Calculate the derivative of $f(x) = | x|^2.$
$begingroup$
Suppose we have a normed space $E$ with the induced inner product and a map $f:Eto R$ such that $f(x) = |x|^2.$ I want to compute the derivative of this map.
We have that
$$f(x+h)=|x|^2+|h|^2+2langle x,hrangle=f(x)+o(|h|)+df_{x}(h).$$
Clearly the derivative $df_x(h) = 2langle x,hrangle$ is linear in $h.$ We also want to show that $df_x(h)$ is continuous in $h$. This can be done by simply observing that $$|df_x(h)|leq 2|x|cdot |h|$$ and therefore
$$|df_x|leq 2|x|$$ implying that $df_x$ is continuous in $h.$
I also want to show that the derivative $df_x$ is $C^{1}.$ For this we have to show that the map $xto df_x$ is continuous. Therefore observe that for each $hin E$ we have that,
$$|df_x(h)-df_y(h)|=2|langle x-y,hrangle|leq |x-y|cdot2|h|.$$
I am not sure how to proceed after this step. Any hints will be much appreciated.
real-analysis functional-analysis
edited Nov 26 '18 at 18:18
Martin Sleziak
44.7k9117272
asked Nov 26 '18 at 17:41
Hello_WorldHello_World
4,12621731
$endgroup$
add a comment |
$begingroup$
Suppose we have a normed space $E$ with the induced inner product and a map $f:Eto R$ such that $f(x) = |x|^2.$ I want to compute the derivative of this map.
We have that
$$f(x+h)=|x|^2+|h|^2+2langle x,hrangle=f(x)+o(|h|)+df_{x}(h).$$
Clearly the derivative $df_x(h) = 2langle x,hrangle$ is linear in $h.$ We also want to show that $df_x(h)$ is continuous in $h$. This can be done by simply observing that $$|df_x(h)|leq 2|x|cdot |h|$$ and therefore
$$|df_x|leq 2|x|$$ implying that $df_x$ is continuous in $h.$
I also want to show that the derivative $df_x$ is $C^{1}.$ For this we have to show that the map $xto df_x$ is continuous. Therefore observe that for each $hin E$ we have that,
$$|df_x(h)-df_y(h)|=2|langle x-y,hrangle|leq |x-y|cdot2|h|.$$
I am not sure how to proceed after this step. Any hints will be much appreciated.
real-analysis functional-analysis
edited Nov 26 '18 at 18:18
Martin Sleziak
44.7k9117272
asked Nov 26 '18 at 17:41
Hello_WorldHello_World
4,12621731
$endgroup$
1
$begingroup$
What is a inner product induced by a norm? Most norms don't come from inner products.
$endgroup$
– José Carlos Santos
Nov 26 '18 at 17:43
1
$begingroup$
Oh I think I meant the other way around.
$endgroup$
– Hello_World
Nov 26 '18 at 18:01
$begingroup$
If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
$endgroup$
– Gibbs
Nov 26 '18 at 18:15
add a comment |
$begingroup$
Suppose we have a normed space $E$ with the induced inner product and a map $f:Eto R$ such that $f(x) = |x|^2.$ I want to compute the derivative of this map.
We have that
$$f(x+h)=|x|^2+|h|^2+2langle x,hrangle=f(x)+o(|h|)+df_{x}(h).$$
Clearly the derivative $df_x(h) = 2langle x,hrangle$ is linear in $h.$ We also want to show that $df_x(h)$ is continuous in $h$. This can be done by simply observing that $$|df_x(h)|leq 2|x|cdot |h|$$ and therefore
$$|df_x|leq 2|x|$$ implying that $df_x$ is continuous in $h.$
I also want to show that the derivative $df_x$ is $C^{1}.$ For this we have to show that the map $xto df_x$ is continuous. Therefore observe that for each $hin E$ we have that,
$$|df_x(h)-df_y(h)|=2|langle x-y,hrangle|leq |x-y|cdot2|h|.$$
I am not sure how to proceed after this step. Any hints will be much appreciated.
real-analysis functional-analysis
edited Nov 26 '18 at 18:18
Martin Sleziak
44.7k9117272
asked Nov 26 '18 at 17:41
Hello_WorldHello_World
4,12621731
$endgroup$
Suppose we have a normed space $E$ with the induced inner product and a map $f:Eto R$ such that $f(x) = |x|^2.$ I want to compute the derivative of this map.
We have that
$$f(x+h)=|x|^2+|h|^2+2langle x,hrangle=f(x)+o(|h|)+df_{x}(h).$$
Clearly the derivative $df_x(h) = 2langle x,hrangle$ is linear in $h.$ We also want to show that $df_x(h)$ is continuous in $h$. This can be done by simply observing that $$|df_x(h)|leq 2|x|cdot |h|$$ and therefore
$$|df_x|leq 2|x|$$ implying that $df_x$ is continuous in $h.$
I also want to show that the derivative $df_x$ is $C^{1}.$ For this we have to show that the map $xto df_x$ is continuous. Therefore observe that for each $hin E$ we have that,
$$|df_x(h)-df_y(h)|=2|langle x-y,hrangle|leq |x-y|cdot2|h|.$$
I am not sure how to proceed after this step. Any hints will be much appreciated.
real-analysis functional-analysis
real-analysis functional-analysis
edited Nov 26 '18 at 18:18
Martin Sleziak
44.7k9117272
asked Nov 26 '18 at 17:41
Hello_WorldHello_World
4,12621731
edited Nov 26 '18 at 18:18
Martin Sleziak
44.7k9117272
asked Nov 26 '18 at 17:41
Hello_WorldHello_World
4,12621731
edited Nov 26 '18 at 18:18
Martin Sleziak
44.7k9117272
edited Nov 26 '18 at 18:18
Martin Sleziak
44.7k9117272
edited Nov 26 '18 at 18:18
Martin Sleziak
44.7k9117272
44.7k9117272
asked Nov 26 '18 at 17:41
Hello_WorldHello_World
4,12621731
asked Nov 26 '18 at 17:41
Hello_WorldHello_World
4,12621731
asked Nov 26 '18 at 17:41
Hello_WorldHello_World
4,12621731
4,12621731
1
$begingroup$
What is a inner product induced by a norm? Most norms don't come from inner products.
$endgroup$
– José Carlos Santos
Nov 26 '18 at 17:43
1
$begingroup$
Oh I think I meant the other way around.
$endgroup$
– Hello_World
Nov 26 '18 at 18:01
$begingroup$
If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
$endgroup$
– Gibbs
Nov 26 '18 at 18:15
add a comment |
1
$begingroup$
What is a inner product induced by a norm? Most norms don't come from inner products.
$endgroup$
– José Carlos Santos
Nov 26 '18 at 17:43
1
$begingroup$
Oh I think I meant the other way around.
$endgroup$
– Hello_World
Nov 26 '18 at 18:01
$begingroup$
If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
$endgroup$
– Gibbs
Nov 26 '18 at 18:15
1
1
$begingroup$
What is a inner product induced by a norm? Most norms don't come from inner products.
$endgroup$
– José Carlos Santos
Nov 26 '18 at 17:43
$begingroup$
What is a inner product induced by a norm? Most norms don't come from inner products.
$endgroup$
– José Carlos Santos
Nov 26 '18 at 17:43
1
1
$begingroup$
Oh I think I meant the other way around.
$endgroup$
– Hello_World
Nov 26 '18 at 18:01
$begingroup$
Oh I think I meant the other way around.
$endgroup$
– Hello_World
Nov 26 '18 at 18:01
$begingroup$
If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
$endgroup$
– Gibbs
Nov 26 '18 at 18:15
$begingroup$
If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
$endgroup$
– Gibbs
Nov 26 '18 at 18:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Alternatively, you can use that the inner product is bilinear and continuous, so $C^1$. Now, write $f$ as a composition:
$$xlongmapsto(x,x)longmapstolangle x,xrangle = f(x)$$
and apply the chain rule.
answered Nov 26 '18 at 18:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
$endgroup$
add a comment |
$begingroup$
I assume you use the norm $|f|=sup |f(x)|$ where the supremum runs over all $|x| = 1.$ (If such set of $x$ is empty, you are dealing with a trivial case; the details are left to you.) So, set $|h| = 1$ and get that the linear transformation $d_xf-d_yf$ has norm $leq 2|x - y|.$ So, when $y to x,$ $d_yf to d_xf.$ Q.E.D.
answered Nov 26 '18 at 18:21
Will M.Will M.
2,460315
$endgroup$
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%2f3014646%2fcalculate-the-derivative-of-fx-x-2%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
$begingroup$
Alternatively, you can use that the inner product is bilinear and continuous, so $C^1$. Now, write $f$ as a composition:
$$xlongmapsto(x,x)longmapstolangle x,xrangle = f(x)$$
and apply the chain rule.
answered Nov 26 '18 at 18:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
$endgroup$
add a comment |
$begingroup$
Alternatively, you can use that the inner product is bilinear and continuous, so $C^1$. Now, write $f$ as a composition:
$$xlongmapsto(x,x)longmapstolangle x,xrangle = f(x)$$
and apply the chain rule.
answered Nov 26 '18 at 18:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
$endgroup$
add a comment |
$begingroup$
Alternatively, you can use that the inner product is bilinear and continuous, so $C^1$. Now, write $f$ as a composition:
$$xlongmapsto(x,x)longmapstolangle x,xrangle = f(x)$$
and apply the chain rule.
answered Nov 26 '18 at 18:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
$endgroup$
Alternatively, you can use that the inner product is bilinear and continuous, so $C^1$. Now, write $f$ as a composition:
$$xlongmapsto(x,x)longmapstolangle x,xrangle = f(x)$$
and apply the chain rule.
answered Nov 26 '18 at 18:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
answered Nov 26 '18 at 18:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
answered Nov 26 '18 at 18:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
answered Nov 26 '18 at 18:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
34.2k42871
add a comment |
add a comment |
$begingroup$
I assume you use the norm $|f|=sup |f(x)|$ where the supremum runs over all $|x| = 1.$ (If such set of $x$ is empty, you are dealing with a trivial case; the details are left to you.) So, set $|h| = 1$ and get that the linear transformation $d_xf-d_yf$ has norm $leq 2|x - y|.$ So, when $y to x,$ $d_yf to d_xf.$ Q.E.D.
answered Nov 26 '18 at 18:21
Will M.Will M.
2,460315
$endgroup$
add a comment |
$begingroup$
I assume you use the norm $|f|=sup |f(x)|$ where the supremum runs over all $|x| = 1.$ (If such set of $x$ is empty, you are dealing with a trivial case; the details are left to you.) So, set $|h| = 1$ and get that the linear transformation $d_xf-d_yf$ has norm $leq 2|x - y|.$ So, when $y to x,$ $d_yf to d_xf.$ Q.E.D.
answered Nov 26 '18 at 18:21
Will M.Will M.
2,460315
$endgroup$
add a comment |
$begingroup$
I assume you use the norm $|f|=sup |f(x)|$ where the supremum runs over all $|x| = 1.$ (If such set of $x$ is empty, you are dealing with a trivial case; the details are left to you.) So, set $|h| = 1$ and get that the linear transformation $d_xf-d_yf$ has norm $leq 2|x - y|.$ So, when $y to x,$ $d_yf to d_xf.$ Q.E.D.
answered Nov 26 '18 at 18:21
Will M.Will M.
2,460315
$endgroup$
I assume you use the norm $|f|=sup |f(x)|$ where the supremum runs over all $|x| = 1.$ (If such set of $x$ is empty, you are dealing with a trivial case; the details are left to you.) So, set $|h| = 1$ and get that the linear transformation $d_xf-d_yf$ has norm $leq 2|x - y|.$ So, when $y to x,$ $d_yf to d_xf.$ Q.E.D.
answered Nov 26 '18 at 18:21
Will M.Will M.
2,460315
answered Nov 26 '18 at 18:21
Will M.Will M.
2,460315
answered Nov 26 '18 at 18:21
Will M.Will M.
2,460315
answered Nov 26 '18 at 18:21
Will M.Will M.
2,460315
2,460315
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%2f3014646%2fcalculate-the-derivative-of-fx-x-2%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
$begingroup$
What is a inner product induced by a norm? Most norms don't come from inner products.
$endgroup$
– José Carlos Santos
Nov 26 '18 at 17:43
1
$begingroup$
Oh I think I meant the other way around.
$endgroup$
– Hello_World
Nov 26 '18 at 18:01
$begingroup$
If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
$endgroup$
– Gibbs
Nov 26 '18 at 18:15