Explaination of the Frechet Derivative
$begingroup$
Let $E$ and $F$ be Banach spaces and let $U$ be an open subset of $E$. Suppose $g:U to F $. $g$ is continuous at $x_0$ if there exists a linear transformation, $T_{x_0}$, such that $$ lim limits_{x_0 to 0}dfrac{||g(x_0+h)-g(x_0)-T_{x_0}(h)||}{||h||}. $$ Then $Dg_{x_0}=T_{x_0}$ is then the Frechet derivative of $g$ at $x_0$.
I have a question about this with two examples from this video (https://www.youtube.com/watch?v=RKHx1vQdZko).
For the first example we compute the Frechet derivative of $f(x)=x^2$ and we get that $T_x(h) = h(2x)$. For the second example we find the Frechet derivative of $f(x,y)=(xy , x+y)$ and here we get that $T_{(x,y)}(h,v)=begin{bmatrix}
y & x \
1 & 1 \
end{bmatrix}begin{bmatrix}
h \
v \
end{bmatrix}$ .
But the definition says that $Df_{x_0}=T_{x_0}$ so does that mean that $Df_{x_0}=T_{x_0}(1)$? It seems to be the case for both of these examples. I believe this is always but am unsure. The reason why I think this is that the transformation is linear so $T_{x_0}(h)=h(Df_{x_0})$ since $T$ linearly maps $h$ so if we let $h=1$ then we are just left with $Df_{x_0}$. Does my reasoning make sense?
frechet-derivative
asked Sep 30 '18 at 3:11
alpastalpast
468314
$endgroup$
add a comment |
$begingroup$
Let $E$ and $F$ be Banach spaces and let $U$ be an open subset of $E$. Suppose $g:U to F $. $g$ is continuous at $x_0$ if there exists a linear transformation, $T_{x_0}$, such that $$ lim limits_{x_0 to 0}dfrac{||g(x_0+h)-g(x_0)-T_{x_0}(h)||}{||h||}. $$ Then $Dg_{x_0}=T_{x_0}$ is then the Frechet derivative of $g$ at $x_0$.
I have a question about this with two examples from this video (https://www.youtube.com/watch?v=RKHx1vQdZko).
For the first example we compute the Frechet derivative of $f(x)=x^2$ and we get that $T_x(h) = h(2x)$. For the second example we find the Frechet derivative of $f(x,y)=(xy , x+y)$ and here we get that $T_{(x,y)}(h,v)=begin{bmatrix}
y & x \
1 & 1 \
end{bmatrix}begin{bmatrix}
h \
v \
end{bmatrix}$ .
But the definition says that $Df_{x_0}=T_{x_0}$ so does that mean that $Df_{x_0}=T_{x_0}(1)$? It seems to be the case for both of these examples. I believe this is always but am unsure. The reason why I think this is that the transformation is linear so $T_{x_0}(h)=h(Df_{x_0})$ since $T$ linearly maps $h$ so if we let $h=1$ then we are just left with $Df_{x_0}$. Does my reasoning make sense?
frechet-derivative
asked Sep 30 '18 at 3:11
alpastalpast
468314
$endgroup$
$begingroup$
No. What does even $h = 1$ mean in Banach spaces?
$endgroup$
– Will M.
Nov 26 '18 at 4:56
add a comment |
$begingroup$
Let $E$ and $F$ be Banach spaces and let $U$ be an open subset of $E$. Suppose $g:U to F $. $g$ is continuous at $x_0$ if there exists a linear transformation, $T_{x_0}$, such that $$ lim limits_{x_0 to 0}dfrac{||g(x_0+h)-g(x_0)-T_{x_0}(h)||}{||h||}. $$ Then $Dg_{x_0}=T_{x_0}$ is then the Frechet derivative of $g$ at $x_0$.
I have a question about this with two examples from this video (https://www.youtube.com/watch?v=RKHx1vQdZko).
For the first example we compute the Frechet derivative of $f(x)=x^2$ and we get that $T_x(h) = h(2x)$. For the second example we find the Frechet derivative of $f(x,y)=(xy , x+y)$ and here we get that $T_{(x,y)}(h,v)=begin{bmatrix}
y & x \
1 & 1 \
end{bmatrix}begin{bmatrix}
h \
v \
end{bmatrix}$ .
But the definition says that $Df_{x_0}=T_{x_0}$ so does that mean that $Df_{x_0}=T_{x_0}(1)$? It seems to be the case for both of these examples. I believe this is always but am unsure. The reason why I think this is that the transformation is linear so $T_{x_0}(h)=h(Df_{x_0})$ since $T$ linearly maps $h$ so if we let $h=1$ then we are just left with $Df_{x_0}$. Does my reasoning make sense?
frechet-derivative
asked Sep 30 '18 at 3:11
alpastalpast
468314
$endgroup$
Let $E$ and $F$ be Banach spaces and let $U$ be an open subset of $E$. Suppose $g:U to F $. $g$ is continuous at $x_0$ if there exists a linear transformation, $T_{x_0}$, such that $$ lim limits_{x_0 to 0}dfrac{||g(x_0+h)-g(x_0)-T_{x_0}(h)||}{||h||}. $$ Then $Dg_{x_0}=T_{x_0}$ is then the Frechet derivative of $g$ at $x_0$.
I have a question about this with two examples from this video (https://www.youtube.com/watch?v=RKHx1vQdZko).
For the first example we compute the Frechet derivative of $f(x)=x^2$ and we get that $T_x(h) = h(2x)$. For the second example we find the Frechet derivative of $f(x,y)=(xy , x+y)$ and here we get that $T_{(x,y)}(h,v)=begin{bmatrix}
y & x \
1 & 1 \
end{bmatrix}begin{bmatrix}
h \
v \
end{bmatrix}$ .
But the definition says that $Df_{x_0}=T_{x_0}$ so does that mean that $Df_{x_0}=T_{x_0}(1)$? It seems to be the case for both of these examples. I believe this is always but am unsure. The reason why I think this is that the transformation is linear so $T_{x_0}(h)=h(Df_{x_0})$ since $T$ linearly maps $h$ so if we let $h=1$ then we are just left with $Df_{x_0}$. Does my reasoning make sense?
frechet-derivative
frechet-derivative
asked Sep 30 '18 at 3:11
alpastalpast
468314
asked Sep 30 '18 at 3:11
alpastalpast
468314
asked Sep 30 '18 at 3:11
alpastalpast
468314
asked Sep 30 '18 at 3:11
alpastalpast
468314
asked Sep 30 '18 at 3:11
alpastalpast
468314
468314
$begingroup$
No. What does even $h = 1$ mean in Banach spaces?
$endgroup$
– Will M.
Nov 26 '18 at 4:56
add a comment |
$begingroup$
No. What does even $h = 1$ mean in Banach spaces?
$endgroup$
– Will M.
Nov 26 '18 at 4:56
$begingroup$
No. What does even $h = 1$ mean in Banach spaces?
$endgroup$
– Will M.
Nov 26 '18 at 4:56
$begingroup$
No. What does even $h = 1$ mean in Banach spaces?
$endgroup$
– Will M.
Nov 26 '18 at 4:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, it makes sense in case $E = mathbb{K}$ where $mathbb{K}$ is $mathbb{R}$ or $mathbb{C}$, i.e. what you said makes sense for your first example. The comment of @WillM. is key here, as $h=1$ in Banach spaces only means something if the Banach space is a field too. Note that your equation $T_{x_0}(h) = h(Df_{x_0})$ doesn't make sense in any other case.
More precisely: $mathcal{L}(mathbb{K},F)$ is isometrically isomorphic to $F$ via $varphicolonmathcal{L}(mathbb{K}, F)to F,, Amapsto A1$.
To see this, note that $varphi$ obviously is linear and injective. Next, consider, given $vin F$, the definition $(A_vx:=xcdot v,; xinmathbb{K})$. That way we get $A_vinmathcal{L}(mathbb{K}, F)$ and $A_v1 = v$, i.e. $varphi$ is surjective. Finally,
$$ Vert AxVert_F = vert xvert Vert A1Vert_Fleq Vert A1Vert_f,quad vert xvert leq 1,; Ainmathcal{L}(mathbb{K}, F),$$
which easily shows that $Vert AVert_{mathcal{L}(mathbb{K},F)} = Vert A1Vert_F$.
In conclusion, you may identify $mathcal{L}(mathbb{K}, F) = F$.
answered Dec 3 '18 at 18:32
RamenRamen
463412
$endgroup$
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%2f2936202%2fexplaination-of-the-frechet-derivative%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, it makes sense in case $E = mathbb{K}$ where $mathbb{K}$ is $mathbb{R}$ or $mathbb{C}$, i.e. what you said makes sense for your first example. The comment of @WillM. is key here, as $h=1$ in Banach spaces only means something if the Banach space is a field too. Note that your equation $T_{x_0}(h) = h(Df_{x_0})$ doesn't make sense in any other case.
More precisely: $mathcal{L}(mathbb{K},F)$ is isometrically isomorphic to $F$ via $varphicolonmathcal{L}(mathbb{K}, F)to F,, Amapsto A1$.
To see this, note that $varphi$ obviously is linear and injective. Next, consider, given $vin F$, the definition $(A_vx:=xcdot v,; xinmathbb{K})$. That way we get $A_vinmathcal{L}(mathbb{K}, F)$ and $A_v1 = v$, i.e. $varphi$ is surjective. Finally,
$$ Vert AxVert_F = vert xvert Vert A1Vert_Fleq Vert A1Vert_f,quad vert xvert leq 1,; Ainmathcal{L}(mathbb{K}, F),$$
which easily shows that $Vert AVert_{mathcal{L}(mathbb{K},F)} = Vert A1Vert_F$.
In conclusion, you may identify $mathcal{L}(mathbb{K}, F) = F$.
answered Dec 3 '18 at 18:32
RamenRamen
463412
$endgroup$
add a comment |
$begingroup$
Yes, it makes sense in case $E = mathbb{K}$ where $mathbb{K}$ is $mathbb{R}$ or $mathbb{C}$, i.e. what you said makes sense for your first example. The comment of @WillM. is key here, as $h=1$ in Banach spaces only means something if the Banach space is a field too. Note that your equation $T_{x_0}(h) = h(Df_{x_0})$ doesn't make sense in any other case.
More precisely: $mathcal{L}(mathbb{K},F)$ is isometrically isomorphic to $F$ via $varphicolonmathcal{L}(mathbb{K}, F)to F,, Amapsto A1$.
To see this, note that $varphi$ obviously is linear and injective. Next, consider, given $vin F$, the definition $(A_vx:=xcdot v,; xinmathbb{K})$. That way we get $A_vinmathcal{L}(mathbb{K}, F)$ and $A_v1 = v$, i.e. $varphi$ is surjective. Finally,
$$ Vert AxVert_F = vert xvert Vert A1Vert_Fleq Vert A1Vert_f,quad vert xvert leq 1,; Ainmathcal{L}(mathbb{K}, F),$$
which easily shows that $Vert AVert_{mathcal{L}(mathbb{K},F)} = Vert A1Vert_F$.
In conclusion, you may identify $mathcal{L}(mathbb{K}, F) = F$.
answered Dec 3 '18 at 18:32
RamenRamen
463412
$endgroup$
add a comment |
$begingroup$
Yes, it makes sense in case $E = mathbb{K}$ where $mathbb{K}$ is $mathbb{R}$ or $mathbb{C}$, i.e. what you said makes sense for your first example. The comment of @WillM. is key here, as $h=1$ in Banach spaces only means something if the Banach space is a field too. Note that your equation $T_{x_0}(h) = h(Df_{x_0})$ doesn't make sense in any other case.
More precisely: $mathcal{L}(mathbb{K},F)$ is isometrically isomorphic to $F$ via $varphicolonmathcal{L}(mathbb{K}, F)to F,, Amapsto A1$.
To see this, note that $varphi$ obviously is linear and injective. Next, consider, given $vin F$, the definition $(A_vx:=xcdot v,; xinmathbb{K})$. That way we get $A_vinmathcal{L}(mathbb{K}, F)$ and $A_v1 = v$, i.e. $varphi$ is surjective. Finally,
$$ Vert AxVert_F = vert xvert Vert A1Vert_Fleq Vert A1Vert_f,quad vert xvert leq 1,; Ainmathcal{L}(mathbb{K}, F),$$
which easily shows that $Vert AVert_{mathcal{L}(mathbb{K},F)} = Vert A1Vert_F$.
In conclusion, you may identify $mathcal{L}(mathbb{K}, F) = F$.
answered Dec 3 '18 at 18:32
RamenRamen
463412
$endgroup$
Yes, it makes sense in case $E = mathbb{K}$ where $mathbb{K}$ is $mathbb{R}$ or $mathbb{C}$, i.e. what you said makes sense for your first example. The comment of @WillM. is key here, as $h=1$ in Banach spaces only means something if the Banach space is a field too. Note that your equation $T_{x_0}(h) = h(Df_{x_0})$ doesn't make sense in any other case.
More precisely: $mathcal{L}(mathbb{K},F)$ is isometrically isomorphic to $F$ via $varphicolonmathcal{L}(mathbb{K}, F)to F,, Amapsto A1$.
To see this, note that $varphi$ obviously is linear and injective. Next, consider, given $vin F$, the definition $(A_vx:=xcdot v,; xinmathbb{K})$. That way we get $A_vinmathcal{L}(mathbb{K}, F)$ and $A_v1 = v$, i.e. $varphi$ is surjective. Finally,
$$ Vert AxVert_F = vert xvert Vert A1Vert_Fleq Vert A1Vert_f,quad vert xvert leq 1,; Ainmathcal{L}(mathbb{K}, F),$$
which easily shows that $Vert AVert_{mathcal{L}(mathbb{K},F)} = Vert A1Vert_F$.
In conclusion, you may identify $mathcal{L}(mathbb{K}, F) = F$.
answered Dec 3 '18 at 18:32
RamenRamen
463412
edited Dec 3 '18 at 18:44
answered Dec 3 '18 at 18:32
RamenRamen
463412
answered Dec 3 '18 at 18:32
RamenRamen
463412
answered Dec 3 '18 at 18:32
RamenRamen
463412
463412
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%2f2936202%2fexplaination-of-the-frechet-derivative%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
$begingroup$
No. What does even $h = 1$ mean in Banach spaces?
$endgroup$
– Will M.
Nov 26 '18 at 4:56