Differentiability, linear operators The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Non-completeness of the space of bounded linear operatorsProb. 8, Sec. 4.5 in Kreyszig's functional analysis book: The inverse of the adjoint operator is the adjoint of the inverse operatorAre invertible linear operators of bounded linear operators also bounded?$X$ be a real normed linear space ; if $mathcal L(X,X)$ is complete then is $X$ also complete?A Question on Bounded linear functionals and operatorsIf $Gin C^2(H,mathfrak L(U,H))$, what's the Fréchet derivative of $Phi(x)(u,v):=(rm DG(x)(G(x)u))v$?Any examples of unbounded linear operators between $ell^infty$ and $ell^infty$? $ell^p$ and $ell^p$? $ell^p$ and $ell^infty$?Any Examples Of Unbounded Linear Maps Between Normed Spaces Apart From The Differentiation Operator?Differentiability of $a:Kto mathcalB(E,F)$?Does the linear bounded operator A = √B exist?
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Differentiability, linear operators
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Non-completeness of the space of bounded linear operatorsProb. 8, Sec. 4.5 in Kreyszig's functional analysis book: The inverse of the adjoint operator is the adjoint of the inverse operatorAre invertible linear operators of bounded linear operators also bounded?$X$ be a real normed linear space ; if $mathcal L(X,X)$ is complete then is $X$ also complete?A Question on Bounded linear functionals and operatorsIf $Gin C^2(H,mathfrak L(U,H))$, what's the Fréchet derivative of $Phi(x)(u,v):=(rm DG(x)(G(x)u))v$?Any examples of unbounded linear operators between $ell^infty$ and $ell^infty$? $ell^p$ and $ell^p$? $ell^p$ and $ell^infty$?Any Examples Of Unbounded Linear Maps Between Normed Spaces Apart From The Differentiation Operator?Differentiability of $a:Kto mathcalB(E,F)$?Does the linear bounded operator A = √B exist?
$begingroup$
Let $Y$ be a complete normed linear space, and let $M$ denote the
space of bounded linear operators from $Y$ to itself. Let $L : M → M$ be the map
defined by
$L(A) := A^2$.
I am supposed to show that L is differentiable at each $Ain M$, and then find the derivatives.
Tried to find the derivatives like this:
$L'(A) = lim_h to 0 fracL(A+h)-L(A)h= lim_h to 0 frac(A+h)^2-A^2h= lim_h to 0 fracA^2+2Ah-h^2-A^2h = lim_h to 0 2A+h = underline2A$
If this is correct, all I have to do now is to show that $L$ is differentiable at all $Ain M$, and I'm wondering if I can do that by saying that $L$ is a composition of $A$ and $A$ since $A$ is a linear operator, and all linear operators are differentiable?
real-analysis derivatives normed-spaces frechet-derivative
edited Apr 8 at 17:58
José Carlos Santos
174k23134243
asked Apr 8 at 12:41
Samsam22Samsam22
224
New contributor
Samsam22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let $Y$ be a complete normed linear space, and let $M$ denote the
space of bounded linear operators from $Y$ to itself. Let $L : M → M$ be the map
defined by
$L(A) := A^2$.
I am supposed to show that L is differentiable at each $Ain M$, and then find the derivatives.
Tried to find the derivatives like this:
$L'(A) = lim_h to 0 fracL(A+h)-L(A)h= lim_h to 0 frac(A+h)^2-A^2h= lim_h to 0 fracA^2+2Ah-h^2-A^2h = lim_h to 0 2A+h = underline2A$
If this is correct, all I have to do now is to show that $L$ is differentiable at all $Ain M$, and I'm wondering if I can do that by saying that $L$ is a composition of $A$ and $A$ since $A$ is a linear operator, and all linear operators are differentiable?
real-analysis derivatives normed-spaces frechet-derivative
edited Apr 8 at 17:58
José Carlos Santos
174k23134243
asked Apr 8 at 12:41
Samsam22Samsam22
224
New contributor
Samsam22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
The way you write things is strange... first, what is the meaning of $fracL(A+h)-L(A)h$ in $M$ ?
$endgroup$
– user657324
Apr 8 at 12:48
add a comment |
$begingroup$
Let $Y$ be a complete normed linear space, and let $M$ denote the
space of bounded linear operators from $Y$ to itself. Let $L : M → M$ be the map
defined by
$L(A) := A^2$.
I am supposed to show that L is differentiable at each $Ain M$, and then find the derivatives.
Tried to find the derivatives like this:
$L'(A) = lim_h to 0 fracL(A+h)-L(A)h= lim_h to 0 frac(A+h)^2-A^2h= lim_h to 0 fracA^2+2Ah-h^2-A^2h = lim_h to 0 2A+h = underline2A$
If this is correct, all I have to do now is to show that $L$ is differentiable at all $Ain M$, and I'm wondering if I can do that by saying that $L$ is a composition of $A$ and $A$ since $A$ is a linear operator, and all linear operators are differentiable?
real-analysis derivatives normed-spaces frechet-derivative
edited Apr 8 at 17:58
José Carlos Santos
174k23134243
asked Apr 8 at 12:41
Samsam22Samsam22
224
New contributor
Samsam22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $Y$ be a complete normed linear space, and let $M$ denote the
space of bounded linear operators from $Y$ to itself. Let $L : M → M$ be the map
defined by
$L(A) := A^2$.
I am supposed to show that L is differentiable at each $Ain M$, and then find the derivatives.
Tried to find the derivatives like this:
$L'(A) = lim_h to 0 fracL(A+h)-L(A)h= lim_h to 0 frac(A+h)^2-A^2h= lim_h to 0 fracA^2+2Ah-h^2-A^2h = lim_h to 0 2A+h = underline2A$
If this is correct, all I have to do now is to show that $L$ is differentiable at all $Ain M$, and I'm wondering if I can do that by saying that $L$ is a composition of $A$ and $A$ since $A$ is a linear operator, and all linear operators are differentiable?
real-analysis derivatives normed-spaces frechet-derivative
real-analysis derivatives normed-spaces frechet-derivative
edited Apr 8 at 17:58
José Carlos Santos
174k23134243
asked Apr 8 at 12:41
Samsam22Samsam22
224
New contributor
Samsam22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 8 at 17:58
José Carlos Santos
174k23134243
asked Apr 8 at 12:41
Samsam22Samsam22
224
New contributor
Samsam22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 8 at 17:58
José Carlos Santos
174k23134243
edited Apr 8 at 17:58
José Carlos Santos
174k23134243
edited Apr 8 at 17:58
José Carlos Santos
174k23134243
174k23134243
asked Apr 8 at 12:41
Samsam22Samsam22
224
New contributor
Samsam22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 12:41
Samsam22Samsam22
224
asked Apr 8 at 12:41
Samsam22Samsam22
224
224
New contributor
Samsam22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Samsam22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Samsam22 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
The way you write things is strange... first, what is the meaning of $fracL(A+h)-L(A)h$ in $M$ ?
$endgroup$
– user657324
Apr 8 at 12:48
add a comment |
$begingroup$
The way you write things is strange... first, what is the meaning of $fracL(A+h)-L(A)h$ in $M$ ?
$endgroup$
– user657324
Apr 8 at 12:48
$begingroup$
The way you write things is strange... first, what is the meaning of $fracL(A+h)-L(A)h$ in $M$ ?
$endgroup$
– user657324
Apr 8 at 12:48
$begingroup$
The way you write things is strange... first, what is the meaning of $fracL(A+h)-L(A)h$ in $M$ ?
$endgroup$
– user657324
Apr 8 at 12:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The derivative of $L$ at an operator $A$ is a linear map. In this case, it is the map $varphi$ defined by $varphi(M)=AM+MA$. In factbeginalignlim_Mto AfracbigllVert L(M)-L(A)-varphi(M-A)bigrrVertlVert M-ArVert&=lim_Mto AfracbigllVert M^2-A^2-A(M-A)-(M-A)AbigrrVertlVert M-ArVert\&=lim_Mto AfraclVert M^2-AM-MA+A^2rVertlVert M-ArVert\&=lim_Mto AfracbigllVert(M-A)^2bigrrVertlVert M-ArVert\&=0.endalign
answered Apr 8 at 12:51
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
$begingroup$
Can we use the notation $lim_Mto A$ whereas $M$ and $A$ are operator ?
$endgroup$
– user657324
Apr 8 at 12:53
$begingroup$
We can use it in any metric space.
$endgroup$
– José Carlos Santos
Apr 8 at 12:56
$begingroup$
No, you can't write it as $2AM$ because, in general, linear operators do not commute. And I guessed that the derivative is $varphi$ by analysing $M^2-A^2$ and trying to determine what I should subtract from it so that$$lim_Mto AfracbigllVert M^2-A^2-varphi(M-A)bigrrVertlVert M-ArVert=0.$$
$endgroup$
– José Carlos Santos
Apr 8 at 13:14
$begingroup$
I got $AM+MA$ by doing this: $lim_alpha to 0 fracL(A+alpha M)-L(A)alpha = lim_alpha to 0 frac(A+alpha M)^2-A^2alpha = lim_alpha to 0 frac(A^2+Aalpha M + alpha M A + alpha ^2 M^2 - A^2alpha = lim_alpha to 0 AM+MA+alpha M^2 = AM+MA$. But I'm not sure if using the $alpha$ makes any sense
$endgroup$
– Samsam22
Apr 8 at 13:25
1
$begingroup$
beginalignlVert AM+MArVert&leqslantlVert AMrVert+lVert MArVert\&leqslantlVert ArVert.lVert MrVert+lVert ArVert.lVert MrVert\&=2lVert ArVert.lVert MrVert.endalign
$endgroup$
– José Carlos Santos
2 days ago
|
show 2 more comments
$begingroup$
In order to discover (and at the same time prove that $D_A(H):=AH+HA$ is the differential of function $L$, which is different from the derivative), just expand
$$L(A+H)=(A+H)^2=underbraceA^2_L(A)+underbrace(AH+HA)_D_A(H)+underbraceH^2_2nd order$$
and collect all the first degree terms.
Do you see the connection with Taylor expansion :
$$f(a+h)=f(a)+underbracef'(a).h_differential+textterms in h^2, h^3, textetc. ?$$
answered Apr 8 at 13:16
Jean MarieJean Marie
31.5k42355
$endgroup$
$begingroup$
We can proceed like this as long as we are in a Banach algebra which is the largest "reasonable" algebraic-topological framework in which differential calculus can be defined.
$endgroup$
– Jean Marie
Apr 9 at 23:51
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The derivative of $L$ at an operator $A$ is a linear map. In this case, it is the map $varphi$ defined by $varphi(M)=AM+MA$. In factbeginalignlim_Mto AfracbigllVert L(M)-L(A)-varphi(M-A)bigrrVertlVert M-ArVert&=lim_Mto AfracbigllVert M^2-A^2-A(M-A)-(M-A)AbigrrVertlVert M-ArVert\&=lim_Mto AfraclVert M^2-AM-MA+A^2rVertlVert M-ArVert\&=lim_Mto AfracbigllVert(M-A)^2bigrrVertlVert M-ArVert\&=0.endalign
answered Apr 8 at 12:51
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
$begingroup$
Can we use the notation $lim_Mto A$ whereas $M$ and $A$ are operator ?
$endgroup$
– user657324
Apr 8 at 12:53
$begingroup$
We can use it in any metric space.
$endgroup$
– José Carlos Santos
Apr 8 at 12:56
$begingroup$
No, you can't write it as $2AM$ because, in general, linear operators do not commute. And I guessed that the derivative is $varphi$ by analysing $M^2-A^2$ and trying to determine what I should subtract from it so that$$lim_Mto AfracbigllVert M^2-A^2-varphi(M-A)bigrrVertlVert M-ArVert=0.$$
$endgroup$
– José Carlos Santos
Apr 8 at 13:14
$begingroup$
I got $AM+MA$ by doing this: $lim_alpha to 0 fracL(A+alpha M)-L(A)alpha = lim_alpha to 0 frac(A+alpha M)^2-A^2alpha = lim_alpha to 0 frac(A^2+Aalpha M + alpha M A + alpha ^2 M^2 - A^2alpha = lim_alpha to 0 AM+MA+alpha M^2 = AM+MA$. But I'm not sure if using the $alpha$ makes any sense
$endgroup$
– Samsam22
Apr 8 at 13:25
1
$begingroup$
beginalignlVert AM+MArVert&leqslantlVert AMrVert+lVert MArVert\&leqslantlVert ArVert.lVert MrVert+lVert ArVert.lVert MrVert\&=2lVert ArVert.lVert MrVert.endalign
$endgroup$
– José Carlos Santos
2 days ago
|
show 2 more comments
$begingroup$
The derivative of $L$ at an operator $A$ is a linear map. In this case, it is the map $varphi$ defined by $varphi(M)=AM+MA$. In factbeginalignlim_Mto AfracbigllVert L(M)-L(A)-varphi(M-A)bigrrVertlVert M-ArVert&=lim_Mto AfracbigllVert M^2-A^2-A(M-A)-(M-A)AbigrrVertlVert M-ArVert\&=lim_Mto AfraclVert M^2-AM-MA+A^2rVertlVert M-ArVert\&=lim_Mto AfracbigllVert(M-A)^2bigrrVertlVert M-ArVert\&=0.endalign
answered Apr 8 at 12:51
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
$begingroup$
Can we use the notation $lim_Mto A$ whereas $M$ and $A$ are operator ?
$endgroup$
– user657324
Apr 8 at 12:53
$begingroup$
We can use it in any metric space.
$endgroup$
– José Carlos Santos
Apr 8 at 12:56
$begingroup$
No, you can't write it as $2AM$ because, in general, linear operators do not commute. And I guessed that the derivative is $varphi$ by analysing $M^2-A^2$ and trying to determine what I should subtract from it so that$$lim_Mto AfracbigllVert M^2-A^2-varphi(M-A)bigrrVertlVert M-ArVert=0.$$
$endgroup$
– José Carlos Santos
Apr 8 at 13:14
$begingroup$
I got $AM+MA$ by doing this: $lim_alpha to 0 fracL(A+alpha M)-L(A)alpha = lim_alpha to 0 frac(A+alpha M)^2-A^2alpha = lim_alpha to 0 frac(A^2+Aalpha M + alpha M A + alpha ^2 M^2 - A^2alpha = lim_alpha to 0 AM+MA+alpha M^2 = AM+MA$. But I'm not sure if using the $alpha$ makes any sense
$endgroup$
– Samsam22
Apr 8 at 13:25
1
$begingroup$
beginalignlVert AM+MArVert&leqslantlVert AMrVert+lVert MArVert\&leqslantlVert ArVert.lVert MrVert+lVert ArVert.lVert MrVert\&=2lVert ArVert.lVert MrVert.endalign
$endgroup$
– José Carlos Santos
2 days ago
|
show 2 more comments
$begingroup$
The derivative of $L$ at an operator $A$ is a linear map. In this case, it is the map $varphi$ defined by $varphi(M)=AM+MA$. In factbeginalignlim_Mto AfracbigllVert L(M)-L(A)-varphi(M-A)bigrrVertlVert M-ArVert&=lim_Mto AfracbigllVert M^2-A^2-A(M-A)-(M-A)AbigrrVertlVert M-ArVert\&=lim_Mto AfraclVert M^2-AM-MA+A^2rVertlVert M-ArVert\&=lim_Mto AfracbigllVert(M-A)^2bigrrVertlVert M-ArVert\&=0.endalign
answered Apr 8 at 12:51
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
The derivative of $L$ at an operator $A$ is a linear map. In this case, it is the map $varphi$ defined by $varphi(M)=AM+MA$. In factbeginalignlim_Mto AfracbigllVert L(M)-L(A)-varphi(M-A)bigrrVertlVert M-ArVert&=lim_Mto AfracbigllVert M^2-A^2-A(M-A)-(M-A)AbigrrVertlVert M-ArVert\&=lim_Mto AfraclVert M^2-AM-MA+A^2rVertlVert M-ArVert\&=lim_Mto AfracbigllVert(M-A)^2bigrrVertlVert M-ArVert\&=0.endalign
answered Apr 8 at 12:51
José Carlos SantosJosé Carlos Santos
174k23134243
answered Apr 8 at 12:51
José Carlos SantosJosé Carlos Santos
174k23134243
answered Apr 8 at 12:51
José Carlos SantosJosé Carlos Santos
174k23134243
answered Apr 8 at 12:51
José Carlos SantosJosé Carlos Santos
174k23134243
174k23134243
$begingroup$
Can we use the notation $lim_Mto A$ whereas $M$ and $A$ are operator ?
$endgroup$
– user657324
Apr 8 at 12:53
$begingroup$
We can use it in any metric space.
$endgroup$
– José Carlos Santos
Apr 8 at 12:56
$begingroup$
No, you can't write it as $2AM$ because, in general, linear operators do not commute. And I guessed that the derivative is $varphi$ by analysing $M^2-A^2$ and trying to determine what I should subtract from it so that$$lim_Mto AfracbigllVert M^2-A^2-varphi(M-A)bigrrVertlVert M-ArVert=0.$$
$endgroup$
– José Carlos Santos
Apr 8 at 13:14
$begingroup$
I got $AM+MA$ by doing this: $lim_alpha to 0 fracL(A+alpha M)-L(A)alpha = lim_alpha to 0 frac(A+alpha M)^2-A^2alpha = lim_alpha to 0 frac(A^2+Aalpha M + alpha M A + alpha ^2 M^2 - A^2alpha = lim_alpha to 0 AM+MA+alpha M^2 = AM+MA$. But I'm not sure if using the $alpha$ makes any sense
$endgroup$
– Samsam22
Apr 8 at 13:25
1
$begingroup$
beginalignlVert AM+MArVert&leqslantlVert AMrVert+lVert MArVert\&leqslantlVert ArVert.lVert MrVert+lVert ArVert.lVert MrVert\&=2lVert ArVert.lVert MrVert.endalign
$endgroup$
– José Carlos Santos
2 days ago
|
show 2 more comments
$begingroup$
Can we use the notation $lim_Mto A$ whereas $M$ and $A$ are operator ?
$endgroup$
– user657324
Apr 8 at 12:53
$begingroup$
We can use it in any metric space.
$endgroup$
– José Carlos Santos
Apr 8 at 12:56
$begingroup$
No, you can't write it as $2AM$ because, in general, linear operators do not commute. And I guessed that the derivative is $varphi$ by analysing $M^2-A^2$ and trying to determine what I should subtract from it so that$$lim_Mto AfracbigllVert M^2-A^2-varphi(M-A)bigrrVertlVert M-ArVert=0.$$
$endgroup$
– José Carlos Santos
Apr 8 at 13:14
$begingroup$
I got $AM+MA$ by doing this: $lim_alpha to 0 fracL(A+alpha M)-L(A)alpha = lim_alpha to 0 frac(A+alpha M)^2-A^2alpha = lim_alpha to 0 frac(A^2+Aalpha M + alpha M A + alpha ^2 M^2 - A^2alpha = lim_alpha to 0 AM+MA+alpha M^2 = AM+MA$. But I'm not sure if using the $alpha$ makes any sense
$endgroup$
– Samsam22
Apr 8 at 13:25
1
$begingroup$
beginalignlVert AM+MArVert&leqslantlVert AMrVert+lVert MArVert\&leqslantlVert ArVert.lVert MrVert+lVert ArVert.lVert MrVert\&=2lVert ArVert.lVert MrVert.endalign
$endgroup$
– José Carlos Santos
2 days ago
$begingroup$
Can we use the notation $lim_Mto A$ whereas $M$ and $A$ are operator ?
$endgroup$
– user657324
Apr 8 at 12:53
$begingroup$
Can we use the notation $lim_Mto A$ whereas $M$ and $A$ are operator ?
$endgroup$
– user657324
Apr 8 at 12:53
$begingroup$
We can use it in any metric space.
$endgroup$
– José Carlos Santos
Apr 8 at 12:56
$begingroup$
We can use it in any metric space.
$endgroup$
– José Carlos Santos
Apr 8 at 12:56
$begingroup$
No, you can't write it as $2AM$ because, in general, linear operators do not commute. And I guessed that the derivative is $varphi$ by analysing $M^2-A^2$ and trying to determine what I should subtract from it so that$$lim_Mto AfracbigllVert M^2-A^2-varphi(M-A)bigrrVertlVert M-ArVert=0.$$
$endgroup$
– José Carlos Santos
Apr 8 at 13:14
$begingroup$
No, you can't write it as $2AM$ because, in general, linear operators do not commute. And I guessed that the derivative is $varphi$ by analysing $M^2-A^2$ and trying to determine what I should subtract from it so that$$lim_Mto AfracbigllVert M^2-A^2-varphi(M-A)bigrrVertlVert M-ArVert=0.$$
$endgroup$
– José Carlos Santos
Apr 8 at 13:14
$begingroup$
I got $AM+MA$ by doing this: $lim_alpha to 0 fracL(A+alpha M)-L(A)alpha = lim_alpha to 0 frac(A+alpha M)^2-A^2alpha = lim_alpha to 0 frac(A^2+Aalpha M + alpha M A + alpha ^2 M^2 - A^2alpha = lim_alpha to 0 AM+MA+alpha M^2 = AM+MA$. But I'm not sure if using the $alpha$ makes any sense
$endgroup$
– Samsam22
Apr 8 at 13:25
$begingroup$
I got $AM+MA$ by doing this: $lim_alpha to 0 fracL(A+alpha M)-L(A)alpha = lim_alpha to 0 frac(A+alpha M)^2-A^2alpha = lim_alpha to 0 frac(A^2+Aalpha M + alpha M A + alpha ^2 M^2 - A^2alpha = lim_alpha to 0 AM+MA+alpha M^2 = AM+MA$. But I'm not sure if using the $alpha$ makes any sense
$endgroup$
– Samsam22
Apr 8 at 13:25
1
1
$begingroup$
beginalignlVert AM+MArVert&leqslantlVert AMrVert+lVert MArVert\&leqslantlVert ArVert.lVert MrVert+lVert ArVert.lVert MrVert\&=2lVert ArVert.lVert MrVert.endalign
$endgroup$
– José Carlos Santos
2 days ago
$begingroup$
beginalignlVert AM+MArVert&leqslantlVert AMrVert+lVert MArVert\&leqslantlVert ArVert.lVert MrVert+lVert ArVert.lVert MrVert\&=2lVert ArVert.lVert MrVert.endalign
$endgroup$
– José Carlos Santos
2 days ago
|
show 2 more comments
$begingroup$
In order to discover (and at the same time prove that $D_A(H):=AH+HA$ is the differential of function $L$, which is different from the derivative), just expand
$$L(A+H)=(A+H)^2=underbraceA^2_L(A)+underbrace(AH+HA)_D_A(H)+underbraceH^2_2nd order$$
and collect all the first degree terms.
Do you see the connection with Taylor expansion :
$$f(a+h)=f(a)+underbracef'(a).h_differential+textterms in h^2, h^3, textetc. ?$$
answered Apr 8 at 13:16
Jean MarieJean Marie
31.5k42355
$endgroup$
$begingroup$
We can proceed like this as long as we are in a Banach algebra which is the largest "reasonable" algebraic-topological framework in which differential calculus can be defined.
$endgroup$
– Jean Marie
Apr 9 at 23:51
add a comment |
$begingroup$
In order to discover (and at the same time prove that $D_A(H):=AH+HA$ is the differential of function $L$, which is different from the derivative), just expand
$$L(A+H)=(A+H)^2=underbraceA^2_L(A)+underbrace(AH+HA)_D_A(H)+underbraceH^2_2nd order$$
and collect all the first degree terms.
Do you see the connection with Taylor expansion :
$$f(a+h)=f(a)+underbracef'(a).h_differential+textterms in h^2, h^3, textetc. ?$$
answered Apr 8 at 13:16
Jean MarieJean Marie
31.5k42355
$endgroup$
$begingroup$
We can proceed like this as long as we are in a Banach algebra which is the largest "reasonable" algebraic-topological framework in which differential calculus can be defined.
$endgroup$
– Jean Marie
Apr 9 at 23:51
add a comment |
$begingroup$
In order to discover (and at the same time prove that $D_A(H):=AH+HA$ is the differential of function $L$, which is different from the derivative), just expand
$$L(A+H)=(A+H)^2=underbraceA^2_L(A)+underbrace(AH+HA)_D_A(H)+underbraceH^2_2nd order$$
and collect all the first degree terms.
Do you see the connection with Taylor expansion :
$$f(a+h)=f(a)+underbracef'(a).h_differential+textterms in h^2, h^3, textetc. ?$$
answered Apr 8 at 13:16
Jean MarieJean Marie
31.5k42355
$endgroup$
In order to discover (and at the same time prove that $D_A(H):=AH+HA$ is the differential of function $L$, which is different from the derivative), just expand
$$L(A+H)=(A+H)^2=underbraceA^2_L(A)+underbrace(AH+HA)_D_A(H)+underbraceH^2_2nd order$$
and collect all the first degree terms.
Do you see the connection with Taylor expansion :
$$f(a+h)=f(a)+underbracef'(a).h_differential+textterms in h^2, h^3, textetc. ?$$
answered Apr 8 at 13:16
Jean MarieJean Marie
31.5k42355
answered Apr 8 at 13:16
Jean MarieJean Marie
31.5k42355
answered Apr 8 at 13:16
Jean MarieJean Marie
31.5k42355
answered Apr 8 at 13:16
Jean MarieJean Marie
31.5k42355
31.5k42355
$begingroup$
We can proceed like this as long as we are in a Banach algebra which is the largest "reasonable" algebraic-topological framework in which differential calculus can be defined.
$endgroup$
– Jean Marie
Apr 9 at 23:51
add a comment |
$begingroup$
We can proceed like this as long as we are in a Banach algebra which is the largest "reasonable" algebraic-topological framework in which differential calculus can be defined.
$endgroup$
– Jean Marie
Apr 9 at 23:51
$begingroup$
We can proceed like this as long as we are in a Banach algebra which is the largest "reasonable" algebraic-topological framework in which differential calculus can be defined.
$endgroup$
– Jean Marie
Apr 9 at 23:51
$begingroup$
We can proceed like this as long as we are in a Banach algebra which is the largest "reasonable" algebraic-topological framework in which differential calculus can be defined.
$endgroup$
– Jean Marie
Apr 9 at 23:51
add a comment |
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$begingroup$
The way you write things is strange... first, what is the meaning of $fracL(A+h)-L(A)h$ in $M$ ?
$endgroup$
– user657324
Apr 8 at 12:48