Let $U$ be a real Hilbert space with orthonormal basis $\{e_n\}_{n \in \mathbb{N}}$. We have a compact linear operator $F'(x)(e_n) : U \to U$ by $$F'(x)(e_n) = 2^{-n} e_n$$ Now i want to find out the function $F$ i.e. how to find out $F$ from its Frechet derivative? Any idea how to go back?
After some ideas, by hit and trial can i say that $$F(x) = \sum_{n=1}^{\infty} \zeta_n^2 2^{-(n+1)} e_n$$ where $x = \sum_{n=1}^{\infty} \zeta_n e_n$ is the $F$?
Note that if you define $$ G:U \to U: x=\sum_{n=1}^\infty x_n e_n \mapsto \sum_{n=1}^\infty \frac{x_n^2}{2^{n+1}} e_n, $$ then for any $x \in U$, $h >0$ and $n \in \mathbb{N}$ we have $$ G(x+he_n)-G(x) = \frac{2x_n h +h^2}{2^{n+1}}e_n. $$ So the Fréchet derivative will actually be $G'(x)(e_n) =x_n 2^{-n} e_n$, which is quite different from $F'(x)(e_n) = 2^{-n}e_n$.
The proper solution is actually simply the linear map given by $F(e_n) = 2^{-n}e_n$. Because a linear map is its own derivative, it's easily seen that this works.