Ok. I'll replace it by a more adequate one.

I thought that if such a result exits, it'd be possible to be best known for more especific spaces like $\mathcal{S}(\mathbb{R}^{n})$.

Thanks for your answer and reference! Just to clarify, it seems that my notion of functional derivative should be this so-called Bastiani derivative and representation (2) will follow from Schwartz Kernel Theorem?

Thanks for your answer! Can I clarify some points? First, does your comment imply that my second order derivative should be a measure such as $\eta(\omega\times \Omega):=\int_{\Omega}f(\omega, \omega')\beta(\omega')d\omega'$ so that $\omega \to \eta(\{\omega\}\times \Omega)$ is the 'function' to be considered in (8)? Second, how exactly should I extend $\langle \cdot, \cdot \rangle$. Does $\eta$ defined above extend my pairing (4), for instance?

@DanieleTampieri thank you so much! It helps a lot!

Yes, forgot the bounded condition. It makes very much sense to me but how can I assure I can write $Df[\alpha+\beta](\gamma)-Df[\alpha](\beta)$ as $\langle \mathcal{L}[\alpha](\beta),\gamma\rangle$? I got confused at this point. I mean, the factos $\delta f/\delta \alpha$ are not derivatives per se, they're elements of $F$. Did you use any representation theorem or something or it is easier than I think?

@DanieleTampieri I just read you answer and it helps me a lot! Let me just clarify one thing: your $\mathcal{L}(\alpha)$ is a linear operator which is the first order derivative of $f$ at $\alpha$?

Thanks for the answer and comment guys! I didn't read it yet (gonna do it soon) but have you seen my latest question? It's an extension of this one and you might be able to contribute a lot! mathoverflow.net/questions/355279/…

Yes! This one is more elaborated.