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Let

  • $E$ be a normed space
  • $(H,\langle\;\cdot\;,\;\cdot\;\rangle)$ be a Hilbert space
  • $f:E\to H$ be Fréchet differentiable
  • $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $H$ and $$f_n:=\langle f,e_n\rangle\;\;\;\text{for }n\in\mathbb N$$

Can we find an expression for the Fréchet derivative of $f_n$?

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Let $T$ be the Fréchet derivative of $f$ at $e\in E$. Then $$ \frac{|f_n(e+h)-f_n(e)-\langle Th,e_n\rangle|}{\|h\|} =\frac{|\langle f(e+h),e_n\rangle-\langle f(e),e_n\rangle-\langle Th,e_n\rangle}{\|h\|} =\frac{|\langle f(e+h)- f(e)- Th,e_n\rangle}{\|h\|} \leq\frac{\| f(e+h)- f(e)- Th\|}{\|h\|}\to0. $$ So the derivative of $f_n$ is $h\longmapsto \langle Th,e_n\rangle$.

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  • $\begingroup$ Yes, you're right. It's even easier when we mention that $L_n:=\langle\;\cdot\;,e_n\rangle$ is a bounded, linear operator and hence its Fréchet derivative ${\rm D}L_n$ at $u\in H$ is equal to $L_n$. Then, we can conclude by the chain rule that $${\rm D}f_n(x)=L_n\circ{\rm D}f(x)=\langle{\rm D}f(x),e_n\rangle\;.$$ But thank you very much for your answer anyway. $\endgroup$ – 0xbadf00d Apr 27 '16 at 17:33
  • $\begingroup$ What about the second derivative? $\endgroup$ – 0xbadf00d Apr 27 '16 at 17:38
  • $\begingroup$ You would need to tell me how you define it. As "the derivative of the derivative" you get nothing new, since the derivative is already linear. $\endgroup$ – Martin Argerami Apr 27 '16 at 17:41
  • $\begingroup$ ${\rm D}f_n(x)$ is a bounded, linear operator, that's true. But ${\rm D}f_n$ itself is not linear. It's a (continuous) mapping $E\to\mathfrak L(E,H)$, where $\mathfrak L(E,H)$ is the space of bounded linear operators from $E$ to $H$. So, the second derivative of $f_n$ is the derivative of ${\rm D}f$ where $\mathfrak L(E,H)$ is equipped with the usual operator norm. $\endgroup$ – 0xbadf00d Apr 27 '16 at 17:51
  • $\begingroup$ My guess is $${\rm D^2}f_n(x)={\rm D}({\rm D}f_n)(x)={\rm D}(L_n\circ{\rm D}f)(x)={\rm D}L_n\left({\rm D}f\left(x\right)\right)\circ{\rm D^2}f(x)=L_n\circ{\rm D^2}f(x)\;,$$ but I'm unsure. $\endgroup$ – 0xbadf00d Apr 27 '16 at 18:16

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