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












6












$begingroup$


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




    $begingroup$
    do you want to specifically know if $|cdot|^2$ is convex? then you should make this more clear, also in the title
    $endgroup$
    – supinf
    Nov 30 '18 at 10:40










  • $begingroup$
    @supinf: I made some edits!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 11:16










  • $begingroup$
    Does your vector space also provide inner product?
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 11:54
















6












$begingroup$


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











$endgroup$








  • 3




    $begingroup$
    do you want to specifically know if $|cdot|^2$ is convex? then you should make this more clear, also in the title
    $endgroup$
    – supinf
    Nov 30 '18 at 10:40










  • $begingroup$
    @supinf: I made some edits!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 11:16










  • $begingroup$
    Does your vector space also provide inner product?
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 11:54














6












6








6


1



$begingroup$


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











$endgroup$




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 '18 at 11:25







Omojola Micheal

















asked Nov 30 '18 at 10:30









Omojola MichealOmojola Micheal

1,753324




1,753324








  • 3




    $begingroup$
    do you want to specifically know if $|cdot|^2$ is convex? then you should make this more clear, also in the title
    $endgroup$
    – supinf
    Nov 30 '18 at 10:40










  • $begingroup$
    @supinf: I made some edits!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 11:16










  • $begingroup$
    Does your vector space also provide inner product?
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 11:54














  • 3




    $begingroup$
    do you want to specifically know if $|cdot|^2$ is convex? then you should make this more clear, also in the title
    $endgroup$
    – supinf
    Nov 30 '18 at 10:40










  • $begingroup$
    @supinf: I made some edits!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 11:16










  • $begingroup$
    Does your vector space also provide inner product?
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 11:54








3




3




$begingroup$
do you want to specifically know if $|cdot|^2$ is convex? then you should make this more clear, also in the title
$endgroup$
– supinf
Nov 30 '18 at 10:40




$begingroup$
do you want to specifically know if $|cdot|^2$ is convex? then you should make this more clear, also in the title
$endgroup$
– supinf
Nov 30 '18 at 10:40












$begingroup$
@supinf: I made some edits!
$endgroup$
– Omojola Micheal
Nov 30 '18 at 11:16




$begingroup$
@supinf: I made some edits!
$endgroup$
– Omojola Micheal
Nov 30 '18 at 11:16












$begingroup$
Does your vector space also provide inner product?
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 11:54




$begingroup$
Does your vector space also provide inner product?
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 11:54










3 Answers
3






active

oldest

votes


















2












$begingroup$

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









$endgroup$













  • $begingroup$
    This is good! I like it!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 11:31



















4












$begingroup$

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











$endgroup$













  • $begingroup$
    Nice! (+1)......
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 11:53



















1












$begingroup$

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









$endgroup$













  • $begingroup$
    That's fine too!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 11:31










  • $begingroup$
    Thank you. Good luck!
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 11:33










  • $begingroup$
    This only works if the norm comes from a scalar product, however.
    $endgroup$
    – Giuseppe Negro
    Nov 30 '18 at 11:50











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3 Answers
3






active

oldest

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3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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









$endgroup$













  • $begingroup$
    This is good! I like it!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 11:31
















2












$begingroup$

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









$endgroup$













  • $begingroup$
    This is good! I like it!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 11:31














2












2








2





$begingroup$

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









$endgroup$



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 '18 at 11:05









EricEric

2068




2068












  • $begingroup$
    This is good! I like it!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 11:31


















  • $begingroup$
    This is good! I like it!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 11:31
















$begingroup$
This is good! I like it!
$endgroup$
– Omojola Micheal
Nov 30 '18 at 11:31




$begingroup$
This is good! I like it!
$endgroup$
– Omojola Micheal
Nov 30 '18 at 11:31











4












$begingroup$

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











$endgroup$













  • $begingroup$
    Nice! (+1)......
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 11:53
















4












$begingroup$

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











$endgroup$













  • $begingroup$
    Nice! (+1)......
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 11:53














4












4








4





$begingroup$

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











$endgroup$



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 '18 at 11:20

























answered Nov 30 '18 at 11:15









Omojola MichealOmojola Micheal

1,753324




1,753324












  • $begingroup$
    Nice! (+1)......
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 11:53


















  • $begingroup$
    Nice! (+1)......
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 11:53
















$begingroup$
Nice! (+1)......
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 11:53




$begingroup$
Nice! (+1)......
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 11:53











1












$begingroup$

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









$endgroup$













  • $begingroup$
    That's fine too!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 11:31










  • $begingroup$
    Thank you. Good luck!
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 11:33










  • $begingroup$
    This only works if the norm comes from a scalar product, however.
    $endgroup$
    – Giuseppe Negro
    Nov 30 '18 at 11:50
















1












$begingroup$

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









$endgroup$













  • $begingroup$
    That's fine too!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 11:31










  • $begingroup$
    Thank you. Good luck!
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 11:33










  • $begingroup$
    This only works if the norm comes from a scalar product, however.
    $endgroup$
    – Giuseppe Negro
    Nov 30 '18 at 11:50














1












1








1





$begingroup$

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









$endgroup$



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 '18 at 11:25









Mostafa AyazMostafa Ayaz

14.7k3938




14.7k3938












  • $begingroup$
    That's fine too!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 11:31










  • $begingroup$
    Thank you. Good luck!
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 11:33










  • $begingroup$
    This only works if the norm comes from a scalar product, however.
    $endgroup$
    – Giuseppe Negro
    Nov 30 '18 at 11:50


















  • $begingroup$
    That's fine too!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 11:31










  • $begingroup$
    Thank you. Good luck!
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 11:33










  • $begingroup$
    This only works if the norm comes from a scalar product, however.
    $endgroup$
    – Giuseppe Negro
    Nov 30 '18 at 11:50
















$begingroup$
That's fine too!
$endgroup$
– Omojola Micheal
Nov 30 '18 at 11:31




$begingroup$
That's fine too!
$endgroup$
– Omojola Micheal
Nov 30 '18 at 11:31












$begingroup$
Thank you. Good luck!
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 11:33




$begingroup$
Thank you. Good luck!
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 11:33












$begingroup$
This only works if the norm comes from a scalar product, however.
$endgroup$
– Giuseppe Negro
Nov 30 '18 at 11:50




$begingroup$
This only works if the norm comes from a scalar product, however.
$endgroup$
– Giuseppe Negro
Nov 30 '18 at 11:50


















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Le Mesnil-Réaume

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Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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web3.py web3.isConnected() returns false always

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1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
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