just studying some variational calculus, but i got stuck in a question about linear normal spaces. I know that it may seem obvious that for a R $\in$ $L(X,Y)$ (which is the normal vector space for all linear and continous transformations between X,Y) we have that $R''=0$, but I just don't get it. How can I arrive to that result? Thanks for reading.
P.S R'', refers to the second Fréchet derivative.
This is easy to demonstrate from first principles. If we take the definition of the Gâteaux derivative and apply it to a linear operator $R\in {\cal L}(X,Y)$ at some point $x\in X$ we have $$R'(x)(h) =: \lim_{t\rightarrow 0} \frac{R(x+th)-R(x)}{t} $$ for all $h\in X$. Since $R$ is linear we see that $R(x+th) = R(x)+tR(h)$ and so substituting that in above, we get $$ \begin{eqnarray*} R'(x)(h) &=& \lim_{t\rightarrow 0} \frac{R(x)+tR(h)-R(x)}{t} \\ {}& = & \lim_{t\rightarrow 0}R(h) \\ {}& = & R(h) \end{eqnarray*} $$ Thus $R'(x)$ is constant (as mentioned in the comments). Taking the derivative of a constant is zero, so $R''(x) = 0 \ \forall x \in X$.
This leaves the case of the Fréchet derivative for you to do yourself, but it's very similar.