Dynamic programming problem: Optimal growth with linear utility This is a question from Stokey-Lucas-Prescott "Recursive Methods in Economic Dynamics", question 5.3: Optimal Growth with Linear Utility. 
Q. Consider the following model-
(1) $\underset{x_{t+1}}{max} \sum_{t=0}^{\infty} \beta^t U[f(k_t) - k_{t+1}]$
st $0 \leq k_{t+1} \leq f(k^t), t=0,1,2 \dots$
given $k_0 \geq 0$ 
(2) $v(x) = \underset{0 \leq y \leq f(x)}{max} U[f(x) - y] + \beta v(y)$
We are given U(c) = c, so $U[f(k_t) - k_{t+1}] = f(k_t) - k_{t+1}$.
We are also given that:
(T1) f is continuous
(T2)f(0) = 0 and for some $\bar{x} >0: x \leq f(x) \leq \bar{x}$, all $0 \leq x \leq \bar{x}$ and $f(x) < x$, all $x > \bar{x}$
(T3) f is strictly increasing
(T4) f is weakly concave
(T5) f is continuously differentiable
b. Define $k* = \underset{k \geq 0}{max} [\beta f(k) - k]$. Show that for some $\epsilon > 0$, $|k - k*| < \epsilon$ implies $v(k) = f(k) - k* + \beta\frac{[f(k*) - k*]}{1-\beta}$ 
I'm stuck at part b. I've worked out that if I set up my $\hat{v}= \frac{f(k) - k}{1-\beta} $, then I can get $v* = \lim_{n \rightarrow \infty} T^n\hat{v}$. But I am having trouble theoretically explaining why $\hat{v}$ should be of this form for $|k - k*| < \epsilon$.
Any help would be appreciated.
 A: Resurrecting this old thread here since no one actually solved this problem and the solution isn't online. The idea behind the value function is that you want to start with a "guess" function and then use the Bellman Operator to iterate, figure out a pattern, and then figure out what the value function is.
In this case, the Bellman Operator on a function $g(x)$ would be given by:
$$T(g)(x) = \max_{y \in [0,f(x)]} f(x) - y + \beta g(y)$$
Usually you have a sup instead of a max, but since the feasible set is compact valued and everything is continuous, you can just use the max.
Now, think back to the problem. $k^*$ is such that it solves the equation $\beta f'(k) -1 = 0$ (as a critical point of the strictly concave function $\beta f(k) - k$), i.e. so $f'(k^*) = \frac{1}{\beta}$. Moreover, for $k$ substantially close to $k^*$, $f'(k) \approx \frac{1}{\beta}$. This is what the problem means for $|k-k^*| < \epsilon$. I think SLP don't necessarily want you to rigorously show what that $\epsilon$, the idea is that if you're close to $k^*$, the proper conditions hold.
The approach you should have to this problem is to start with a smart "guess" for the value function and then iterate. Look at the proposed value function $v(k) = f(k) - k^* + \frac{\beta}{1-\beta}(f(k^*) - k*)$. If you look at it long enough, you can come up with a pretty good guess for what the value function is. If you haven't figured it out what it is, check the spoiler below and I'll just tell you.

 Try the initial guess $g(x) = f(x) - k^*$. The first iteration under the Bellman operator $T$ will be: $$T(g)(x) = \max_{y \in [0,f(x)]} f(x) - y + \beta g (y) = \max_{y \in [0,f(x)]} f(x) - y + \beta (f(y) - k^*)$$ If you use calculus to maximize this function, you'll get the first derivative w.r.t $y$ to be $$- 1 + \beta f(y) = 0$$ Doesn't that look familiar? This is precisely the condition that gave us $k^*$, so the optimal choice of $y$ here is $y = k^*$. In turn, we have $$T(g)(x) = f(x) - k^* + \beta(f(k^*) - k^*)$$ Long story short, if you apply  $T$ again, you use the same trick and get $$T^2(g)(x) = f(x) - k^* + \beta(f(k^*) - k^*) + \beta^2(f(k^*) - k^*)$$ In general for $T^n$, you should get $$T^n(g)(x) = f(x) - k^* + \sum_{i=1}^n \beta^i(f(k^*) - k^*)$$ As $n \to \infty$, you'll converge to the value function $v(x)$, which is $$\lim_{n\to\infty} T^n(g)(x) = f(x) - k^* + \sum_{i=1}^\infty \beta^i(f(k^*) - k^*) = f(x) - k^* + \frac{\beta}{1-\beta}(f(k^*)-k^*)$$

Please excuse any typos and let me know if anything looks off!
