I want to think of $L^2[0, 1]$ as a generalization of the finite-dimensional $\mathbb{R}^n$.

In this case, the gradient of a function $f \colon \mathbb{R}^n \rightarrow \mathbb{R}$ is $$ \nabla f = \left( \partial_1 f, \partial_2 f, ..., \partial_n f \right). $$

The question is: what is $\nabla f$ for the case $f \colon L^2[0, 1] \rightarrow \mathbb{R}$?

The ultimate purpose is to be able to take directional derivatives of maps $L^2[0, 1] \rightarrow \Bbb R$ with respect to elements of $L^2[0, 1]$ by using the inner product on $L^2[0,1]$.

$$\nabla_\mathbf{u} f = \langle \nabla f , \mathbf{u} \rangle $$

Knowing a simple form for the "gradient" would help make calculations of the covariant derivative easy.

The functional derivative seems to be an option, but the calculation doesn't seem to be straightforward because of the use of the Riesz Theorem.

  • 3
    $\begingroup$ en.wikipedia.org/wiki/Fr%C3%A9chet_derivative $\endgroup$ Nov 7, 2017 at 18:18
  • $\begingroup$ Also if by $L^2[0,1]^*$ you mean the dual space consisting of linear functionals, then by any sensible definition we should have $\nabla_u f = f(u)$ for any $u \in L^2[0,1]$. This is because the derivative of a map is the best linear approximation to that map, so the derivative of a linear map would be the very same linear map. $\endgroup$
    – Demophilus
    Nov 7, 2017 at 23:16
  • $\begingroup$ @Demophilus You're right, I just mean (differentiable?) functions on $L^2[0, 1]$. $\endgroup$ Nov 7, 2017 at 23:46
  • $\begingroup$ Well then the Fréchet derivative is probably what you're looking for, as Giuseppe has suggested. $\endgroup$
    – Demophilus
    Nov 7, 2017 at 23:48


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.