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Operator norm of differential
Clash Royale CLAN TAG#URR8PPP
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Definition: $|D|_mathrmop=sup_xinmathbbR^n/0 frac Dx x$
Where $|cdot|$ is the Euclidean norm.
We have sub-multiplicativity, $|AB|leq|A|cdot|B|$.
Now say I have some $f:mathbbR^m to mathbbR^n$ differentiable and every partial derivative is bounded. So I have $|Dx|<M $ for all $xinmathbbR^m$.
Now I`m interested in showing $|D|_mathrmop$ exists , and moreover, bound it with $M$.
I have no idea how to approach this,since I don`t know much about operators and operators norms.
I'd appriciate it if anyone can provide an elementary proof.
the motivation behind the question is to easily show that bounded partials implies local lipchitz:
beginalign
& |f(x+h)-f(x)|=|D_f(x)+o(|h|)|leq|D_f(x)h|+|o(|h|)| \[10pt]
leq & |D_f(x)|_mathrmop |h|+||o(|h|)leq hat M|h| +o(|h|)to 0
endalign
P.S : I wasn`t sure under what tags my question belongs, so I've entered the main topic I'm studying right now.
calculus real-analysis multivariable-calculus
edited 15 hours ago
Michael Hardy
204k23185460
asked 17 hours ago
Sar
3529
add a comment |Â
up vote
0
down vote
favorite
Definition: $|D|_mathrmop=sup_xinmathbbR^n/0 frac Dx x$
Where $|cdot|$ is the Euclidean norm.
We have sub-multiplicativity, $|AB|leq|A|cdot|B|$.
Now say I have some $f:mathbbR^m to mathbbR^n$ differentiable and every partial derivative is bounded. So I have $|Dx|<M $ for all $xinmathbbR^m$.
Now I`m interested in showing $|D|_mathrmop$ exists , and moreover, bound it with $M$.
I have no idea how to approach this,since I don`t know much about operators and operators norms.
I'd appriciate it if anyone can provide an elementary proof.
the motivation behind the question is to easily show that bounded partials implies local lipchitz:
beginalign
& |f(x+h)-f(x)|=|D_f(x)+o(|h|)|leq|D_f(x)h|+|o(|h|)| \[10pt]
leq & |D_f(x)|_mathrmop |h|+||o(|h|)leq hat M|h| +o(|h|)to 0
endalign
P.S : I wasn`t sure under what tags my question belongs, so I've entered the main topic I'm studying right now.
calculus real-analysis multivariable-calculus
edited 15 hours ago
Michael Hardy
204k23185460
asked 17 hours ago
Sar
3529
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Definition: $|D|_mathrmop=sup_xinmathbbR^n/0 frac Dx x$
Where $|cdot|$ is the Euclidean norm.
We have sub-multiplicativity, $|AB|leq|A|cdot|B|$.
Now say I have some $f:mathbbR^m to mathbbR^n$ differentiable and every partial derivative is bounded. So I have $|Dx|<M $ for all $xinmathbbR^m$.
Now I`m interested in showing $|D|_mathrmop$ exists , and moreover, bound it with $M$.
I have no idea how to approach this,since I don`t know much about operators and operators norms.
I'd appriciate it if anyone can provide an elementary proof.
the motivation behind the question is to easily show that bounded partials implies local lipchitz:
beginalign
& |f(x+h)-f(x)|=|D_f(x)+o(|h|)|leq|D_f(x)h|+|o(|h|)| \[10pt]
leq & |D_f(x)|_mathrmop |h|+||o(|h|)leq hat M|h| +o(|h|)to 0
endalign
P.S : I wasn`t sure under what tags my question belongs, so I've entered the main topic I'm studying right now.
calculus real-analysis multivariable-calculus
edited 15 hours ago
Michael Hardy
204k23185460
asked 17 hours ago
Sar
3529
Definition: $|D|_mathrmop=sup_xinmathbbR^n/0 frac Dx x$
Where $|cdot|$ is the Euclidean norm.
We have sub-multiplicativity, $|AB|leq|A|cdot|B|$.
Now say I have some $f:mathbbR^m to mathbbR^n$ differentiable and every partial derivative is bounded. So I have $|Dx|<M $ for all $xinmathbbR^m$.
Now I`m interested in showing $|D|_mathrmop$ exists , and moreover, bound it with $M$.
I have no idea how to approach this,since I don`t know much about operators and operators norms.
I'd appriciate it if anyone can provide an elementary proof.
the motivation behind the question is to easily show that bounded partials implies local lipchitz:
beginalign
& |f(x+h)-f(x)|=|D_f(x)+o(|h|)|leq|D_f(x)h|+|o(|h|)| \[10pt]
leq & |D_f(x)|_mathrmop |h|+||o(|h|)leq hat M|h| +o(|h|)to 0
endalign
P.S : I wasn`t sure under what tags my question belongs, so I've entered the main topic I'm studying right now.
calculus real-analysis multivariable-calculus
edited 15 hours ago
Michael Hardy
204k23185460
asked 17 hours ago
Sar
3529
edited 15 hours ago
Michael Hardy
204k23185460
edited 15 hours ago
Michael Hardy
204k23185460
edited 15 hours ago
Michael Hardy
204k23185460
204k23185460
asked 17 hours ago
Sar
3529
asked 17 hours ago
Sar
3529
asked 17 hours ago
Sar
3529
3529
add a comment |Â
add a comment |Â
1 Answer
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There is not much to prove here. Your troubles stem from the unprecise notation. I'm using $|x|$ for the euclidean norm of vectors $xin mathbb R^n$, whatever $ngeq1$. For a constant linear map $A:>mathbb R^ntomathbb R^m$ one then defines the operator norm
$$|A|:=sup_xne0 .$$
Now, given a differentiable function $f:>mathbb R^ntomathbb R^m$, this $f$ has at every single point $xinmathbb R^n$ a derivative $$df(x):quadmathbb R^ntomathbb R^m>,$$
which is an element of $cal L(mathbb R^n,mathbb R^m)$, and is characterized by
$$f(x+X)-f(x)=df(x).X+obigl(|X|bigr)qquad(Xto0) .$$
The matrix of this $df(x)$ is the matrix of partial derivatives $f_i.k(x):=partial f_ioverpartial x_k(x)$. Therefore the operator norm $|df(x)|$ can be estimated in terms of these partial derivatives. A simple estimate is
$$|df(x)|leqsqrtf_i.k(x) .$$
answered 13 hours ago
Christian Blatter
163k7105305
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
There is not much to prove here. Your troubles stem from the unprecise notation. I'm using $|x|$ for the euclidean norm of vectors $xin mathbb R^n$, whatever $ngeq1$. For a constant linear map $A:>mathbb R^ntomathbb R^m$ one then defines the operator norm
$$|A|:=sup_xne0 .$$
Now, given a differentiable function $f:>mathbb R^ntomathbb R^m$, this $f$ has at every single point $xinmathbb R^n$ a derivative $$df(x):quadmathbb R^ntomathbb R^m>,$$
which is an element of $cal L(mathbb R^n,mathbb R^m)$, and is characterized by
$$f(x+X)-f(x)=df(x).X+obigl(|X|bigr)qquad(Xto0) .$$
The matrix of this $df(x)$ is the matrix of partial derivatives $f_i.k(x):=partial f_ioverpartial x_k(x)$. Therefore the operator norm $|df(x)|$ can be estimated in terms of these partial derivatives. A simple estimate is
$$|df(x)|leqsqrtf_i.k(x) .$$
answered 13 hours ago
Christian Blatter
163k7105305
add a comment |Â
up vote
0
down vote
There is not much to prove here. Your troubles stem from the unprecise notation. I'm using $|x|$ for the euclidean norm of vectors $xin mathbb R^n$, whatever $ngeq1$. For a constant linear map $A:>mathbb R^ntomathbb R^m$ one then defines the operator norm
$$|A|:=sup_xne0 .$$
Now, given a differentiable function $f:>mathbb R^ntomathbb R^m$, this $f$ has at every single point $xinmathbb R^n$ a derivative $$df(x):quadmathbb R^ntomathbb R^m>,$$
which is an element of $cal L(mathbb R^n,mathbb R^m)$, and is characterized by
$$f(x+X)-f(x)=df(x).X+obigl(|X|bigr)qquad(Xto0) .$$
The matrix of this $df(x)$ is the matrix of partial derivatives $f_i.k(x):=partial f_ioverpartial x_k(x)$. Therefore the operator norm $|df(x)|$ can be estimated in terms of these partial derivatives. A simple estimate is
$$|df(x)|leqsqrtf_i.k(x) .$$
answered 13 hours ago
Christian Blatter
163k7105305
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There is not much to prove here. Your troubles stem from the unprecise notation. I'm using $|x|$ for the euclidean norm of vectors $xin mathbb R^n$, whatever $ngeq1$. For a constant linear map $A:>mathbb R^ntomathbb R^m$ one then defines the operator norm
$$|A|:=sup_xne0 .$$
Now, given a differentiable function $f:>mathbb R^ntomathbb R^m$, this $f$ has at every single point $xinmathbb R^n$ a derivative $$df(x):quadmathbb R^ntomathbb R^m>,$$
which is an element of $cal L(mathbb R^n,mathbb R^m)$, and is characterized by
$$f(x+X)-f(x)=df(x).X+obigl(|X|bigr)qquad(Xto0) .$$
The matrix of this $df(x)$ is the matrix of partial derivatives $f_i.k(x):=partial f_ioverpartial x_k(x)$. Therefore the operator norm $|df(x)|$ can be estimated in terms of these partial derivatives. A simple estimate is
$$|df(x)|leqsqrtf_i.k(x) .$$
answered 13 hours ago
Christian Blatter
163k7105305
There is not much to prove here. Your troubles stem from the unprecise notation. I'm using $|x|$ for the euclidean norm of vectors $xin mathbb R^n$, whatever $ngeq1$. For a constant linear map $A:>mathbb R^ntomathbb R^m$ one then defines the operator norm
$$|A|:=sup_xne0 .$$
Now, given a differentiable function $f:>mathbb R^ntomathbb R^m$, this $f$ has at every single point $xinmathbb R^n$ a derivative $$df(x):quadmathbb R^ntomathbb R^m>,$$
which is an element of $cal L(mathbb R^n,mathbb R^m)$, and is characterized by
$$f(x+X)-f(x)=df(x).X+obigl(|X|bigr)qquad(Xto0) .$$
The matrix of this $df(x)$ is the matrix of partial derivatives $f_i.k(x):=partial f_ioverpartial x_k(x)$. Therefore the operator norm $|df(x)|$ can be estimated in terms of these partial derivatives. A simple estimate is
$$|df(x)|leqsqrtf_i.k(x) .$$
answered 13 hours ago
Christian Blatter
163k7105305
answered 13 hours ago
Christian Blatter
163k7105305
answered 13 hours ago
Christian Blatter
163k7105305
answered 13 hours ago
Christian Blatter
163k7105305
163k7105305
add a comment |Â
add a comment |Â
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