Let $Y$ be uniformly smooth Banach space. Consider the convex $C^1$ functional $\Phi:Y \to \mathbb{R}$ defined $$\Phi(y) = \frac{1}{q}\Vert y \Vert^q_{Y}.$$
Its derivative $\varphi:Y \to Y'$ is a monotone operator satisfying $$\langle \varphi(y), y \rangle = \Vert y \Vert^q_Y.$$
How to prove this? I presume this refers to the Gateaux derivative but I cannot get anywhere after forming the directional derivative expression. Is the proof hard?
Let $u \in L^p_{loc}(0,T;X) \cap L^q_{loc}(0,T;Y)$ be such that $\varphi(u) \in L^{q'}_{loc}(0,T;Y')$ is weakly differentiable with $(\varphi(u))' \in L^{p'}_{loc}(0,T;X')$. Then $$\frac{d}{dt}\frac{1}{q'}\Vert u(t) \Vert^q_Y = \langle (\varphi(u))', u \rangle$$ holds in the sense of distributions on $(0,T)$.
I don't obtain this: I don't get the factor of $\frac{1}{q'}$. I get: $$-\int_0^T \psi(t)\langle (\varphi(u))', u \rangle = \int_0^T \psi'(t)\langle (\varphi(u)), u \rangle = \int_0^T \psi'(t)\Vert u(t) \Vert^q$$ by the equation above. So I get $$\frac{d}{dt}\Vert u(t) \Vert^q_Y = \langle (\varphi(u))', u \rangle.$$ Where did I go wrong? Thanks for any help.
EDIT Source is http://link.springer.com/article/10.1007%2Fs10492-012-0004-0, section A.3 in the appendix.