Mixing
inequality with differential norm
$begingroup$
I'm trying to prove that $sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|} = sup_{x in B} |D_f(x)|$
Where $f:B to mathbb R^n$ is differentiable and $B subset mathbb R^n$.
I have difficulties.
this is what I tried -
Let $x_0 in B$, $epsilon > 0$, $t >0$ and $|v| = 1$.
$f(x_0+tv) = f(x_0)+D_f(x_0)tv + frac{E(tv)}{t}$, so $D_f(x_0)v = frac{f(x_0+tv)-f(x_0)}{t}-frac{E(tv)}{t}$
Hence $|D_f(x_0)| leq frac{|f(x_0+tv)-f(x_0)|}{t}+frac{|E(tv)|}{t} leq sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|} + frac{|E(tv)|}{t}$
If $t$ is small enough, then $sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|} + frac{|E(tv)|}{t} leq sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|} +epsilon$
But this is true for all $epsilon$, so overall $|D_f(x_0)| leq sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|}$
That's as far as I got, not sure how to continue.
calculus supremum-and-infimum
asked Jan 17 at 20:54
Rick JokerRick Joker
399111
$endgroup$
add a comment |
$begingroup$
I'm trying to prove that $sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|} = sup_{x in B} |D_f(x)|$
Where $f:B to mathbb R^n$ is differentiable and $B subset mathbb R^n$.
I have difficulties.
this is what I tried -
Let $x_0 in B$, $epsilon > 0$, $t >0$ and $|v| = 1$.
$f(x_0+tv) = f(x_0)+D_f(x_0)tv + frac{E(tv)}{t}$, so $D_f(x_0)v = frac{f(x_0+tv)-f(x_0)}{t}-frac{E(tv)}{t}$
Hence $|D_f(x_0)| leq frac{|f(x_0+tv)-f(x_0)|}{t}+frac{|E(tv)|}{t} leq sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|} + frac{|E(tv)|}{t}$
If $t$ is small enough, then $sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|} + frac{|E(tv)|}{t} leq sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|} +epsilon$
But this is true for all $epsilon$, so overall $|D_f(x_0)| leq sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|}$
That's as far as I got, not sure how to continue.
calculus supremum-and-infimum
asked Jan 17 at 20:54
Rick JokerRick Joker
399111
$endgroup$
$begingroup$
This might help.
$endgroup$
– user635162
Jan 17 at 23:59
add a comment |
$begingroup$
I'm trying to prove that $sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|} = sup_{x in B} |D_f(x)|$
Where $f:B to mathbb R^n$ is differentiable and $B subset mathbb R^n$.
I have difficulties.
this is what I tried -
Let $x_0 in B$, $epsilon > 0$, $t >0$ and $|v| = 1$.
$f(x_0+tv) = f(x_0)+D_f(x_0)tv + frac{E(tv)}{t}$, so $D_f(x_0)v = frac{f(x_0+tv)-f(x_0)}{t}-frac{E(tv)}{t}$
Hence $|D_f(x_0)| leq frac{|f(x_0+tv)-f(x_0)|}{t}+frac{|E(tv)|}{t} leq sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|} + frac{|E(tv)|}{t}$
If $t$ is small enough, then $sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|} + frac{|E(tv)|}{t} leq sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|} +epsilon$
But this is true for all $epsilon$, so overall $|D_f(x_0)| leq sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|}$
That's as far as I got, not sure how to continue.
calculus supremum-and-infimum
asked Jan 17 at 20:54
Rick JokerRick Joker
399111
$endgroup$
I'm trying to prove that $sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|} = sup_{x in B} |D_f(x)|$
Where $f:B to mathbb R^n$ is differentiable and $B subset mathbb R^n$.
I have difficulties.
this is what I tried -
Let $x_0 in B$, $epsilon > 0$, $t >0$ and $|v| = 1$.
$f(x_0+tv) = f(x_0)+D_f(x_0)tv + frac{E(tv)}{t}$, so $D_f(x_0)v = frac{f(x_0+tv)-f(x_0)}{t}-frac{E(tv)}{t}$
Hence $|D_f(x_0)| leq frac{|f(x_0+tv)-f(x_0)|}{t}+frac{|E(tv)|}{t} leq sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|} + frac{|E(tv)|}{t}$
If $t$ is small enough, then $sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|} + frac{|E(tv)|}{t} leq sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|} +epsilon$
But this is true for all $epsilon$, so overall $|D_f(x_0)| leq sup_{x,y in B}frac{|f(x)-f(y)|}{|x-y|}$
That's as far as I got, not sure how to continue.
calculus supremum-and-infimum
calculus supremum-and-infimum
asked Jan 17 at 20:54
Rick JokerRick Joker
399111
asked Jan 17 at 20:54
Rick JokerRick Joker
399111
edited Jan 17 at 21:08
Rick Joker
asked Jan 17 at 20:54
Rick JokerRick Joker
399111
asked Jan 17 at 20:54
Rick JokerRick Joker
399111
asked Jan 17 at 20:54
Rick JokerRick Joker
399111
399111
$begingroup$
This might help.
$endgroup$
– user635162
Jan 17 at 23:59
add a comment |
$begingroup$
This might help.
$endgroup$
– user635162
Jan 17 at 23:59
$begingroup$
This might help.
$endgroup$
– user635162
Jan 17 at 23:59
$begingroup$
This might help.
$endgroup$
– user635162
Jan 17 at 23:59
add a comment |
0
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',
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%2f3077498%2finequality-with-differential-norm%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3077498%2finequality-with-differential-norm%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
$begingroup$
This might help.
$endgroup$
– user635162
Jan 17 at 23:59