Show that $|f(a+tu)-f(a)|leq max_{1leq ileq n}|f(apm te_i )- f(a)|$ where $f:mathbb{R}^nto mathbb{R}$ is...
$begingroup$
Show that $|f(a+tu)-f(a)|leq max_{1leq ileq n}|f(apm te_i )- f(a)|$ where $f:mathbb{R}^nto mathbb{R}$ is function convex and $(e_1,e_2,cdots ,e_n)$ is the canonical basis of $mathbb{R}^n$ and $x,u,ain mathbb{R}^n$ with $||u||_1=sum_{1}^{n}|u_i|=1.$ Also $tin [0,1].$
I tried to create something of the form $f((1-t)x+ ty)$ but nothing seems to work. Maybe there is some other fact about convex functions that I need to use here. Perhaps someone can give a hint for this problem.
real-analysis
asked Dec 14 '18 at 7:21
Hello_WorldHello_World
4,13121831
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add a comment |
$begingroup$
Show that $|f(a+tu)-f(a)|leq max_{1leq ileq n}|f(apm te_i )- f(a)|$ where $f:mathbb{R}^nto mathbb{R}$ is function convex and $(e_1,e_2,cdots ,e_n)$ is the canonical basis of $mathbb{R}^n$ and $x,u,ain mathbb{R}^n$ with $||u||_1=sum_{1}^{n}|u_i|=1.$ Also $tin [0,1].$
I tried to create something of the form $f((1-t)x+ ty)$ but nothing seems to work. Maybe there is some other fact about convex functions that I need to use here. Perhaps someone can give a hint for this problem.
real-analysis
asked Dec 14 '18 at 7:21
Hello_WorldHello_World
4,13121831
$endgroup$
add a comment |
$begingroup$
Show that $|f(a+tu)-f(a)|leq max_{1leq ileq n}|f(apm te_i )- f(a)|$ where $f:mathbb{R}^nto mathbb{R}$ is function convex and $(e_1,e_2,cdots ,e_n)$ is the canonical basis of $mathbb{R}^n$ and $x,u,ain mathbb{R}^n$ with $||u||_1=sum_{1}^{n}|u_i|=1.$ Also $tin [0,1].$
I tried to create something of the form $f((1-t)x+ ty)$ but nothing seems to work. Maybe there is some other fact about convex functions that I need to use here. Perhaps someone can give a hint for this problem.
real-analysis
asked Dec 14 '18 at 7:21
Hello_WorldHello_World
4,13121831
$endgroup$
Show that $|f(a+tu)-f(a)|leq max_{1leq ileq n}|f(apm te_i )- f(a)|$ where $f:mathbb{R}^nto mathbb{R}$ is function convex and $(e_1,e_2,cdots ,e_n)$ is the canonical basis of $mathbb{R}^n$ and $x,u,ain mathbb{R}^n$ with $||u||_1=sum_{1}^{n}|u_i|=1.$ Also $tin [0,1].$
I tried to create something of the form $f((1-t)x+ ty)$ but nothing seems to work. Maybe there is some other fact about convex functions that I need to use here. Perhaps someone can give a hint for this problem.
real-analysis
real-analysis
asked Dec 14 '18 at 7:21
Hello_WorldHello_World
4,13121831
asked Dec 14 '18 at 7:21
Hello_WorldHello_World
4,13121831
asked Dec 14 '18 at 7:21
Hello_WorldHello_World
4,13121831
asked Dec 14 '18 at 7:21
Hello_WorldHello_World
4,13121831
asked Dec 14 '18 at 7:21
Hello_WorldHello_World
4,13121831
4,13121831
add a comment |
add a comment |
1 Answer
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Try to use $a + t u = sum_i^n |u_i| (a + sign(u_i)t e_i)$, i.e. $a+t u$ is a convex combination of $(a + sign(u_i)t e_i)$ ($i=1,..,n$).
answered Dec 14 '18 at 7:40
VictorZurkowskiVictorZurkowski
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1 Answer
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1 Answer
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$begingroup$
Try to use $a + t u = sum_i^n |u_i| (a + sign(u_i)t e_i)$, i.e. $a+t u$ is a convex combination of $(a + sign(u_i)t e_i)$ ($i=1,..,n$).
answered Dec 14 '18 at 7:40
VictorZurkowskiVictorZurkowski
1,314412
$endgroup$
add a comment |
$begingroup$
Try to use $a + t u = sum_i^n |u_i| (a + sign(u_i)t e_i)$, i.e. $a+t u$ is a convex combination of $(a + sign(u_i)t e_i)$ ($i=1,..,n$).
answered Dec 14 '18 at 7:40
VictorZurkowskiVictorZurkowski
1,314412
$endgroup$
add a comment |
$begingroup$
Try to use $a + t u = sum_i^n |u_i| (a + sign(u_i)t e_i)$, i.e. $a+t u$ is a convex combination of $(a + sign(u_i)t e_i)$ ($i=1,..,n$).
answered Dec 14 '18 at 7:40
VictorZurkowskiVictorZurkowski
1,314412
$endgroup$
Try to use $a + t u = sum_i^n |u_i| (a + sign(u_i)t e_i)$, i.e. $a+t u$ is a convex combination of $(a + sign(u_i)t e_i)$ ($i=1,..,n$).
answered Dec 14 '18 at 7:40
VictorZurkowskiVictorZurkowski
1,314412
answered Dec 14 '18 at 7:40
VictorZurkowskiVictorZurkowski
1,314412
answered Dec 14 '18 at 7:40
VictorZurkowskiVictorZurkowski
1,314412
answered Dec 14 '18 at 7:40
VictorZurkowskiVictorZurkowski
1,314412
1,314412
add a comment |
add a comment |
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