Let $U$ be a Banach space and $\Theta$ be a convex function. Let $$\partial\Theta(x):=\{\zeta\in U^*: \Theta(x')-\Theta(x)-\langle \zeta, x'-x\rangle\geq 0\ \forall\ x'\in U\}.$$ Here $\langle \zeta, x\rangle=\zeta(x)$ denotes the duality pairing for $\zeta \in U^*$ (dual space) of a Banach space $U$.
For $\zeta\in \partial\Theta(x)$, we define a distance function $$D(\zeta, x', x)=\Theta(x')-\Theta(x)-\langle \zeta, x'-x\rangle.$$ By definition, it is clear that $D(\zeta, x', x)\geq 0$.
Next, for a nonlinear operator $F:U\to V$ given by $F(x)=y$, consider the iterative sequence $\{x_n\}$, $n\geq 1$ for a given $x_0$, where $x_n$ is such that $$\frac1r\|F(x_n)-y\|^r+\alpha D(\zeta_{n-1}, x_n, x_{n-1})\leq \frac1r\|F(x)-y\|^r+\alpha D(\zeta_{n-1}, x, x_{n-1})\ \ \ \ \ (1)$$ for all $x\in U$, $r>1$ and $\alpha>0$. Here $\zeta_{n-1}\in \partial\Theta(x_{n-1})$ which means that $D(\zeta_{n-1}, x, x_{n-1})\geq 0$ for all $x$. Also, it is known that $$\zeta_n=\zeta_{n-1}-\frac{1}{\alpha}F'(x_n)^*J_r(F(x_n)-y),\ \ \ (2)$$ Here $J_r$ is the duality mapping that fulfill $$\langle J_r(x), x\rangle =\|x\|^r, \ \ \|J_r(x)\|=\|x\|^{r-1}.$$
Also, we know that $$D(\zeta_{n-1}, x, x_{n-1})-D(\zeta_{n-1}, x_n, x_{n-1})=D(\zeta_{n}, x, x_{n})+\langle \zeta_n-\zeta_{n-1}, x-x_n\rangle.\ \ (3)$$
To prove: Use $(1)$, $(2)$ and $(3)$ to show that $$D(\zeta_{n}, x, x_{n})\geq 0\ \forall x.$$
Please help. I have putted many efforts in it but unable to solve it. Any hint or help is appreciated.