Prove that $Vertcdot Vert^2:Xto Bbb{R},$ where $X$ is a vector space, is convex












6














Let $X$ be a vector space. I was able to prove that $Vertcdot Vert:Xto Bbb{R},$ is a convex function, i.e., for all $x,yin X$ and $lambda in [0,1],$



begin{align} Vert lambda x+(1-lambda)y Vert leq lambda Vert xVert+(1-lambda)Vert y Vertend{align}



Now, I want to prove that $Vertcdot Vert^2:Xto Bbb{R},$ where $X$ is a vector space, is convex. So, here's what I've done!



MY WORK



begin{align} Vert lambda x+(1-lambda)y Vert^2 leq left( lambda Vert xVert+(1-lambda)Vert y Vertright)^2,;;text{for all};; x,yin X;; text{and};; lambda in [0,1].end{align}



So, any help please on how to proceed?






functional-analysis analysis convex-analysis norm normed-spaces







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  • 3




    do you want to specifically know if $|cdot|^2$ is convex? then you should make this more clear, also in the title
    – supinf
    Nov 30 at 10:40










  • @supinf: I made some edits!
    – Mike
    Nov 30 at 11:16










  • Does your vector space also provide inner product?
    – Mostafa Ayaz
    Nov 30 at 11:54
















6














Let $X$ be a vector space. I was able to prove that $Vertcdot Vert:Xto Bbb{R},$ is a convex function, i.e., for all $x,yin X$ and $lambda in [0,1],$



begin{align} Vert lambda x+(1-lambda)y Vert leq lambda Vert xVert+(1-lambda)Vert y Vertend{align}



Now, I want to prove that $Vertcdot Vert^2:Xto Bbb{R},$ where $X$ is a vector space, is convex. So, here's what I've done!



MY WORK



begin{align} Vert lambda x+(1-lambda)y Vert^2 leq left( lambda Vert xVert+(1-lambda)Vert y Vertright)^2,;;text{for all};; x,yin X;; text{and};; lambda in [0,1].end{align}



So, any help please on how to proceed?






functional-analysis analysis convex-analysis norm normed-spaces







share|cite|improve this question




















  • 3




    do you want to specifically know if $|cdot|^2$ is convex? then you should make this more clear, also in the title
    – supinf
    Nov 30 at 10:40










  • @supinf: I made some edits!
    – Mike
    Nov 30 at 11:16










  • Does your vector space also provide inner product?
    – Mostafa Ayaz
    Nov 30 at 11:54














6












6








6


1





Let $X$ be a vector space. I was able to prove that $Vertcdot Vert:Xto Bbb{R},$ is a convex function, i.e., for all $x,yin X$ and $lambda in [0,1],$



begin{align} Vert lambda x+(1-lambda)y Vert leq lambda Vert xVert+(1-lambda)Vert y Vertend{align}



Now, I want to prove that $Vertcdot Vert^2:Xto Bbb{R},$ where $X$ is a vector space, is convex. So, here's what I've done!



MY WORK



begin{align} Vert lambda x+(1-lambda)y Vert^2 leq left( lambda Vert xVert+(1-lambda)Vert y Vertright)^2,;;text{for all};; x,yin X;; text{and};; lambda in [0,1].end{align}



So, any help please on how to proceed?






functional-analysis analysis convex-analysis norm normed-spaces







share|cite|improve this question















Let $X$ be a vector space. I was able to prove that $Vertcdot Vert:Xto Bbb{R},$ is a convex function, i.e., for all $x,yin X$ and $lambda in [0,1],$



begin{align} Vert lambda x+(1-lambda)y Vert leq lambda Vert xVert+(1-lambda)Vert y Vertend{align}



Now, I want to prove that $Vertcdot Vert^2:Xto Bbb{R},$ where $X$ is a vector space, is convex. So, here's what I've done!



MY WORK



begin{align} Vert lambda x+(1-lambda)y Vert^2 leq left( lambda Vert xVert+(1-lambda)Vert y Vertright)^2,;;text{for all};; x,yin X;; text{and};; lambda in [0,1].end{align}



So, any help please on how to proceed?






functional-analysis analysis convex-analysis norm normed-spaces




functional-analysis analysis convex-analysis norm normed-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 at 11:25

























asked Nov 30 at 10:30









Mike

1,117218




1,117218








  • 3




    do you want to specifically know if $|cdot|^2$ is convex? then you should make this more clear, also in the title
    – supinf
    Nov 30 at 10:40










  • @supinf: I made some edits!
    – Mike
    Nov 30 at 11:16










  • Does your vector space also provide inner product?
    – Mostafa Ayaz
    Nov 30 at 11:54














  • 3




    do you want to specifically know if $|cdot|^2$ is convex? then you should make this more clear, also in the title
    – supinf
    Nov 30 at 10:40










  • @supinf: I made some edits!
    – Mike
    Nov 30 at 11:16










  • Does your vector space also provide inner product?
    – Mostafa Ayaz
    Nov 30 at 11:54








3




3




do you want to specifically know if $|cdot|^2$ is convex? then you should make this more clear, also in the title
– supinf
Nov 30 at 10:40




do you want to specifically know if $|cdot|^2$ is convex? then you should make this more clear, also in the title
– supinf
Nov 30 at 10:40












@supinf: I made some edits!
– Mike
Nov 30 at 11:16




@supinf: I made some edits!
– Mike
Nov 30 at 11:16












Does your vector space also provide inner product?
– Mostafa Ayaz
Nov 30 at 11:54




Does your vector space also provide inner product?
– Mostafa Ayaz
Nov 30 at 11:54










3 Answers
3






active

oldest

votes


















2














In general if $f$ is a convex function and $g$ is a convex nondecreasing function then the composition $g circ f$ is a convex function. Let $f(cdot)=|cdot |$ which maps to $mathbb{R}_{geq 0}$ and let $g(x)=x^2$ which is a nondecreasing convex function on $mathbb{R}_{geq 0}$. If follows that $g circ f (cdot)=| cdot |^2$ is a convex function.



See The composition of two convex functions is convex for the original claim.






share|cite|improve this answer





















  • This is good! I like it!
    – Mike
    Nov 30 at 11:31



















4














Just solved and thought to share it for the sake of future readers.
begin{align} Vert lambda x+(1-lambda)y Vert^2 &leq left( lambda Vert xVert+(1-lambda)Vert y Vertright)^2\ &leq lambda^2 Vert xVert^2+2lambda(1-lambda)Vert xVertVert yVert+ (1-lambda)^2Vert yVert^2\ &= lambda^2 Vert xVert^2+2lambda(1-lambda)Vert xVertVert yVert+ (1-lambda)^2Vert yVert^2 -lambdaVert xVert^2 -(1-lambda)Vert yVert^2\&quad+lambdaVert xVert^2 +(1-lambda)Vert yVert^2,;;text{adding and substracting};lambdaVert xVert^2 +(1-lambda)Vert yVert^2 \ &= -lambda (1-lambda)left(Vert xVert-Vert yVertright)^2+lambdaVert xVert^2 +(1-lambda)Vert yVert^2\ &leq lambdaVert xVert^2 +(1-lambda)Vert yVert^2,;;text{since};-lambda (1-lambda)left(Vert xVert-Vert yVertright)^2leq 0.end{align}
Hence, $VertcdotVert^2$ is a convex function.






share|cite|improve this answer























  • Nice! (+1)......
    – Mostafa Ayaz
    Nov 30 at 11:53



















1














Define $p=lambda x$ and $q=(1-lambda)y$, therefore we need to show that $$||p+q||^2le (||p||+||q||)^2$$which reduces to $$pcdot qle ||p||cdot ||q||$$which is the same famous Cauchy-Schwartz inequality. Therefore $||.||^2$ is convex.






share|cite|improve this answer





















  • That's fine too!
    – Mike
    Nov 30 at 11:31










  • Thank you. Good luck!
    – Mostafa Ayaz
    Nov 30 at 11:33










  • This only works if the norm comes from a scalar product, however.
    – Giuseppe Negro
    Nov 30 at 11:50











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3 Answers
3






active

oldest

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3 Answers
3






active

oldest

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active

oldest

votes






active

oldest

votes









2














In general if $f$ is a convex function and $g$ is a convex nondecreasing function then the composition $g circ f$ is a convex function. Let $f(cdot)=|cdot |$ which maps to $mathbb{R}_{geq 0}$ and let $g(x)=x^2$ which is a nondecreasing convex function on $mathbb{R}_{geq 0}$. If follows that $g circ f (cdot)=| cdot |^2$ is a convex function.



See The composition of two convex functions is convex for the original claim.






share|cite|improve this answer





















  • This is good! I like it!
    – Mike
    Nov 30 at 11:31
















2














In general if $f$ is a convex function and $g$ is a convex nondecreasing function then the composition $g circ f$ is a convex function. Let $f(cdot)=|cdot |$ which maps to $mathbb{R}_{geq 0}$ and let $g(x)=x^2$ which is a nondecreasing convex function on $mathbb{R}_{geq 0}$. If follows that $g circ f (cdot)=| cdot |^2$ is a convex function.



See The composition of two convex functions is convex for the original claim.






share|cite|improve this answer





















  • This is good! I like it!
    – Mike
    Nov 30 at 11:31














2












2








2






In general if $f$ is a convex function and $g$ is a convex nondecreasing function then the composition $g circ f$ is a convex function. Let $f(cdot)=|cdot |$ which maps to $mathbb{R}_{geq 0}$ and let $g(x)=x^2$ which is a nondecreasing convex function on $mathbb{R}_{geq 0}$. If follows that $g circ f (cdot)=| cdot |^2$ is a convex function.



See The composition of two convex functions is convex for the original claim.






share|cite|improve this answer












In general if $f$ is a convex function and $g$ is a convex nondecreasing function then the composition $g circ f$ is a convex function. Let $f(cdot)=|cdot |$ which maps to $mathbb{R}_{geq 0}$ and let $g(x)=x^2$ which is a nondecreasing convex function on $mathbb{R}_{geq 0}$. If follows that $g circ f (cdot)=| cdot |^2$ is a convex function.



See The composition of two convex functions is convex for the original claim.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 30 at 11:05









Eric

2088




2088












  • This is good! I like it!
    – Mike
    Nov 30 at 11:31


















  • This is good! I like it!
    – Mike
    Nov 30 at 11:31
















This is good! I like it!
– Mike
Nov 30 at 11:31




This is good! I like it!
– Mike
Nov 30 at 11:31











4














Just solved and thought to share it for the sake of future readers.
begin{align} Vert lambda x+(1-lambda)y Vert^2 &leq left( lambda Vert xVert+(1-lambda)Vert y Vertright)^2\ &leq lambda^2 Vert xVert^2+2lambda(1-lambda)Vert xVertVert yVert+ (1-lambda)^2Vert yVert^2\ &= lambda^2 Vert xVert^2+2lambda(1-lambda)Vert xVertVert yVert+ (1-lambda)^2Vert yVert^2 -lambdaVert xVert^2 -(1-lambda)Vert yVert^2\&quad+lambdaVert xVert^2 +(1-lambda)Vert yVert^2,;;text{adding and substracting};lambdaVert xVert^2 +(1-lambda)Vert yVert^2 \ &= -lambda (1-lambda)left(Vert xVert-Vert yVertright)^2+lambdaVert xVert^2 +(1-lambda)Vert yVert^2\ &leq lambdaVert xVert^2 +(1-lambda)Vert yVert^2,;;text{since};-lambda (1-lambda)left(Vert xVert-Vert yVertright)^2leq 0.end{align}
Hence, $VertcdotVert^2$ is a convex function.






share|cite|improve this answer























  • Nice! (+1)......
    – Mostafa Ayaz
    Nov 30 at 11:53
















4














Just solved and thought to share it for the sake of future readers.
begin{align} Vert lambda x+(1-lambda)y Vert^2 &leq left( lambda Vert xVert+(1-lambda)Vert y Vertright)^2\ &leq lambda^2 Vert xVert^2+2lambda(1-lambda)Vert xVertVert yVert+ (1-lambda)^2Vert yVert^2\ &= lambda^2 Vert xVert^2+2lambda(1-lambda)Vert xVertVert yVert+ (1-lambda)^2Vert yVert^2 -lambdaVert xVert^2 -(1-lambda)Vert yVert^2\&quad+lambdaVert xVert^2 +(1-lambda)Vert yVert^2,;;text{adding and substracting};lambdaVert xVert^2 +(1-lambda)Vert yVert^2 \ &= -lambda (1-lambda)left(Vert xVert-Vert yVertright)^2+lambdaVert xVert^2 +(1-lambda)Vert yVert^2\ &leq lambdaVert xVert^2 +(1-lambda)Vert yVert^2,;;text{since};-lambda (1-lambda)left(Vert xVert-Vert yVertright)^2leq 0.end{align}
Hence, $VertcdotVert^2$ is a convex function.






share|cite|improve this answer























  • Nice! (+1)......
    – Mostafa Ayaz
    Nov 30 at 11:53














4












4








4






Just solved and thought to share it for the sake of future readers.
begin{align} Vert lambda x+(1-lambda)y Vert^2 &leq left( lambda Vert xVert+(1-lambda)Vert y Vertright)^2\ &leq lambda^2 Vert xVert^2+2lambda(1-lambda)Vert xVertVert yVert+ (1-lambda)^2Vert yVert^2\ &= lambda^2 Vert xVert^2+2lambda(1-lambda)Vert xVertVert yVert+ (1-lambda)^2Vert yVert^2 -lambdaVert xVert^2 -(1-lambda)Vert yVert^2\&quad+lambdaVert xVert^2 +(1-lambda)Vert yVert^2,;;text{adding and substracting};lambdaVert xVert^2 +(1-lambda)Vert yVert^2 \ &= -lambda (1-lambda)left(Vert xVert-Vert yVertright)^2+lambdaVert xVert^2 +(1-lambda)Vert yVert^2\ &leq lambdaVert xVert^2 +(1-lambda)Vert yVert^2,;;text{since};-lambda (1-lambda)left(Vert xVert-Vert yVertright)^2leq 0.end{align}
Hence, $VertcdotVert^2$ is a convex function.






share|cite|improve this answer














Just solved and thought to share it for the sake of future readers.
begin{align} Vert lambda x+(1-lambda)y Vert^2 &leq left( lambda Vert xVert+(1-lambda)Vert y Vertright)^2\ &leq lambda^2 Vert xVert^2+2lambda(1-lambda)Vert xVertVert yVert+ (1-lambda)^2Vert yVert^2\ &= lambda^2 Vert xVert^2+2lambda(1-lambda)Vert xVertVert yVert+ (1-lambda)^2Vert yVert^2 -lambdaVert xVert^2 -(1-lambda)Vert yVert^2\&quad+lambdaVert xVert^2 +(1-lambda)Vert yVert^2,;;text{adding and substracting};lambdaVert xVert^2 +(1-lambda)Vert yVert^2 \ &= -lambda (1-lambda)left(Vert xVert-Vert yVertright)^2+lambdaVert xVert^2 +(1-lambda)Vert yVert^2\ &leq lambdaVert xVert^2 +(1-lambda)Vert yVert^2,;;text{since};-lambda (1-lambda)left(Vert xVert-Vert yVertright)^2leq 0.end{align}
Hence, $VertcdotVert^2$ is a convex function.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 at 11:20

























answered Nov 30 at 11:15









Mike

1,117218




1,117218












  • Nice! (+1)......
    – Mostafa Ayaz
    Nov 30 at 11:53


















  • Nice! (+1)......
    – Mostafa Ayaz
    Nov 30 at 11:53
















Nice! (+1)......
– Mostafa Ayaz
Nov 30 at 11:53




Nice! (+1)......
– Mostafa Ayaz
Nov 30 at 11:53











1














Define $p=lambda x$ and $q=(1-lambda)y$, therefore we need to show that $$||p+q||^2le (||p||+||q||)^2$$which reduces to $$pcdot qle ||p||cdot ||q||$$which is the same famous Cauchy-Schwartz inequality. Therefore $||.||^2$ is convex.






share|cite|improve this answer





















  • That's fine too!
    – Mike
    Nov 30 at 11:31










  • Thank you. Good luck!
    – Mostafa Ayaz
    Nov 30 at 11:33










  • This only works if the norm comes from a scalar product, however.
    – Giuseppe Negro
    Nov 30 at 11:50
















1














Define $p=lambda x$ and $q=(1-lambda)y$, therefore we need to show that $$||p+q||^2le (||p||+||q||)^2$$which reduces to $$pcdot qle ||p||cdot ||q||$$which is the same famous Cauchy-Schwartz inequality. Therefore $||.||^2$ is convex.






share|cite|improve this answer





















  • That's fine too!
    – Mike
    Nov 30 at 11:31










  • Thank you. Good luck!
    – Mostafa Ayaz
    Nov 30 at 11:33










  • This only works if the norm comes from a scalar product, however.
    – Giuseppe Negro
    Nov 30 at 11:50














1












1








1






Define $p=lambda x$ and $q=(1-lambda)y$, therefore we need to show that $$||p+q||^2le (||p||+||q||)^2$$which reduces to $$pcdot qle ||p||cdot ||q||$$which is the same famous Cauchy-Schwartz inequality. Therefore $||.||^2$ is convex.






share|cite|improve this answer












Define $p=lambda x$ and $q=(1-lambda)y$, therefore we need to show that $$||p+q||^2le (||p||+||q||)^2$$which reduces to $$pcdot qle ||p||cdot ||q||$$which is the same famous Cauchy-Schwartz inequality. Therefore $||.||^2$ is convex.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 30 at 11:25









Mostafa Ayaz

13.7k3836




13.7k3836












  • That's fine too!
    – Mike
    Nov 30 at 11:31










  • Thank you. Good luck!
    – Mostafa Ayaz
    Nov 30 at 11:33










  • This only works if the norm comes from a scalar product, however.
    – Giuseppe Negro
    Nov 30 at 11:50


















  • That's fine too!
    – Mike
    Nov 30 at 11:31










  • Thank you. Good luck!
    – Mostafa Ayaz
    Nov 30 at 11:33










  • This only works if the norm comes from a scalar product, however.
    – Giuseppe Negro
    Nov 30 at 11:50
















That's fine too!
– Mike
Nov 30 at 11:31




That's fine too!
– Mike
Nov 30 at 11:31












Thank you. Good luck!
– Mostafa Ayaz
Nov 30 at 11:33




Thank you. Good luck!
– Mostafa Ayaz
Nov 30 at 11:33












This only works if the norm comes from a scalar product, however.
– Giuseppe Negro
Nov 30 at 11:50




This only works if the norm comes from a scalar product, however.
– Giuseppe Negro
Nov 30 at 11:50


















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