Prove that $|mathbfx+mathbfy|le|mathbfx|+|mathbfy|$

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











up vote
2
down vote

favorite












Let $l_infty$ be the vector space of bounded sequences of real numbers. Let $|mathbfx|=sup_nge 1|x_n|$. Prove that $|mathbfx+mathbfy|le|mathbfx|+|mathbfy|$.



My attempt:



Let $mathbfx, mathbfyin l_infty$. Then $mathbfx=(x_1, x_2, ldots)$ and $mathbfy=(y_1, y_2, ldots)$. $mathbfx+mathbfy=(x_1+y_1, x_2+y_2, ldots)$



$|mathbfx|=sup_nge 1|x_n|implies |x_n|le|mathbfx| forall nge 1$



$|mathbfy|=sup_nge 1|y_n|implies |y_n|le|mathbfy| forall nge 1$



$|mathbfx+mathbfy|=sup_nge 1|x_n+y_n|implies |x_n+y_n|le|mathbfx+mathbfy| forall nge 1$



$|x_n+y_n|le|x_n|+|y_n| forall nge 1$ (Triangle inequality)



But $|x_n|le|mathbfx|$ and $|y_n|le|mathbfy| forall nge 1$



So, $|x_n+y_n|le|mathbfx|+|mathbfy| forall nge 1$



So, $|mathbfx|+|mathbfy|$ is an upper bound for $|x_n+y_n|$. --------------- (1)



So, $sup_nge 1|x_n+y_n|le|mathbfx|+|mathbfy|$



$Leftrightarrow|mathbfx+mathbfy|le |mathbfx|+|mathbfy|$



QED






real-analysis norm







share|cite|improve this question



















  • 1




    Is this even true? What if $mathbfx = (1,0,1,0,1,0,dots)$ and $mathbfy = (0,1,0,1,0,1,dots)$?
    – Matthew Leingang
    Sep 11 at 2:42






  • 1




    Am I missing something? Take any not-constantly-zero sequence $(x_n)$ and put $y_n = -x_n$; doesn't this provide an immediate counter-example to equality? Where did you see this claim?
    – mathguy
    Sep 11 at 2:43











  • What if $mathbfy=-mathbfx$ and $mathbfx$ is nonzero?
    – edm
    Sep 11 at 2:46







  • 2




    @Callus you're talking about $limsup$. I think. In any case, the OP's definition of $sup$ is correct.
    – Theoretical Economist
    Sep 11 at 3:17






  • 1




    It looks right, or simply: $$sup_n |x_n + y_n| leq sup_n (|x_n| + |y_n|) leq sup_n |x_n| + sup_n |y_n|$$
    – user3669039
    Sep 11 at 14:07














up vote
2
down vote

favorite












Let $l_infty$ be the vector space of bounded sequences of real numbers. Let $|mathbfx|=sup_nge 1|x_n|$. Prove that $|mathbfx+mathbfy|le|mathbfx|+|mathbfy|$.



My attempt:



Let $mathbfx, mathbfyin l_infty$. Then $mathbfx=(x_1, x_2, ldots)$ and $mathbfy=(y_1, y_2, ldots)$. $mathbfx+mathbfy=(x_1+y_1, x_2+y_2, ldots)$



$|mathbfx|=sup_nge 1|x_n|implies |x_n|le|mathbfx| forall nge 1$



$|mathbfy|=sup_nge 1|y_n|implies |y_n|le|mathbfy| forall nge 1$



$|mathbfx+mathbfy|=sup_nge 1|x_n+y_n|implies |x_n+y_n|le|mathbfx+mathbfy| forall nge 1$



$|x_n+y_n|le|x_n|+|y_n| forall nge 1$ (Triangle inequality)



But $|x_n|le|mathbfx|$ and $|y_n|le|mathbfy| forall nge 1$



So, $|x_n+y_n|le|mathbfx|+|mathbfy| forall nge 1$



So, $|mathbfx|+|mathbfy|$ is an upper bound for $|x_n+y_n|$. --------------- (1)



So, $sup_nge 1|x_n+y_n|le|mathbfx|+|mathbfy|$



$Leftrightarrow|mathbfx+mathbfy|le |mathbfx|+|mathbfy|$



QED






real-analysis norm







share|cite|improve this question



















  • 1




    Is this even true? What if $mathbfx = (1,0,1,0,1,0,dots)$ and $mathbfy = (0,1,0,1,0,1,dots)$?
    – Matthew Leingang
    Sep 11 at 2:42






  • 1




    Am I missing something? Take any not-constantly-zero sequence $(x_n)$ and put $y_n = -x_n$; doesn't this provide an immediate counter-example to equality? Where did you see this claim?
    – mathguy
    Sep 11 at 2:43











  • What if $mathbfy=-mathbfx$ and $mathbfx$ is nonzero?
    – edm
    Sep 11 at 2:46







  • 2




    @Callus you're talking about $limsup$. I think. In any case, the OP's definition of $sup$ is correct.
    – Theoretical Economist
    Sep 11 at 3:17






  • 1




    It looks right, or simply: $$sup_n |x_n + y_n| leq sup_n (|x_n| + |y_n|) leq sup_n |x_n| + sup_n |y_n|$$
    – user3669039
    Sep 11 at 14:07












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Let $l_infty$ be the vector space of bounded sequences of real numbers. Let $|mathbfx|=sup_nge 1|x_n|$. Prove that $|mathbfx+mathbfy|le|mathbfx|+|mathbfy|$.



My attempt:



Let $mathbfx, mathbfyin l_infty$. Then $mathbfx=(x_1, x_2, ldots)$ and $mathbfy=(y_1, y_2, ldots)$. $mathbfx+mathbfy=(x_1+y_1, x_2+y_2, ldots)$



$|mathbfx|=sup_nge 1|x_n|implies |x_n|le|mathbfx| forall nge 1$



$|mathbfy|=sup_nge 1|y_n|implies |y_n|le|mathbfy| forall nge 1$



$|mathbfx+mathbfy|=sup_nge 1|x_n+y_n|implies |x_n+y_n|le|mathbfx+mathbfy| forall nge 1$



$|x_n+y_n|le|x_n|+|y_n| forall nge 1$ (Triangle inequality)



But $|x_n|le|mathbfx|$ and $|y_n|le|mathbfy| forall nge 1$



So, $|x_n+y_n|le|mathbfx|+|mathbfy| forall nge 1$



So, $|mathbfx|+|mathbfy|$ is an upper bound for $|x_n+y_n|$. --------------- (1)



So, $sup_nge 1|x_n+y_n|le|mathbfx|+|mathbfy|$



$Leftrightarrow|mathbfx+mathbfy|le |mathbfx|+|mathbfy|$



QED






real-analysis norm







share|cite|improve this question















Let $l_infty$ be the vector space of bounded sequences of real numbers. Let $|mathbfx|=sup_nge 1|x_n|$. Prove that $|mathbfx+mathbfy|le|mathbfx|+|mathbfy|$.



My attempt:



Let $mathbfx, mathbfyin l_infty$. Then $mathbfx=(x_1, x_2, ldots)$ and $mathbfy=(y_1, y_2, ldots)$. $mathbfx+mathbfy=(x_1+y_1, x_2+y_2, ldots)$



$|mathbfx|=sup_nge 1|x_n|implies |x_n|le|mathbfx| forall nge 1$



$|mathbfy|=sup_nge 1|y_n|implies |y_n|le|mathbfy| forall nge 1$



$|mathbfx+mathbfy|=sup_nge 1|x_n+y_n|implies |x_n+y_n|le|mathbfx+mathbfy| forall nge 1$



$|x_n+y_n|le|x_n|+|y_n| forall nge 1$ (Triangle inequality)



But $|x_n|le|mathbfx|$ and $|y_n|le|mathbfy| forall nge 1$



So, $|x_n+y_n|le|mathbfx|+|mathbfy| forall nge 1$



So, $|mathbfx|+|mathbfy|$ is an upper bound for $|x_n+y_n|$. --------------- (1)



So, $sup_nge 1|x_n+y_n|le|mathbfx|+|mathbfy|$



$Leftrightarrow|mathbfx+mathbfy|le |mathbfx|+|mathbfy|$



QED






real-analysis norm




real-analysis norm






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 11 at 13:11

























asked Sep 11 at 2:39









Thomas

672415




672415







  • 1




    Is this even true? What if $mathbfx = (1,0,1,0,1,0,dots)$ and $mathbfy = (0,1,0,1,0,1,dots)$?
    – Matthew Leingang
    Sep 11 at 2:42






  • 1




    Am I missing something? Take any not-constantly-zero sequence $(x_n)$ and put $y_n = -x_n$; doesn't this provide an immediate counter-example to equality? Where did you see this claim?
    – mathguy
    Sep 11 at 2:43











  • What if $mathbfy=-mathbfx$ and $mathbfx$ is nonzero?
    – edm
    Sep 11 at 2:46







  • 2




    @Callus you're talking about $limsup$. I think. In any case, the OP's definition of $sup$ is correct.
    – Theoretical Economist
    Sep 11 at 3:17






  • 1




    It looks right, or simply: $$sup_n |x_n + y_n| leq sup_n (|x_n| + |y_n|) leq sup_n |x_n| + sup_n |y_n|$$
    – user3669039
    Sep 11 at 14:07












  • 1




    Is this even true? What if $mathbfx = (1,0,1,0,1,0,dots)$ and $mathbfy = (0,1,0,1,0,1,dots)$?
    – Matthew Leingang
    Sep 11 at 2:42






  • 1




    Am I missing something? Take any not-constantly-zero sequence $(x_n)$ and put $y_n = -x_n$; doesn't this provide an immediate counter-example to equality? Where did you see this claim?
    – mathguy
    Sep 11 at 2:43











  • What if $mathbfy=-mathbfx$ and $mathbfx$ is nonzero?
    – edm
    Sep 11 at 2:46







  • 2




    @Callus you're talking about $limsup$. I think. In any case, the OP's definition of $sup$ is correct.
    – Theoretical Economist
    Sep 11 at 3:17






  • 1




    It looks right, or simply: $$sup_n |x_n + y_n| leq sup_n (|x_n| + |y_n|) leq sup_n |x_n| + sup_n |y_n|$$
    – user3669039
    Sep 11 at 14:07







1




1




Is this even true? What if $mathbfx = (1,0,1,0,1,0,dots)$ and $mathbfy = (0,1,0,1,0,1,dots)$?
– Matthew Leingang
Sep 11 at 2:42




Is this even true? What if $mathbfx = (1,0,1,0,1,0,dots)$ and $mathbfy = (0,1,0,1,0,1,dots)$?
– Matthew Leingang
Sep 11 at 2:42




1




1




Am I missing something? Take any not-constantly-zero sequence $(x_n)$ and put $y_n = -x_n$; doesn't this provide an immediate counter-example to equality? Where did you see this claim?
– mathguy
Sep 11 at 2:43





Am I missing something? Take any not-constantly-zero sequence $(x_n)$ and put $y_n = -x_n$; doesn't this provide an immediate counter-example to equality? Where did you see this claim?
– mathguy
Sep 11 at 2:43













What if $mathbfy=-mathbfx$ and $mathbfx$ is nonzero?
– edm
Sep 11 at 2:46





What if $mathbfy=-mathbfx$ and $mathbfx$ is nonzero?
– edm
Sep 11 at 2:46





2




2




@Callus you're talking about $limsup$. I think. In any case, the OP's definition of $sup$ is correct.
– Theoretical Economist
Sep 11 at 3:17




@Callus you're talking about $limsup$. I think. In any case, the OP's definition of $sup$ is correct.
– Theoretical Economist
Sep 11 at 3:17




1




1




It looks right, or simply: $$sup_n |x_n + y_n| leq sup_n (|x_n| + |y_n|) leq sup_n |x_n| + sup_n |y_n|$$
– user3669039
Sep 11 at 14:07




It looks right, or simply: $$sup_n |x_n + y_n| leq sup_n (|x_n| + |y_n|) leq sup_n |x_n| + sup_n |y_n|$$
– user3669039
Sep 11 at 14:07















active

oldest

votes











Your Answer





StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2912674%2fprove-that-mathbfx-mathbfy-le-mathbfx-mathbfy%23new-answer', 'question_page');

);

Post as a guest






















active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2912674%2fprove-that-mathbfx-mathbfy-le-mathbfx-mathbfy%23new-answer', 'question_page');

);

Post as a guest
































































-

這個網誌中的熱門文章

How to combine Bézier curves to a surface?

圖片
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
閱讀完整內容

Mutual Information Always Non-negative

圖片
Clash Royale CLAN TAG #URR8PPP up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 '12 at 15:38 Cameron Buie 83.6k 7 71 154 asked Jun 17 '12 at 15:30 Omri 359 3 13 3 Convexity of the function $tmapsto tlog t$. – Did Jun 17 '12 at 15:33 In addition, the convexity properties require the coefficients in the linear combination sum 1. Then, as p(x,y) is a probability distribution, it fullfits such condition. – Francisco Javier Delgado Ceped Aug 10 at 18:44 add a comment  |  up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 ...
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