Gradient of a function $f \colon L^2[0, 1] \rightarrow \mathbb{R}$.

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.

• en.wikipedia.org/wiki/Fr%C3%A9chet_derivative – Giuseppe Negro Nov 7 '17 at 18:18
• 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. – Demophilus Nov 7 '17 at 23:16
• @Demophilus You're right, I just mean (differentiable?) functions on $L^2[0, 1]$. – Open Season Nov 7 '17 at 23:46
• Well then the Fréchet derivative is probably what you're looking for, as Giuseppe has suggested. – Demophilus Nov 7 '17 at 23:48