Proving $mid parallel x parallel - parallel y parallel mid$ $leqq$ $parallel x-y parallel$? [duplicate]

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
-2
down vote

favorite













This question already has an answer here:



  • Reverse Triangle Inequality Proof

    3 answers



Let $V$ be a normed vector space and I need to prove the follwing inequality $mid parallel x parallel - parallel y parallel mid$ $leqq$ $parallel x-y parallel$ containing the norm and the absolute value of the real numbers. However things just get twisted and I cannot see how to prove it.... Could anyone please tell me how to prove it?




inequality vector-spaces norm




share|cite|improve this question












marked as duplicate by Jyrki Lahtonen, Martin R, Nosrati, Arnaud D., amWhy Aug 17 at 11:13


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • The question that presents itself : what can you infer from this inequality ?
    – Neil hawking
    Aug 17 at 6:54










  • Search for triangle inequality.
    – Jyrki Lahtonen
    Aug 17 at 7:00














up vote
-2
down vote

favorite













This question already has an answer here:



  • Reverse Triangle Inequality Proof

    3 answers



Let $V$ be a normed vector space and I need to prove the follwing inequality $mid parallel x parallel - parallel y parallel mid$ $leqq$ $parallel x-y parallel$ containing the norm and the absolute value of the real numbers. However things just get twisted and I cannot see how to prove it.... Could anyone please tell me how to prove it?




inequality vector-spaces norm




share|cite|improve this question












marked as duplicate by Jyrki Lahtonen, Martin R, Nosrati, Arnaud D., amWhy Aug 17 at 11:13


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • The question that presents itself : what can you infer from this inequality ?
    – Neil hawking
    Aug 17 at 6:54










  • Search for triangle inequality.
    – Jyrki Lahtonen
    Aug 17 at 7:00












up vote
-2
down vote

favorite









up vote
-2
down vote

favorite












This question already has an answer here:



  • Reverse Triangle Inequality Proof

    3 answers



Let $V$ be a normed vector space and I need to prove the follwing inequality $mid parallel x parallel - parallel y parallel mid$ $leqq$ $parallel x-y parallel$ containing the norm and the absolute value of the real numbers. However things just get twisted and I cannot see how to prove it.... Could anyone please tell me how to prove it?




inequality vector-spaces norm




share|cite|improve this question













This question already has an answer here:



  • Reverse Triangle Inequality Proof

    3 answers



Let $V$ be a normed vector space and I need to prove the follwing inequality $mid parallel x parallel - parallel y parallel mid$ $leqq$ $parallel x-y parallel$ containing the norm and the absolute value of the real numbers. However things just get twisted and I cannot see how to prove it.... Could anyone please tell me how to prove it?





This question already has an answer here:



  • Reverse Triangle Inequality Proof

    3 answers





inequality vector-spaces norm





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 17 at 6:44









Keith

1,350720




1,350720




marked as duplicate by Jyrki Lahtonen, Martin R, Nosrati, Arnaud D., amWhy Aug 17 at 11:13


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Jyrki Lahtonen, Martin R, Nosrati, Arnaud D., amWhy Aug 17 at 11:13


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • The question that presents itself : what can you infer from this inequality ?
    – Neil hawking
    Aug 17 at 6:54










  • Search for triangle inequality.
    – Jyrki Lahtonen
    Aug 17 at 7:00
















  • The question that presents itself : what can you infer from this inequality ?
    – Neil hawking
    Aug 17 at 6:54










  • Search for triangle inequality.
    – Jyrki Lahtonen
    Aug 17 at 7:00















The question that presents itself : what can you infer from this inequality ?
– Neil hawking
Aug 17 at 6:54




The question that presents itself : what can you infer from this inequality ?
– Neil hawking
Aug 17 at 6:54












Search for triangle inequality.
– Jyrki Lahtonen
Aug 17 at 7:00




Search for triangle inequality.
– Jyrki Lahtonen
Aug 17 at 7:00










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










$||x||=||x-y+y|| le ||x-y||+||y||$, hence



$(1)$ $||x||-||y|| le ||x-y||$.



In a similar way we get



$(2)$ $||y||-||x|| le ||y-x||$.



Since $||y-x||=||x-y||$, $(1)$ and $(2)$ give the result.






share|cite|improve this answer




















  • Oh.............Thank you for your answer....
    – Keith
    Aug 17 at 6:52










  • Why the downvote ??????????????????????????????
    – Fred
    Aug 17 at 7:14






  • 1




    The downvote is because it looks like it didn't occur to you to that may be, during its 8 years of existence, the site has already handled the triangle inequality. IMO it is not useful to repeat material already explained multiples of times. And "downvote" = "this answer is not useful" right click it to see the explanation.
    – Jyrki Lahtonen
    Aug 17 at 10:16











  • Yes, my sheriff !
    – Fred
    Aug 17 at 10:20

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










$||x||=||x-y+y|| le ||x-y||+||y||$, hence



$(1)$ $||x||-||y|| le ||x-y||$.



In a similar way we get



$(2)$ $||y||-||x|| le ||y-x||$.



Since $||y-x||=||x-y||$, $(1)$ and $(2)$ give the result.






share|cite|improve this answer




















  • Oh.............Thank you for your answer....
    – Keith
    Aug 17 at 6:52










  • Why the downvote ??????????????????????????????
    – Fred
    Aug 17 at 7:14






  • 1




    The downvote is because it looks like it didn't occur to you to that may be, during its 8 years of existence, the site has already handled the triangle inequality. IMO it is not useful to repeat material already explained multiples of times. And "downvote" = "this answer is not useful" right click it to see the explanation.
    – Jyrki Lahtonen
    Aug 17 at 10:16











  • Yes, my sheriff !
    – Fred
    Aug 17 at 10:20














up vote
1
down vote



accepted










$||x||=||x-y+y|| le ||x-y||+||y||$, hence



$(1)$ $||x||-||y|| le ||x-y||$.



In a similar way we get



$(2)$ $||y||-||x|| le ||y-x||$.



Since $||y-x||=||x-y||$, $(1)$ and $(2)$ give the result.






share|cite|improve this answer




















  • Oh.............Thank you for your answer....
    – Keith
    Aug 17 at 6:52










  • Why the downvote ??????????????????????????????
    – Fred
    Aug 17 at 7:14






  • 1




    The downvote is because it looks like it didn't occur to you to that may be, during its 8 years of existence, the site has already handled the triangle inequality. IMO it is not useful to repeat material already explained multiples of times. And "downvote" = "this answer is not useful" right click it to see the explanation.
    – Jyrki Lahtonen
    Aug 17 at 10:16











  • Yes, my sheriff !
    – Fred
    Aug 17 at 10:20












up vote
1
down vote



accepted







up vote
1
down vote



accepted






$||x||=||x-y+y|| le ||x-y||+||y||$, hence



$(1)$ $||x||-||y|| le ||x-y||$.



In a similar way we get



$(2)$ $||y||-||x|| le ||y-x||$.



Since $||y-x||=||x-y||$, $(1)$ and $(2)$ give the result.






share|cite|improve this answer












$||x||=||x-y+y|| le ||x-y||+||y||$, hence



$(1)$ $||x||-||y|| le ||x-y||$.



In a similar way we get



$(2)$ $||y||-||x|| le ||y-x||$.



Since $||y-x||=||x-y||$, $(1)$ and $(2)$ give the result.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 17 at 6:51









Fred

38.2k1238




38.2k1238











  • Oh.............Thank you for your answer....
    – Keith
    Aug 17 at 6:52










  • Why the downvote ??????????????????????????????
    – Fred
    Aug 17 at 7:14






  • 1




    The downvote is because it looks like it didn't occur to you to that may be, during its 8 years of existence, the site has already handled the triangle inequality. IMO it is not useful to repeat material already explained multiples of times. And "downvote" = "this answer is not useful" right click it to see the explanation.
    – Jyrki Lahtonen
    Aug 17 at 10:16











  • Yes, my sheriff !
    – Fred
    Aug 17 at 10:20
















  • Oh.............Thank you for your answer....
    – Keith
    Aug 17 at 6:52










  • Why the downvote ??????????????????????????????
    – Fred
    Aug 17 at 7:14






  • 1




    The downvote is because it looks like it didn't occur to you to that may be, during its 8 years of existence, the site has already handled the triangle inequality. IMO it is not useful to repeat material already explained multiples of times. And "downvote" = "this answer is not useful" right click it to see the explanation.
    – Jyrki Lahtonen
    Aug 17 at 10:16











  • Yes, my sheriff !
    – Fred
    Aug 17 at 10:20















Oh.............Thank you for your answer....
– Keith
Aug 17 at 6:52




Oh.............Thank you for your answer....
– Keith
Aug 17 at 6:52












Why the downvote ??????????????????????????????
– Fred
Aug 17 at 7:14




Why the downvote ??????????????????????????????
– Fred
Aug 17 at 7:14




1




1




The downvote is because it looks like it didn't occur to you to that may be, during its 8 years of existence, the site has already handled the triangle inequality. IMO it is not useful to repeat material already explained multiples of times. And "downvote" = "this answer is not useful" right click it to see the explanation.
– Jyrki Lahtonen
Aug 17 at 10:16





The downvote is because it looks like it didn't occur to you to that may be, during its 8 years of existence, the site has already handled the triangle inequality. IMO it is not useful to repeat material already explained multiples of times. And "downvote" = "this answer is not useful" right click it to see the explanation.
– Jyrki Lahtonen
Aug 17 at 10:16













Yes, my sheriff !
– Fred
Aug 17 at 10:20




Yes, my sheriff !
– Fred
Aug 17 at 10:20


-

這個網誌中的熱門文章

Is there any way to eliminate the singular point to solve this integral by hand or by approximations?

圖片
Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite 1 $$int_0^1 (1+(x^-p-1)^1-p)^frac1pdx, pin mathbbR, pge 1$$ I find that $x=0,1$ are singular points(discontinuous points of this integral), is there a way to eliminate those singular points to make sure that this integral can be calculated by hand or by Taylor approximations. In either cases, what are those steps of calculating the integral? calculus real-analysis analysis approximation share | cite | improve this question edited Sep 11 at 5:07 asked Sep 11 at 4:48 Mclalalala 126 8 1 What is $p$? Please give more information to make it complete. – xbh Sep 11 at 5:02 @xbh already fixed it, thanks for reminding. – Mclalalala Sep 11 at 5:10 Seems that $0$ is not singular here… [I might compute wrong] – xbh Sep 11 at 5:30 @xbh can you show me how to get the close formula for the integral above? – Mclalalala Sep 11 at 5:33
閱讀完整內容

Why am i infinitely getting the same tweet with the Twitter Search API?

圖片
up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia
閱讀完整內容

Carbon dioxide

圖片
"CO2" redirects here. For other uses, see CO2 (disambiguation). Carbon dioxide Names Other names Carbonic acid gas Carbonic anhydride Carbonic oxide Carbon oxide Carbon(IV) oxide Dry ice (solid phase) Identifiers CAS Number 124-38-9   Y 3D model (JSmol) Interactive image Interactive image 3DMet B01131 Beilstein Reference 1900390 ChEBI CHEBI:16526   Y ChEMBL ChEMBL1231871   N ChemSpider 274   Y ECHA InfoCard 100.004.271 EC Number 204-696-9 E number E290 (preservatives) Gmelin Reference 989 KEGG D00004   Y MeSH Carbon+dioxide PubChem CID 280 RTECS number FF6400000 UNII 142M471B3J   Y UN number 1013 (gas), 1845 (solid) InChI InChI=1S/CO2/c2-1-3   Y Key: CURLTUGMZLYLDI-UHFFFAOYSA-N   Y InChI=1/CO2/c2-1-3 Key: CURLTUGMZLYLDI-UHFFFAOYAO SMILES O=C=O C(=O)=O Properties Chemical formula C O 2 Molar mass 44.01 g·mol −1 Appearance Colorless gas Odor Low concentrations: none High concentrations: sharp; acidic [1] Density 1562 kg/m 3 (solid at 1 atm and −78.5 °C) 1101 kg/m 3 (liquid at
閱讀完整內容