Fréchet derivative: dependency of the choice of norm
Let $ f_1: L^2(Omega)rightarrowmathbb{R}, umapsto Vert uVert_{L^2 (Omega)}^2 $, $ f_2: H^1(Omega)rightarrowmathbb{R}, umapsto Vert uVert_{L^2 (Omega)}^2 $ and $f_2: H^1(Omega)rightarrowmathbb{R}, umapsto Vert uVert_{H^1 (Omega)}^2$.
Now it is possible to calculate the Gateaux-Derivatives:
$f_1'(u)h=(2u,h)_{L^2(Omega)}$- $f_2'(u)h=(2u,h)_{L^2(Omega)}$
- $f_3'(u)h=(2u,h)_{H^1(Omega)}$
Is the following statement true?
$f_2$ is not Fréchet differentiable with respect to $Vert cdot Vert _{H^1 (Omega)}$.
My current calculations are:
$frac{{vert f_2(u+h)-f_2(u)-(2u,h)_{L^2} vert}}{{Vert hVert_{H^1}}}=frac{{vert (h,h)_{L^2} vert}}{{Vert hVert_{H^1}}}$
functional-analysis normed-spaces frechet-derivative
asked Nov 27 '18 at 16:24
sandmath
112
add a comment |
Let $ f_1: L^2(Omega)rightarrowmathbb{R}, umapsto Vert uVert_{L^2 (Omega)}^2 $, $ f_2: H^1(Omega)rightarrowmathbb{R}, umapsto Vert uVert_{L^2 (Omega)}^2 $ and $f_2: H^1(Omega)rightarrowmathbb{R}, umapsto Vert uVert_{H^1 (Omega)}^2$.
Now it is possible to calculate the Gateaux-Derivatives:
$f_1'(u)h=(2u,h)_{L^2(Omega)}$- $f_2'(u)h=(2u,h)_{L^2(Omega)}$
- $f_3'(u)h=(2u,h)_{H^1(Omega)}$
Is the following statement true?
$f_2$ is not Fréchet differentiable with respect to $Vert cdot Vert _{H^1 (Omega)}$.
My current calculations are:
$frac{{vert f_2(u+h)-f_2(u)-(2u,h)_{L^2} vert}}{{Vert hVert_{H^1}}}=frac{{vert (h,h)_{L^2} vert}}{{Vert hVert_{H^1}}}$
functional-analysis normed-spaces frechet-derivative
asked Nov 27 '18 at 16:24
sandmath
112
All these $f$'s should be squares of norms. Then all functions are Frechet.
– daw
Nov 27 '18 at 17:52
To calculate, expand the squares.
– daw
Nov 27 '18 at 17:53
Of course i have to take the squared norm. Sorry, that was a typing mistake. So by $frac{{vert f_2(u+h)-f_2(u)-(2u,h)_{L^2} vert}}{{Vert hVert_{H^1}}}=frac{{vert (h,h)_{L^2} vert}}{{Vert hVert_{H^1}}}leq frac{{vert (h,h)_{L^2} + (nabla h, nabla h)_{L^2} vert}}{{Vert hVert_{H^1}}}$ I am done!
– sandmath
Nov 28 '18 at 7:26
No norm is differentiable (with respect to itself and the absolute value) at zero. If the norm comes from an inner product, it is differentiable at all nonzero vectors. Derivatives are independent of the choice of equivalent norms. Cheers
– Will M.
Dec 11 '18 at 5:16
add a comment |
Let $ f_1: L^2(Omega)rightarrowmathbb{R}, umapsto Vert uVert_{L^2 (Omega)}^2 $, $ f_2: H^1(Omega)rightarrowmathbb{R}, umapsto Vert uVert_{L^2 (Omega)}^2 $ and $f_2: H^1(Omega)rightarrowmathbb{R}, umapsto Vert uVert_{H^1 (Omega)}^2$.
Now it is possible to calculate the Gateaux-Derivatives:
$f_1'(u)h=(2u,h)_{L^2(Omega)}$- $f_2'(u)h=(2u,h)_{L^2(Omega)}$
- $f_3'(u)h=(2u,h)_{H^1(Omega)}$
Is the following statement true?
$f_2$ is not Fréchet differentiable with respect to $Vert cdot Vert _{H^1 (Omega)}$.
My current calculations are:
$frac{{vert f_2(u+h)-f_2(u)-(2u,h)_{L^2} vert}}{{Vert hVert_{H^1}}}=frac{{vert (h,h)_{L^2} vert}}{{Vert hVert_{H^1}}}$
functional-analysis normed-spaces frechet-derivative
asked Nov 27 '18 at 16:24
sandmath
112
Let $ f_1: L^2(Omega)rightarrowmathbb{R}, umapsto Vert uVert_{L^2 (Omega)}^2 $, $ f_2: H^1(Omega)rightarrowmathbb{R}, umapsto Vert uVert_{L^2 (Omega)}^2 $ and $f_2: H^1(Omega)rightarrowmathbb{R}, umapsto Vert uVert_{H^1 (Omega)}^2$.
Now it is possible to calculate the Gateaux-Derivatives:
$f_1'(u)h=(2u,h)_{L^2(Omega)}$- $f_2'(u)h=(2u,h)_{L^2(Omega)}$
- $f_3'(u)h=(2u,h)_{H^1(Omega)}$
Is the following statement true?
$f_2$ is not Fréchet differentiable with respect to $Vert cdot Vert _{H^1 (Omega)}$.
My current calculations are:
$frac{{vert f_2(u+h)-f_2(u)-(2u,h)_{L^2} vert}}{{Vert hVert_{H^1}}}=frac{{vert (h,h)_{L^2} vert}}{{Vert hVert_{H^1}}}$
functional-analysis normed-spaces frechet-derivative
functional-analysis normed-spaces frechet-derivative
asked Nov 27 '18 at 16:24
sandmath
112
asked Nov 27 '18 at 16:24
sandmath
112
edited Nov 28 '18 at 7:30
asked Nov 27 '18 at 16:24
sandmath
112
asked Nov 27 '18 at 16:24
sandmath
112
asked Nov 27 '18 at 16:24
sandmath
112
112
All these $f$'s should be squares of norms. Then all functions are Frechet.
– daw
Nov 27 '18 at 17:52
To calculate, expand the squares.
– daw
Nov 27 '18 at 17:53
Of course i have to take the squared norm. Sorry, that was a typing mistake. So by $frac{{vert f_2(u+h)-f_2(u)-(2u,h)_{L^2} vert}}{{Vert hVert_{H^1}}}=frac{{vert (h,h)_{L^2} vert}}{{Vert hVert_{H^1}}}leq frac{{vert (h,h)_{L^2} + (nabla h, nabla h)_{L^2} vert}}{{Vert hVert_{H^1}}}$ I am done!
– sandmath
Nov 28 '18 at 7:26
No norm is differentiable (with respect to itself and the absolute value) at zero. If the norm comes from an inner product, it is differentiable at all nonzero vectors. Derivatives are independent of the choice of equivalent norms. Cheers
– Will M.
Dec 11 '18 at 5:16
add a comment |
All these $f$'s should be squares of norms. Then all functions are Frechet.
– daw
Nov 27 '18 at 17:52
To calculate, expand the squares.
– daw
Nov 27 '18 at 17:53
Of course i have to take the squared norm. Sorry, that was a typing mistake. So by $frac{{vert f_2(u+h)-f_2(u)-(2u,h)_{L^2} vert}}{{Vert hVert_{H^1}}}=frac{{vert (h,h)_{L^2} vert}}{{Vert hVert_{H^1}}}leq frac{{vert (h,h)_{L^2} + (nabla h, nabla h)_{L^2} vert}}{{Vert hVert_{H^1}}}$ I am done!
– sandmath
Nov 28 '18 at 7:26
No norm is differentiable (with respect to itself and the absolute value) at zero. If the norm comes from an inner product, it is differentiable at all nonzero vectors. Derivatives are independent of the choice of equivalent norms. Cheers
– Will M.
Dec 11 '18 at 5:16
All these $f$'s should be squares of norms. Then all functions are Frechet.
– daw
Nov 27 '18 at 17:52
All these $f$'s should be squares of norms. Then all functions are Frechet.
– daw
Nov 27 '18 at 17:52
To calculate, expand the squares.
– daw
Nov 27 '18 at 17:53
To calculate, expand the squares.
– daw
Nov 27 '18 at 17:53
Of course i have to take the squared norm. Sorry, that was a typing mistake. So by $frac{{vert f_2(u+h)-f_2(u)-(2u,h)_{L^2} vert}}{{Vert hVert_{H^1}}}=frac{{vert (h,h)_{L^2} vert}}{{Vert hVert_{H^1}}}leq frac{{vert (h,h)_{L^2} + (nabla h, nabla h)_{L^2} vert}}{{Vert hVert_{H^1}}}$ I am done!
– sandmath
Nov 28 '18 at 7:26
Of course i have to take the squared norm. Sorry, that was a typing mistake. So by $frac{{vert f_2(u+h)-f_2(u)-(2u,h)_{L^2} vert}}{{Vert hVert_{H^1}}}=frac{{vert (h,h)_{L^2} vert}}{{Vert hVert_{H^1}}}leq frac{{vert (h,h)_{L^2} + (nabla h, nabla h)_{L^2} vert}}{{Vert hVert_{H^1}}}$ I am done!
– sandmath
Nov 28 '18 at 7:26
No norm is differentiable (with respect to itself and the absolute value) at zero. If the norm comes from an inner product, it is differentiable at all nonzero vectors. Derivatives are independent of the choice of equivalent norms. Cheers
– Will M.
Dec 11 '18 at 5:16
No norm is differentiable (with respect to itself and the absolute value) at zero. If the norm comes from an inner product, it is differentiable at all nonzero vectors. Derivatives are independent of the choice of equivalent norms. Cheers
– Will M.
Dec 11 '18 at 5:16
add a comment |
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All these $f$'s should be squares of norms. Then all functions are Frechet.
– daw
Nov 27 '18 at 17:52
To calculate, expand the squares.
– daw
Nov 27 '18 at 17:53
Of course i have to take the squared norm. Sorry, that was a typing mistake. So by $frac{{vert f_2(u+h)-f_2(u)-(2u,h)_{L^2} vert}}{{Vert hVert_{H^1}}}=frac{{vert (h,h)_{L^2} vert}}{{Vert hVert_{H^1}}}leq frac{{vert (h,h)_{L^2} + (nabla h, nabla h)_{L^2} vert}}{{Vert hVert_{H^1}}}$ I am done!
– sandmath
Nov 28 '18 at 7:26
No norm is differentiable (with respect to itself and the absolute value) at zero. If the norm comes from an inner product, it is differentiable at all nonzero vectors. Derivatives are independent of the choice of equivalent norms. Cheers
– Will M.
Dec 11 '18 at 5:16