$F$ with Gateaux-derivative $A$, then $pF(u)=A(u)(u)$ Let $F:X\to \mathbb{R}$ Gateaux-differentiable with Gateaux-derivative $A:X\to X^*,$ ($X^*$) is the dual space pf $X$. Let $p\in\mathbb{R}$ such that $$F(\lambda u)=\lambda^pF(u)$$for all $u\in X$ and $\lambda >0$. I already proved the equality $A(\lambda u)=\lambda^{p-1}A(u)$ and I now want to prove $pF(u)=A(u)(u)$ and here I'm stuck. One of my first tries was to consider $\frac{\partial{\lambda^pF(u)}}{\partial \lambda}=p\lambda^{p-1}F(u)$, $\lambda=1$ and the definition of the Gateaux-derivative $\lim\limits_{h\to 0}\frac{F(u+hu)-F(u)}{h}=A(u)(u)$, then mix everything together. But I only see what could be needed for a proof, but not how to prove it exactly. 
Could you help me? Regards

Fix $u \in X$. Define $g \colon \mathbf R \to \mathbf R$ by $g(\lambda) = F(\lambda u)$. Then - as $F$ is Gateaux differentiable - by the chain rule 
$$ g'(\lambda) = A(\lambda u)(u) $$
On the other hand $g(\lambda) = \lambda^p F(u)$, hence
$$ g'(\lambda) = p\lambda^{p-1} F(u) $$
Now let $\lambda = 1$.
