I have a question on exercise given in Evans' PDE

Exercise 4.11 (p.248, second edition). Assume that $\left\{ L\left(t\right)\right\} _{t\ge0}$ is a family of symmetric linear operators on some real Hilbert space $H$, satisfying the evolution equation $$ \dot{L}=\left[B,L\right]=BL-LB, $$ for some collection of operators $\left\{ B\left(t\right)\right\} _{t\ge0}$. Suppose also that we have a corresponding family of eigenvalues $\left\{ \lambda\left(t\right)\right\} _{t\ge0}$ and eigenvectors $\left\{ w\left(t\right)\right\} _{t\ge0}:$ \begin{equation} L\left(t\right)w\left(t\right)=\lambda\left(t\right)w\left(t\right).\label{eq:eigenvalue} \end{equation} Assume that $L,B,\lambda$ and $w$ all depend smoothly upon the time parameter $t$. Show that $$\dot{\lambda}=0.$$

Solving this exercise is really easy when I just follow the hint of this exercise. But I have the following question:

"What is the meaning of differentiation?"

Well, $\lambda(t)$ is a real-variable function. But $w:\mathbb{R}\rightarrow H$, $L(t):R \rightarrow Sym(H)$, where $Sym(H)$ denotes the class of symmetric linear operators on real Hilbert space and $B(t):\mathbb{R} \rightarrow \mathcal{L}(H)$.

I have no experience on differentiation for Banach-valued function. What is an appropriate condition or definition of differentiation so that the problem can be solved more safely?

Thanks for in advance.

  • $\begingroup$ When you take a derivative, you look for a linear approximation of the function, but how do you approximate? Possible, you may take the definition of derivative DL at $t_0$ as $\lVert L(t_0+h)-L(t_0)-hDL(t_0)\rVert\to 0$ (operator norm), but usually, this is too strong and we can safely weaken this definition to: for every $v\in H$ we have $\lVert L(t_0+h)v-L(t_0)v-hDL(t_0)v\rVert_{H}\to 0$ (the norm of the space $H$). If this is still too strong, then you can take: for every $v\in H$ and $v^*\in H^*$... As you say, it depends on your experience on the subject to choose a definition. $\endgroup$ – user90189 Aug 5 '17 at 18:13
  • $\begingroup$ The weaker approximation you need to prove your result, the more robust it is. $\endgroup$ – user90189 Aug 5 '17 at 18:19

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