Problem: Let us consider the following minimization problem \begin{align*} &\min_{X,v,w} v \tag{1a}\\ \text{s.t }& X- X_b - (1-v)\Delta t F(X) = 0 \tag{1b}\\ & 1- (1-v)\Delta t F'(X) - \varepsilon - w = 0 \tag{1c}\\ & v \ge 0 \tag{1d} \\ & w \ge 0 \tag{1e} \end{align*} For a brief explanation, this problem comes from the Euler Implicit scheme with $\Delta$ is the time step, $X_b$ is a known value, the function $F$ is continuously differentiable as much as possible, and Lipschitz continuous. I wonder when this problem has at least a solution.
My attempt: To complete the proof, I intend to separate the task into two cases: $F$ is linear and nonlinear. For the linear case, I have done. About the nonlinear case, I have thought about the direction using the Weierstrass theorem for the reason that this set is non-empty (include $(X_b,1,1-\varepsilon)$ and closed. Now, I just have to prove the constraint set $C$ is bounded.
Firstly, by the formulation of the problem, we can assume that there exists a number $v_0 \ge 0$ such that $$0 \le v \le v_0,\ \forall (X,v,w) \in C.$$ Next, using the second equation, we can deduce that $$w = 1-(1-v)\Delta t F'(X) - \varepsilon,$$ which shows that $w$ is bounded because $F'$ is bounded ($F$ is Lipschitz) and $v$ is bounded. Now, the only remaining thing that we are caring about is the boundedness of $X$. I have tried a lot of ways to indicate it, but I failed. That's why I think I need to add more assumptions to the function $F$ to attain boundedness.
Edit 1 (first assumption): If $F$ is bounded, we can deduce from the first equation that $X$ is bounded