(Part of the following can be found at page 36 of "Non-linear Differential Equations of Monotone Types in Banach Spaces" of Viorel Barbu.)
Note for the following, that $A$ is a coercive and maximal monotone subset of $X \times X^*$, where $X$ is a reflexive Banach space.
Let $y_0 \in X^*$ be arbitrary but fixed and that $X,X^*$ are strictly convex. Then, for every $\lambda >0$ the equation $y_0 \in \lambda Jx_\lambda +Ax_\lambda $ has a unique solution $x_\lambda \in D\left(A\right)$.
$A$ is coercive, thus that means that $$\lim_{n \to \infty} \frac{\left\langle x_n - x^0, y_n\right\rangle}{\left\|x_n\right\|} = \infty$$ for some $x^0 \in X$ and all $\left[x_n,y_n\right] \in A$ such that $\left\|x_n\right\| \xrightarrow{n \to \infty} \infty$.
Multiplying in the sense of duality pairing the equation above by $\left( x_\lambda - x^0\right)$, one gets:
$$\left\langle x_\lambda - x^0, y_0 \right\rangle = \left\langle x_\lambda - x^0, \lambda Jx_\lambda \right\rangle + \left\langle x_\lambda - x^0, Ax_\lambda\right\rangle$$ $$\Leftrightarrow$$ $$\lambda\left\|x_\lambda\right\|^2 + \left\langle x_\lambda - x^0, Ax_\lambda\right\rangle = \left\langle x_\lambda - x^0, y_0\right\rangle + \lambda \left\langle x^0, Jx_\lambda\right\rangle$$
I am interested in showing that the sequence $\left\{x_\lambda\right\}_{\lambda > 0}$ is bounded as $\lambda \to 0$. Since we have the coercivity of $A$ in the hypothesis, my intuition evolves around it being an essential part (the book also mentions that by it one can decude the boundedness). Specifically, let's assume that $\left\{x_\lambda\right\}_{\lambda >0}$ was not bounded. Let $\left\{\lambda_n\right\}_{n \in \mathbb N}$ be a sequence, such that $\lambda_n \xrightarrow{n \to \infty} 0$ and $\lambda_n >0, \forall n \in \mathbb N$. Then, the last expression is re-written as:
$$\lambda_n \left\|x_{\lambda_n}\right\|^2 + \left\langle x_{\lambda_n} - x^0, A x_{\lambda_n}\right\rangle = \left\langle x_{\lambda_n} - x^0, y_0\right\rangle + \lambda_n \left\langle x^0, Jx_{\lambda_n}\right\rangle$$ $$\Leftrightarrow$$ $$\left\| x_{\lambda_n}\right\| + \frac{1}{\lambda_n}\frac{\left\langle x_{\lambda_n} - x^0, A x_{\lambda_n}\right\rangle}{\left\|x_{\lambda_n}\right\|} = \frac{1}{\lambda_n}\frac{\left\langle x_{\lambda_n} - x^0, y_0\right\rangle}{\left\|x_{\lambda_n}\right\|} + \frac{\left\langle x^0, Jx_{\lambda_n}\right\rangle}{\left\|x_{\lambda_n}\right\|}$$
Now, by the coercivity condition, we can obviously get that:
$$\lim_{n \to \infty} \frac{\left\langle x_{\lambda_n} - x^0, A x_{\lambda_n}\right\rangle}{\left\|x_{\lambda_n}\right\|} = \infty$$
I want to deduct a contradiction but I am unable to find a way around it.
Any hints or tips will be greatly appreciated.
If one does not divide by $\lambda_n$, the last equation is $$ \lambda_n\left\| x_{\lambda_n}\right\| + \frac{\left\langle x_{\lambda_n} - x^0, A x_{\lambda_n}\right\rangle}{\left\|x_{\lambda_n}\right\|} = \frac{\left\langle x_{\lambda_n} - x^0, y_0\right\rangle}{\left\|x_{\lambda_n}\right\|} +\lambda_n \frac{\left\langle x^0, Jx_{\lambda_n}\right\rangle}{\left\|x_{\lambda_n}\right\|}.$$ By coercivity, the left-hand side tends to $+\infty$, while the right-hand side is bounded. Contradiction.