Pochodna normy

[Sirkami]

Witam!

Mam delikatny problem z policzeniem pewnej pochodnej. Jako że nie studiuje w Polsce i nieszczególnie radzę sobie z polskim nazewnictem sformułuje problem po angielsku. Mam nadzieje że nie łamie w ten sposób regulaminu forum i nikomu to zbytnio nie będzie przeszkadzać.

Setting: (H, <>) stands for a real Hilbert space. The goal is to calculate the Fréchet derivative of the following function

\(\displaystyle{ F: H \to \mathbb{R}\hspace{0.5cm}x\mapsto \left| x \right|_H^p.}\)

Having the chain rule in mind, it can be advatanagous to decompose the function F into

\(\displaystyle{ F = f\circ g}\)

with

\(\displaystyle{ f(x) = x^{p/2},\hspace{0.5cm} g(x) = \left| x\right|^2_H.}\)

It is then clear that

\(\displaystyle{ \partial f(x) = \frac{p}{2}x^{(p-2)/2},\hspace{0.5cm} \partial g_x(\cdot) = 2\left\langle x, \cdot \right\rangle_H.}\)

The chain rule tells us then

\(\displaystyle{ \partial F = \partial f (g) \circ \partial g}\)

and finally we have

\(\displaystyle{ \partial F_x(\cdot ) = p\left| x\right|^{p-2}_H \cdot \left\langle x,\cdot\right\rangle_H,}\)

where the composition became multiplication. As expected (at least by me) our derivative maps into the space of linear bounded operators from H into R. The problem is that the solution I found in some books says that the result should be someting like that

\(\displaystyle{ \partial F_x = p\left| x\right|^{p-2}_H x,}\)

wchich for me is not an operator-valued function and therefore can not be the Frechet derivative of F. At what point am I thinking not correctly?

Z góry dziękuję za każdą pomoc!

[janusz47]

But

\(\displaystyle{ \partial g_{x}(\cdot ) =\frac{ 2\left\langle x, \cdot \right \rangle_{H}}{\parallel x \parallel}.}\)

[bartek118]

finally we have

\(\displaystyle{ \partial F_x(\cdot ) = p\left| x\right|^{p-2}_H \cdot \left\langle x,\cdot\right\rangle_H,}\)

where the composition became multiplication. As expected (at least by me) our derivative maps into the space of linear bounded operators from H into R. The problem is that the solution I found in some books says that the result should be someting like that

\(\displaystyle{ \partial F_x = p\left| x\right|^{p-2}_H x,}\)

wchich for me is not an operator-valued function and therefore can not be the Frechet derivative of F. At what point am I thinking not correctly?

W istocie, to formalnie \(\displaystyle{ \partial F_x(\cdot ) = p\left| x\right|^{p-2}_H \cdot \left\langle x,\cdot\right\rangle_H}\), natomiast to, co mamy tutaj:
\(\displaystyle{ p\left| x\right|^{p-2}_H x}\)
to gradient \(\displaystyle{ F}\). Byłbym skłonny zapisać raczej \(\displaystyle{ \nabla F (x) = p\left| x\right|^{p-2}_H x}\). Wtedy oczywiście zachodzi prawdziwy wzór
\(\displaystyle{ \partial F_x (y) = \langle \nabla F (x), y \rangle}\)

-- 1 paź 2017, o 14:27 --

finally we have

\(\displaystyle{ \partial F_x(\cdot ) = p\left| x\right|^{p-2}_H \cdot \left\langle x,\cdot\right\rangle_H,}\)

where the composition became multiplication. As expected (at least by me) our derivative maps into the space of linear bounded operators from H into R. The problem is that the solution I found in some books says that the result should be someting like that

\(\displaystyle{ \partial F_x = p\left| x\right|^{p-2}_H x,}\)

wchich for me is not an operator-valued function and therefore can not be the Frechet derivative of F. At what point am I thinking not correctly?

W istocie, to formalnie \(\displaystyle{ \partial F_x(\cdot ) = p\left| x\right|^{p-2}_H \cdot \left\langle x,\cdot\right\rangle_H}\), natomiast to, co mamy tutaj:
\(\displaystyle{ p\left| x\right|^{p-2}_H x}\)
to gradient \(\displaystyle{ F}\). Byłbym skłonny zapisać raczej \(\displaystyle{ \nabla F (x) = p\left| x\right|^{p-2}_H x}\). Wtedy oczywiście zachodzi prawdziwy wzór
\(\displaystyle{ \partial F_x (y) = \langle \nabla F (x), y \rangle}\)

Innymi słowy - izomorfizm Riesza pozwala nam na utożsamianie \(\displaystyle{ H^*}\) z \(\displaystyle{ H}\); gradient jest niczym innym jak wektorem uzyskanym jako obraz pochodnej Frechet'a poprzez izomorfizm Riesza.

[Sirkami]

@janusz47: Moglbys rozwinac? Z tego co zrozumialem moj problem polegal bardzej na interpretacji obiektow w rownaniu niz bledach w obliczeniach. Dodatkowo Amann & Escher

Kod: Zaznacz cały

https://math.stackexchange.com/questions/1357979/frechet-derivative-of-squared-norm-x-2

zdaja sie potwierdzac moj rezultat.

@bartek118: To co napisales bardzo mi pomoglo. Idac dalej tym tokiem myslenia chialbym teraz obliczyc \(\displaystyle{ \partial^2F_{xx}}\). Ponownie, pamietajac o regule lancuchowej i Leibniza, zapisujemy te funkcje jako

\(\displaystyle{ \partial (\partial F_x)(h) = \partial \big(p\left| x\right|_H^{p-2}\left\langle x,\cdot\right\rangle_H \big) = p\partial \big(\left| x\right|_H^{p-2}\big)\left\langle x,\cdot\right\rangle_H + p\left| x\right|_H^{p-2}\partial \big( \left\langle x,\cdot\right\rangle_H\big).}\)

Caly czas mamy w pamieci ze \(\displaystyle{ \partial^2F_{xx}: H \to L(H,L(H,\mathbb{R}))= L(H\times H, \mathbb{R})}\). Ponownie korzystajac z reguly lancuchowej i Leibniza obliczamy pierwsza i druga pochodna otrzymujac

\(\displaystyle{ \partial \big(p\left| x\right|_H^{p-2} \big)= \left\langle (p-2)\left| x\right|^{p-4}x,\cdot \right\rangle_H,\hspace{1cm}\partial \big( \left\langle x,\cdot\right\rangle_H\big)= \left\langle x,\cdot\right\rangle_H.}\)

Podstawiajac do pierwszego rownania otrzymujemy wtedy

\(\displaystyle{ \partial^2F_{xx}(x) = p(p-2)\left| x\right|^{p-4}_H\left\langle x, \cdot\right\rangle_H \left\langle x, \cdot\right\rangle_H + p\left| x\right|^{p-2}_H\left\langle x, \cdot\right\rangle_H.}\)

Oczekiwany wyniki prezentuje sie nastepujaco

\(\displaystyle{ \partial^2F_{xx}(x) = p(p-2)\left| x\right|^{p-4}_Hx\otimes x +p\left| x\right|^{p-2}_H Id_H,}\)

gdzie \(\displaystyle{ x\otimes y = x\left\langle y,\cdot\right\rangle}\) oraz (przynajmniej z takiego zalozenia wychodze) \(\displaystyle{ Id_H: H\to H, h\mapsto h}\). I tych roznic nie jestem w stanie sobie wytlumaczyc. Jedyne co przychodzi mi do glowy to fakt ze moga one wynikac z tego ze moje obliczenia produkuja pochodna z wartosciami w \(\displaystyle{ L(H, L(H,\mathbb{R}))}\) (??) a wynik oczekiwany w \(\displaystyle{ L(H\times H, \mathbb{R})}\). Jakies pomysly gdzie moglem popelnic blad?

[janusz47]

Chciałem zwrócić uwagę, że jeżeli

\(\displaystyle{ g(X) = \parallel f(X)\parallel^2 _{2},}\)

to

\(\displaystyle{ \Delta g(X) = \frac{2\sum_{i=1}^{n}f_{i}(X)\Delta f_{i}(X)}{\parallel f(X)\parallel_{2}}.}\)