Show an inequality
$begingroup$
Show that on the Hilbert space: $$xbot y $$ if and only if $$ Vert x + CyVert ge Vert x Vert,$$
$forall Cin mathbb R.$
inequality hilbert-spaces
$endgroup$
add a comment |
$begingroup$
Show that on the Hilbert space: $$xbot y $$ if and only if $$ Vert x + CyVert ge Vert x Vert,$$
$forall Cin mathbb R.$
inequality hilbert-spaces
$endgroup$
1
$begingroup$
Consider $|x+Cy|^2-|x|^2$ for $C$ near zero.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 17:02
add a comment |
$begingroup$
Show that on the Hilbert space: $$xbot y $$ if and only if $$ Vert x + CyVert ge Vert x Vert,$$
$forall Cin mathbb R.$
inequality hilbert-spaces
$endgroup$
Show that on the Hilbert space: $$xbot y $$ if and only if $$ Vert x + CyVert ge Vert x Vert,$$
$forall Cin mathbb R.$
inequality hilbert-spaces
inequality hilbert-spaces
asked Jan 7 at 16:59
PhilipPhilip
826
826
1
$begingroup$
Consider $|x+Cy|^2-|x|^2$ for $C$ near zero.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 17:02
add a comment |
1
$begingroup$
Consider $|x+Cy|^2-|x|^2$ for $C$ near zero.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 17:02
1
1
$begingroup$
Consider $|x+Cy|^2-|x|^2$ for $C$ near zero.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 17:02
$begingroup$
Consider $|x+Cy|^2-|x|^2$ for $C$ near zero.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 17:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Observe that, for given $Cin mathbb{R},$
$$|x+Cy|^2=left<x+Cy,x+Cyright>=|x|^2+C^2|y|^2geq |x|^2 $$
due to the orthogonality ($xperp y implies left<x,y right>=0.$)
Take the square root, then you get
$$|x+Cy|geq |x|$$
$endgroup$
add a comment |
$begingroup$
$||x+Cy||^{2}=||x||^{2}+2Clangle x,yrangle+C^{2}||y||^{2}$
$||x+Cy||^{2}ge ||x||^{2} iff 2Clangle x,yrangle+C^{2}||y||^{2}ge 0$
This associated equation is an quation of second degree has two real roots $C_{1}=0$, $C_{2}=frac{2langle x,yrangle}{||y||^{2}}$. So the inequality is always true for all $Cinmathbb{R} iff C_{1}=C_{2} iff xperp y$
$endgroup$
add a comment |
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',
autoActivateHeartbeat: false,
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065224%2fshow-an-inequality%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Observe that, for given $Cin mathbb{R},$
$$|x+Cy|^2=left<x+Cy,x+Cyright>=|x|^2+C^2|y|^2geq |x|^2 $$
due to the orthogonality ($xperp y implies left<x,y right>=0.$)
Take the square root, then you get
$$|x+Cy|geq |x|$$
$endgroup$
add a comment |
$begingroup$
Observe that, for given $Cin mathbb{R},$
$$|x+Cy|^2=left<x+Cy,x+Cyright>=|x|^2+C^2|y|^2geq |x|^2 $$
due to the orthogonality ($xperp y implies left<x,y right>=0.$)
Take the square root, then you get
$$|x+Cy|geq |x|$$
$endgroup$
add a comment |
$begingroup$
Observe that, for given $Cin mathbb{R},$
$$|x+Cy|^2=left<x+Cy,x+Cyright>=|x|^2+C^2|y|^2geq |x|^2 $$
due to the orthogonality ($xperp y implies left<x,y right>=0.$)
Take the square root, then you get
$$|x+Cy|geq |x|$$
$endgroup$
Observe that, for given $Cin mathbb{R},$
$$|x+Cy|^2=left<x+Cy,x+Cyright>=|x|^2+C^2|y|^2geq |x|^2 $$
due to the orthogonality ($xperp y implies left<x,y right>=0.$)
Take the square root, then you get
$$|x+Cy|geq |x|$$
answered Jan 7 at 17:10
LeBLeB
1,080317
1,080317
add a comment |
add a comment |
$begingroup$
$||x+Cy||^{2}=||x||^{2}+2Clangle x,yrangle+C^{2}||y||^{2}$
$||x+Cy||^{2}ge ||x||^{2} iff 2Clangle x,yrangle+C^{2}||y||^{2}ge 0$
This associated equation is an quation of second degree has two real roots $C_{1}=0$, $C_{2}=frac{2langle x,yrangle}{||y||^{2}}$. So the inequality is always true for all $Cinmathbb{R} iff C_{1}=C_{2} iff xperp y$
$endgroup$
add a comment |
$begingroup$
$||x+Cy||^{2}=||x||^{2}+2Clangle x,yrangle+C^{2}||y||^{2}$
$||x+Cy||^{2}ge ||x||^{2} iff 2Clangle x,yrangle+C^{2}||y||^{2}ge 0$
This associated equation is an quation of second degree has two real roots $C_{1}=0$, $C_{2}=frac{2langle x,yrangle}{||y||^{2}}$. So the inequality is always true for all $Cinmathbb{R} iff C_{1}=C_{2} iff xperp y$
$endgroup$
add a comment |
$begingroup$
$||x+Cy||^{2}=||x||^{2}+2Clangle x,yrangle+C^{2}||y||^{2}$
$||x+Cy||^{2}ge ||x||^{2} iff 2Clangle x,yrangle+C^{2}||y||^{2}ge 0$
This associated equation is an quation of second degree has two real roots $C_{1}=0$, $C_{2}=frac{2langle x,yrangle}{||y||^{2}}$. So the inequality is always true for all $Cinmathbb{R} iff C_{1}=C_{2} iff xperp y$
$endgroup$
$||x+Cy||^{2}=||x||^{2}+2Clangle x,yrangle+C^{2}||y||^{2}$
$||x+Cy||^{2}ge ||x||^{2} iff 2Clangle x,yrangle+C^{2}||y||^{2}ge 0$
This associated equation is an quation of second degree has two real roots $C_{1}=0$, $C_{2}=frac{2langle x,yrangle}{||y||^{2}}$. So the inequality is always true for all $Cinmathbb{R} iff C_{1}=C_{2} iff xperp y$
answered Jan 7 at 17:37
ecrinecrin
3547
3547
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065224%2fshow-an-inequality%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Consider $|x+Cy|^2-|x|^2$ for $C$ near zero.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 17:02