y_1^2)=(xx_1)^2+(xy_1)^2+(x_1 y)^2+(y y_1)^2 geq (x x_1)2+2(x x_1y y_1)+(y y_1)2=(x x_1+y y_1)2.$$
answered Jan 20 at 17:58
Jordan GreenJordan Green
1,133310
$endgroup$
add a comment |
$begingroup$
Yes, you can show the inequality $$x x_1+y y_1 leq z z_1.$$ It is sufficient to show the inequality
$$(x x_1 + y y_1)^2 leq (z z_1)^2.$$ (Note that sufficiency follows from the fact that $z, z_1 geq 0$.)
To show the squared form, multiply the inequalities
$$x^2 + y^2 leq z^2$$
$$x_1^2 + y_1^2 leq z_1^2$$
together. Then use the inequality $a^2 + b^2 geq 2ab$ with
for $a=x y_1$ and
$ b= x_1y$
. This gives
$$z^2 z_1^2 geq (x^2+y^2)(x_1^2+y_1^2)=(xx_1)^2+(xy_1)^2+(x_1 y)^2+(y y_1)^2 geq (x x_1)2+2(x x_1y y_1)+(y y_1)2=(x x_1+y y_1)2.$$
answered Jan 20 at 17:58
Jordan GreenJordan Green
1,133310
$endgroup$
Yes, you can show the inequality $$x x_1+y y_1 leq z z_1.$$ It is sufficient to show the inequality
$$(x x_1 + y y_1)^2 leq (z z_1)^2.$$ (Note that sufficiency follows from the fact that $z, z_1 geq 0$.)
To show the squared form, multiply the inequalities
$$x^2 + y^2 leq z^2$$
$$x_1^2 + y_1^2 leq z_1^2$$
together. Then use the inequality $a^2 + b^2 geq 2ab$ with
for $a=x y_1$ and
$ b= x_1y$
. This gives
$$z^2 z_1^2 geq (x^2+y^2)(x_1^2+y_1^2)=(xx_1)^2+(xy_1)^2+(x_1 y)^2+(y y_1)^2 geq (x x_1)2+2(x x_1y y_1)+(y y_1)2=(x x_1+y y_1)2.$$
answered Jan 20 at 17:58
Jordan GreenJordan Green
1,133310
edited Jan 20 at 18:04
answered Jan 20 at 17:58
Jordan GreenJordan Green
1,133310
answered Jan 20 at 17:58
Jordan GreenJordan Green
1,133310
answered Jan 20 at 17:58
Jordan GreenJordan Green
1,133310
1,133310
add a comment |
add a comment |
$begingroup$
Note that $(x,y,z)in Simplies lambda (x,y,z)in S$ since $$(lambda x)^2+(lambda y)^2=lambda^2(x^2+y^2)le lambda^2z^2=(lambda z)^2$$
Thus $S$ is a cone.
Prove that a cone is convex if and only if it is closed under addition.
Then you see that your condition $xx_1+yy_1le zz_1$ is the closure under addition of $(x,y,z)$ and $(x_1,y_1,z_1)$.
This one is obtained using $able frac {a^2+b^2}2$ since both $begin{cases}x^2+y^2le z^2\{x_1}^2+{y_1}^2le{z_1}^2end{cases}$ are verified for points of $S$.
answered Jan 20 at 18:11
zwimzwim
12.5k831
$endgroup$
add a comment |
$begingroup$
Note that $(x,y,z)in Simplies lambda (x,y,z)in S$ since $$(lambda x)^2+(lambda y)^2=lambda^2(x^2+y^2)le lambda^2z^2=(lambda z)^2$$
Thus $S$ is a cone.
Prove that a cone is convex if and only if it is closed under addition.
Then you see that your condition $xx_1+yy_1le zz_1$ is the closure under addition of $(x,y,z)$ and $(x_1,y_1,z_1)$.
This one is obtained using $able frac {a^2+b^2}2$ since both $begin{cases}x^2+y^2le z^2\{x_1}^2+{y_1}^2le{z_1}^2end{cases}$ are verified for points of $S$.
answered Jan 20 at 18:11
zwimzwim
12.5k831
$endgroup$
add a comment |
$begingroup$
Note that $(x,y,z)in Simplies lambda (x,y,z)in S$ since $$(lambda x)^2+(lambda y)^2=lambda^2(x^2+y^2)le lambda^2z^2=(lambda z)^2$$
Thus $S$ is a cone.
Prove that a cone is convex if and only if it is closed under addition.
Then you see that your condition $xx_1+yy_1le zz_1$ is the closure under addition of $(x,y,z)$ and $(x_1,y_1,z_1)$.
This one is obtained using $able frac {a^2+b^2}2$ since both $begin{cases}x^2+y^2le z^2\{x_1}^2+{y_1}^2le{z_1}^2end{cases}$ are verified for points of $S$.
answered Jan 20 at 18:11
zwimzwim
12.5k831
$endgroup$
Note that $(x,y,z)in Simplies lambda (x,y,z)in S$ since $$(lambda x)^2+(lambda y)^2=lambda^2(x^2+y^2)le lambda^2z^2=(lambda z)^2$$
Thus $S$ is a cone.
Prove that a cone is convex if and only if it is closed under addition.
Then you see that your condition $xx_1+yy_1le zz_1$ is the closure under addition of $(x,y,z)$ and $(x_1,y_1,z_1)$.
This one is obtained using $able frac {a^2+b^2}2$ since both $begin{cases}x^2+y^2le z^2\{x_1}^2+{y_1}^2le{z_1}^2end{cases}$ are verified for points of $S$.
answered Jan 20 at 18:11
zwimzwim
12.5k831
answered Jan 20 at 18:11
zwimzwim
12.5k831
answered Jan 20 at 18:11
zwimzwim
12.5k831
answered Jan 20 at 18:11
zwimzwim
12.5k831
12.5k831
add a comment |
add a comment |
$begingroup$
Here we have
$$
xx_1 + yy_1 le sqrt{(x^2+y^2)(x_1^2+y_1^2)} le sqrt{z^2}sqrt{z_1^2}= zz_1,.
$$
The final equality relies on $z,z_1 ge 0$.
The first inequality is an example of the Cauchy-Schwarz inequality, and can be directly shown in this case. (Note that the inequality is trivial if $xx_1+yy_1 < 0$.)
$$
x^2y_1^2 - 2xyx_1y_1 + y^2x_1^2 = (xy_1-yx_1)^2 ge 0 \
x^2y_1^2 + y^2x_1^2 ge 2xyx_1y_1 \
x^2x_1^2+ x^2y_1^2 + y^2x_1^2 + y^2y_1^2 ge x^2x_1^2+ 2xyx_1y_1 + y^2y_1^2 \
(x^2+y^2)(x_1^2+y_1^2) ge (xx_1+yy_1)^2
$$
edited Jan 21 at 16:00
Martin Sleziak
44.8k10119272
answered Jan 20 at 18:01
Rolf HoyerRolf Hoyer
11.3k31629
$endgroup$
add a comment |
$begingroup$
Here we have
$$
xx_1 + yy_1 le sqrt{(x^2+y^2)(x_1^2+y_1^2)} le sqrt{z^2}sqrt{z_1^2}= zz_1,.
$$
The final equality relies on $z,z_1 ge 0$.
The first inequality is an example of the Cauchy-Schwarz inequality, and can be directly shown in this case. (Note that the inequality is trivial if $xx_1+yy_1 < 0$.)
$$
x^2y_1^2 - 2xyx_1y_1 + y^2x_1^2 = (xy_1-yx_1)^2 ge 0 \
x^2y_1^2 + y^2x_1^2 ge 2xyx_1y_1 \
x^2x_1^2+ x^2y_1^2 + y^2x_1^2 + y^2y_1^2 ge x^2x_1^2+ 2xyx_1y_1 + y^2y_1^2 \
(x^2+y^2)(x_1^2+y_1^2) ge (xx_1+yy_1)^2
$$
edited Jan 21 at 16:00
Martin Sleziak
44.8k10119272
answered Jan 20 at 18:01
Rolf HoyerRolf Hoyer
11.3k31629
$endgroup$
add a comment |
$begingroup$
Here we have
$$
xx_1 + yy_1 le sqrt{(x^2+y^2)(x_1^2+y_1^2)} le sqrt{z^2}sqrt{z_1^2}= zz_1,.
$$
The final equality relies on $z,z_1 ge 0$.
The first inequality is an example of the Cauchy-Schwarz inequality, and can be directly shown in this case. (Note that the inequality is trivial if $xx_1+yy_1 < 0$.)
$$
x^2y_1^2 - 2xyx_1y_1 + y^2x_1^2 = (xy_1-yx_1)^2 ge 0 \
x^2y_1^2 + y^2x_1^2 ge 2xyx_1y_1 \
x^2x_1^2+ x^2y_1^2 + y^2x_1^2 + y^2y_1^2 ge x^2x_1^2+ 2xyx_1y_1 + y^2y_1^2 \
(x^2+y^2)(x_1^2+y_1^2) ge (xx_1+yy_1)^2
$$
edited Jan 21 at 16:00
Martin Sleziak
44.8k10119272
answered Jan 20 at 18:01
Rolf HoyerRolf Hoyer
11.3k31629
$endgroup$
Here we have
$$
xx_1 + yy_1 le sqrt{(x^2+y^2)(x_1^2+y_1^2)} le sqrt{z^2}sqrt{z_1^2}= zz_1,.
$$
The final equality relies on $z,z_1 ge 0$.
The first inequality is an example of the Cauchy-Schwarz inequality, and can be directly shown in this case. (Note that the inequality is trivial if $xx_1+yy_1 < 0$.)
$$
x^2y_1^2 - 2xyx_1y_1 + y^2x_1^2 = (xy_1-yx_1)^2 ge 0 \
x^2y_1^2 + y^2x_1^2 ge 2xyx_1y_1 \
x^2x_1^2+ x^2y_1^2 + y^2x_1^2 + y^2y_1^2 ge x^2x_1^2+ 2xyx_1y_1 + y^2y_1^2 \
(x^2+y^2)(x_1^2+y_1^2) ge (xx_1+yy_1)^2
$$
edited Jan 21 at 16:00
Martin Sleziak
44.8k10119272
answered Jan 20 at 18:01
Rolf HoyerRolf Hoyer
11.3k31629
edited Jan 21 at 16:00
Martin Sleziak
44.8k10119272
edited Jan 21 at 16:00
Martin Sleziak
44.8k10119272
edited Jan 21 at 16:00
Martin Sleziak
44.8k10119272
44.8k10119272
answered Jan 20 at 18:01
Rolf HoyerRolf Hoyer
11.3k31629
answered Jan 20 at 18:01
Rolf HoyerRolf Hoyer
11.3k31629
answered Jan 20 at 18:01
Rolf HoyerRolf Hoyer
11.3k31629
11.3k31629
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$begingroup$
Let us consider the usual norm $|vec a|=sqrt{a_1^2+a_2^2}$ on $mathbb R^2$
Notice also that the condition describing $S$ is exactly $sqrt{x^2+y^2} le z$, so
$$S={(x,y,z); zge0, |(x,y)|le z}.$$
So now if we have two points $(x_1,y_1,z_1),(x_2,y_2,z_2)in S$, then we get for any $lambdain(0,1)$
$$|(lambda x_1+(1-lambda)x_2,lambda y_1+(1-lambda)y_2)| overset{(*)}le
lambda|(x_1,y_1)| + (1-lambda)|(x_2,y_2)| le lambda z_1+(1-lambda)z_2.$$
The inequality $(*)$ follows from triangle inequality applied to vectors $(lambda x_1,lambda y_1)$ and $((1-lambda)x_2,(1-lambda)y_2)$. But we can view it also as convexity of the function $(x,y)mapsto|(x,y)|$. See, for example Why is every $p$-norm convex? and other posts linked there.
We have shown that
$$|(lambda x_1+(1-lambda)x_2,lambda y_1+(1-lambda)y_2)| le lambda z_1+(1-lambda)z_2,$$
which means that also the convex combination $(lambda x_1+(1-lambda)x_2,lambda y_1+(1-lambda)y_2,lambda z_1+(1-lambda)z_2)$ belongs to $S$.
Notice that similar approach would work to prove convexity of $S={(x_1,dots,x_{n+1}); x_{n+1}ge 0, sum_{k=1}^n x_k^2 le x_{n+1}^2}$ in $mathbb R^{n+1}$. (This condition can be equivalently stated as $|(x_1,dots,x_n)|lesqrt{x_{n+1}}$.)
answered Jan 21 at 16:59
Martin SleziakMartin Sleziak
44.8k10119272
$endgroup$
add a comment |
$begingroup$
Let us consider the usual norm $|vec a|=sqrt{a_1^2+a_2^2}$ on $mathbb R^2$
Notice also that the condition describing $S$ is exactly $sqrt{x^2+y^2} le z$, so
$$S={(x,y,z); zge0, |(x,y)|le z}.$$
So now if we have two points $(x_1,y_1,z_1),(x_2,y_2,z_2)in S$, then we get for any $lambdain(0,1)$
$$|(lambda x_1+(1-lambda)x_2,lambda y_1+(1-lambda)y_2)| overset{(*)}le
lambda|(x_1,y_1)| + (1-lambda)|(x_2,y_2)| le lambda z_1+(1-lambda)z_2.$$
The inequality $(*)$ follows from triangle inequality applied to vectors $(lambda x_1,lambda y_1)$ and $((1-lambda)x_2,(1-lambda)y_2)$. But we can view it also as convexity of the function $(x,y)mapsto|(x,y)|$. See, for example Why is every $p$-norm convex? and other posts linked there.
We have shown that
$$|(lambda x_1+(1-lambda)x_2,lambda y_1+(1-lambda)y_2)| le lambda z_1+(1-lambda)z_2,$$
which means that also the convex combination $(lambda x_1+(1-lambda)x_2,lambda y_1+(1-lambda)y_2,lambda z_1+(1-lambda)z_2)$ belongs to $S$.
Notice that similar approach would work to prove convexity of $S={(x_1,dots,x_{n+1}); x_{n+1}ge 0, sum_{k=1}^n x_k^2 le x_{n+1}^2}$ in $mathbb R^{n+1}$. (This condition can be equivalently stated as $|(x_1,dots,x_n)|lesqrt{x_{n+1}}$.)
answered Jan 21 at 16:59
Martin SleziakMartin Sleziak
44.8k10119272
$endgroup$
add a comment |
$begingroup$
Let us consider the usual norm $|vec a|=sqrt{a_1^2+a_2^2}$ on $mathbb R^2$
Notice also that the condition describing $S$ is exactly $sqrt{x^2+y^2} le z$, so
$$S={(x,y,z); zge0, |(x,y)|le z}.$$
So now if we have two points $(x_1,y_1,z_1),(x_2,y_2,z_2)in S$, then we get for any $lambdain(0,1)$
$$|(lambda x_1+(1-lambda)x_2,lambda y_1+(1-lambda)y_2)| overset{(*)}le
lambda|(x_1,y_1)| + (1-lambda)|(x_2,y_2)| le lambda z_1+(1-lambda)z_2.$$
The inequality $(*)$ follows from triangle inequality applied to vectors $(lambda x_1,lambda y_1)$ and $((1-lambda)x_2,(1-lambda)y_2)$. But we can view it also as convexity of the function $(x,y)mapsto|(x,y)|$. See, for example Why is every $p$-norm convex? and other posts linked there.
We have shown that
$$|(lambda x_1+(1-lambda)x_2,lambda y_1+(1-lambda)y_2)| le lambda z_1+(1-lambda)z_2,$$
which means that also the convex combination $(lambda x_1+(1-lambda)x_2,lambda y_1+(1-lambda)y_2,lambda z_1+(1-lambda)z_2)$ belongs to $S$.
Notice that similar approach would work to prove convexity of $S={(x_1,dots,x_{n+1}); x_{n+1}ge 0, sum_{k=1}^n x_k^2 le x_{n+1}^2}$ in $mathbb R^{n+1}$. (This condition can be equivalently stated as $|(x_1,dots,x_n)|lesqrt{x_{n+1}}$.)
answered Jan 21 at 16:59
Martin SleziakMartin Sleziak
44.8k10119272
$endgroup$
Let us consider the usual norm $|vec a|=sqrt{a_1^2+a_2^2}$ on $mathbb R^2$
Notice also that the condition describing $S$ is exactly $sqrt{x^2+y^2} le z$, so
$$S={(x,y,z); zge0, |(x,y)|le z}.$$
So now if we have two points $(x_1,y_1,z_1),(x_2,y_2,z_2)in S$, then we get for any $lambdain(0,1)$
$$|(lambda x_1+(1-lambda)x_2,lambda y_1+(1-lambda)y_2)| overset{(*)}le
lambda|(x_1,y_1)| + (1-lambda)|(x_2,y_2)| le lambda z_1+(1-lambda)z_2.$$
The inequality $(*)$ follows from triangle inequality applied to vectors $(lambda x_1,lambda y_1)$ and $((1-lambda)x_2,(1-lambda)y_2)$. But we can view it also as convexity of the function $(x,y)mapsto|(x,y)|$. See, for example Why is every $p$-norm convex? and other posts linked there.
We have shown that
$$|(lambda x_1+(1-lambda)x_2,lambda y_1+(1-lambda)y_2)| le lambda z_1+(1-lambda)z_2,$$
which means that also the convex combination $(lambda x_1+(1-lambda)x_2,lambda y_1+(1-lambda)y_2,lambda z_1+(1-lambda)z_2)$ belongs to $S$.
Notice that similar approach would work to prove convexity of $S={(x_1,dots,x_{n+1}); x_{n+1}ge 0, sum_{k=1}^n x_k^2 le x_{n+1}^2}$ in $mathbb R^{n+1}$. (This condition can be equivalently stated as $|(x_1,dots,x_n)|lesqrt{x_{n+1}}$.)
answered Jan 21 at 16:59
Martin SleziakMartin Sleziak
44.8k10119272
edited Jan 28 at 9:07
answered Jan 21 at 16:59
Martin SleziakMartin Sleziak
44.8k10119272
answered Jan 21 at 16:59
Martin SleziakMartin Sleziak
44.8k10119272
answered Jan 21 at 16:59
Martin SleziakMartin Sleziak
44.8k10119272
44.8k10119272
add a comment |
add a comment |
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