Showing that reward function is bounded (dynamic programing)

I have the following dynamic programming problem:

$$\max_{\{x_t,y_{t+1}\}_{t=0}^\infty}\sum_{t=0}^\infty \beta^tu(F(x_t)-y_{t+1})\;\;\;\;\;\text{s.t}\;\;\;\;y_{t+1}\in\Gamma(x_t)$$

where $$\Gamma(x)=\{y\in\mathbb{R}_{+}\mid 0\le y\le F(x)\}$$, and the functions $$u$$ and $$F$$ have the following properties:

1. $$u:\mathbb{R}_{+}\to\mathbb{R}$$ is continuous, strictly increasing, and strictly concave;
2. $$F:\mathbb{R}_{+}\to\mathbb{R}$$ is continuous, strictly increasing, and concave;
3. $$F(0)=0$$;
4. And there exists $$\bar{x}>0$$ such that:
• If $$x\in[0,\bar{x}]$$ then $$F(x)\in[x,\bar{x}]$$.

• If $$x>\bar{x}$$ then $$F(x).

I've got to show that the function $$G(x,y)\equiv u(F(x)-y)$$ is bounded over the set $$A\equiv\{(x,y)\in \mathbb{R}_{+}\times\mathbb{R}_{+}\mid y\in\Gamma(x)\}$$.

What I have tried: Let $$(x,y)\in A$$, then $$y\in\Gamma(x)$$. First, by definition the functions $$u$$ and $$F$$ clearly $$G$$ is bounded below. Now in order to show that $$G$$ is bounded above, I noticed that the set of images $$G(x,\Gamma(x))$$ is compact in the second argument because $$G$$ is continuous and $$\Gamma(x)$$ is compact, i.e. given $$x$$ we have that the function $$G(x,\cdot)$$ is bounded in $$\Gamma(x)$$.

Now, trying to show that $$G$$ is bounded above in $$x$$ (i.e. in the first argument) has proven to be more involved, I'm trying to use the fourth property shown above about the existence of $$\bar{x}$$, where we have two cases, if $$x\in[0, \bar{x}]$$ then clearly $$G(x,y)\le u(F(\bar{x})-y)\le u(F(\bar{x}))$$. But when considering $$x>\bar{x}$$ we have $$F(x), thus $$G(x,y) which does not show that $$G$$ is bounded since the "bound" still depends on $$x$$ which is in a unbounded space $$(\bar{x},\infty)$$.

I think there must be a trick with $$F(x) but I have not been able to see it, any help would be appreciated.

Thanks!

The function $$u(F(x)-y)$$ is not necessarily bounded on $$A$$. For example, if $$u(x) = F(x) = \sqrt{x}$$ then: $$u(F(x)-y) = \sqrt{\sqrt{x}-y},$$ Taking $$y = 0$$, this gives $$u(F(x)) = \sqrt{\sqrt{x}}$$, which is clearly unbounded in $$x$$.
The problem you are looking at is the following: $$\max_{(x_t, y_t)_{t = 0}^\infty} \sum_{t = 0}^\infty \beta^t u(F(x_t) - y_{t+1}) \text{ s.t. } y_{t+1} \in [0, F(x_t)].$$ Notice that this problem is not well defined. In particular, given that $$u$$ is strictly increasing, the best choice for $$y_{t+1}$$ is to set it equal to zero at every instance $$t$$. This gives the payoff function $$\sum_{t = 0}^\infty \beta^t u(F(x_t)),$$ which is clearly increasing in $$x_t$$ (as both $$u$$ and $$F$$ are strictly increasing). As such, the best thing to do is to set $$x_t$$ as large as possible (so equal to $$+ \infty$$). This means that your maximisation problem has no solution.
Now, assume that you have a better behaving model, for example, $$y_t = x_t$$. This then gives the problem: $$\max_{(x_t)_{t = 0}^\infty} \sum_{t = 0}^\infty \beta^t u(F(x_t) - x_{t+1}) \text{ s.t. } x_{t+1} \in [0, F(x_t)].$$ The function $$u(F(x)-y)$$ is still unbounded, but now you can show that the objective function $$\sum_{t=0}^\infty \beta^t u(F(x_t) - x_{t+1})$$ is bounded for every feasible path. Let $$\partial u(0)$$ be a supdifferential (derivative) for $$u$$ at zero. Then, using concavity of $$u$$, we have: \begin{align*} \sum_{t = 0}^n \beta^t u(F(x_t) - x_{t+1}) &\le \sum_{t = 0}^n \beta^t \left(u(0) + \partial u(0)[F(x_t) - x_{t+1}]\right),\\ &= u(0) \sum_{t = 0}^n \beta^t + \partial u(0) \underbrace{\sum_{t = 0}^n \beta^t [F(x_t) - x_{t+1}]}_{=A}. \end{align*} Now, $$F(x) \le \overline{x} + x$$ (if $$x \le \overline{x}$$ then $$F(x) \le \overline{x}$$ and if $$x \ge \overline{x}$$, then $$F(x) \le x$$). So, \begin{align*} A &\le \sum_{t = 0}^n \beta^t [\overline{x} + x_t - x_{t+1}],\\ &= \overline{x} \sum_{t = 0}^n \beta^t + \underbrace{\sum_{t = 0}^n \beta^t (x_t - x_{t+1})}_{=B} \end{align*} Then, \begin{align*} B &= \sum_{t = 0}^n \beta^t (x_t - x_{t+1}),\\ &=x_0 \underbrace{- x_1 + \beta x_1}_{\le 0} \underbrace{- \beta x_2 + \beta^2 x_2}_{\le 0} \underbrace{- \beta^2 x_3 + \,\,}_{\le 0} \ldots \underbrace{\,\, + \beta^n x_n}_{\le 0} - \beta^n x_{n+1},\\ &\le x_0 - \beta^n x_{n+1} \le x_0 \end{align*} Putting it all together, we obtain the bound: \begin{align*} \sum_{t = 0}^n \beta^t u(F(x_t) - x_{t+1}) & \le (u(0) + \partial u(0) \overline{x}) \sum_{t = 0}^n \beta^t + \partial u(0) x_0 \end{align*} Taking the limit $$t \to \infty$$, the right hand side converges to $$\frac{u(0) + \partial u(0) \overline{x}}{1 - \beta} + \partial u(0) x_0$$ (a fixed number).