0
$\begingroup$

Let $f\colon L^p(\Omega;X) \to L^q(\Omega;Y)$ and $g\colon L^q(\Omega;Y) \to L^r(\Omega;Z)$ where $X,Y,Z$ are separable Hilbert spaces and $\Omega$ is a smooth and bounded open set (eg. $\Omega = [0,T]$).

If these maps are continuously Frechet differentiable, under what conditions on the exponents do we obtain that $g \circ f \colon L^p(\Omega;X) \to L^r(\Omega;Z)$ is also continuously Frechet differentiable?

Does anyone know of a reference?

$\endgroup$
4
  • 3
    $\begingroup$ This is just the chain rule for Frechet differentiable maps. No (additional) conditions on $p,q,r$ required. $\endgroup$
    – daw
    Commented Apr 28, 2022 at 10:25
  • 3
    $\begingroup$ The chain rule is proved in almost any multivariable calculus text. For example, see Loomis and Sternberg's Advanced Calculus, or Henri Cartan's Differential Calculus, or Dieudonne's Treatise on Analysis Volume I for the Banach space setting. Spivak's Calculus on Manifolds or Munkres' Analysis on Manifolds also have proofs (they're stated in the context of maps $\Bbb{R}^n\to\Bbb{R}^m$ but with very minor modifications the proofs hold for arbitrary Banach spaces). $\endgroup$
    – peek-a-boo
    Commented Apr 28, 2022 at 10:30
  • $\begingroup$ @peek-a-boo all of those references hold for normed spaces (no need of completeness in general, except when dealing with open maps or integral; for chain rule, no need of completeness for sure). $\endgroup$
    – William M.
    Commented Jun 15, 2022 at 17:50
  • $\begingroup$ @WilliamM. sure; much of the basics rules of calculus can be developed for normed spaces (I just didn't mention it due to lack of space in a comment section:) $\endgroup$
    – peek-a-boo
    Commented Jun 15, 2022 at 17:55

0

Reset to default

You must log in to answer this question.