Let $\phi : H \to \mathbb{R}$ ($H$ is a vectorial space) be a convex function $\mathcal{C}^1$.
I have the following inequality, for $\sigma \in H$ fixed, $$\forall \tau\in H, \ (\sigma - \tau \mid \chi-\partial \phi(\tau)\ge 0$$ Can we conclude that $\chi = \partial \phi(\sigma)$ (here with identified the subdifferential and the gradient since $\phi$ is $\mathcal{C}^1$)? If so, how to prove this fact ?
Thank you in advance.
By assumption, you have $$\langle \chi,\sigma- \tau \rangle \geq \langle \partial \phi(\tau),\sigma- \tau \rangle, \ \forall \tau \in H.$$
And by definition $$\phi(\sigma)- \phi(\tau) \geq \langle \zeta, \sigma-\tau \rangle, \ \forall \sigma \in H \Leftrightarrow \zeta \in \partial \phi(\tau),\\ \phi(\tau)-\phi(\sigma) \geq \langle \zeta,\tau-\sigma \rangle, \ \forall \tau \in H \Leftrightarrow \zeta \in \partial \phi(\sigma)$$
Therefore, $$\langle \chi,\tau-\sigma \rangle \leq - \langle \partial \phi(\tau),\sigma-\tau \rangle \leq \phi(\tau)-\phi(\sigma), \ \forall \tau \in H,$$
hence $\chi \in \partial \phi(\sigma)$ or $\chi = \partial \phi(\sigma)$ since subdifferential and gradient can be identified.