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Consider the Gale cake problem with $u(c) = log(c)$, so that the problem becomes:$$\max_{x}\sum_{t=0}^{\infty} \beta^{t}log(x_t-x_{t+1})\: sub\: 0<x_{t+1}\leq x_t,\: x_0>0\ given$$

It is pretty easy to show that the function $v:(0,x_0]\rightarrow\mathbb{R}$ given by $v(x)=A+B\:log(x)$, where $A=\frac{\beta}{(1-\beta)^2}log\beta+\frac{1}{1-\beta}log(1-\beta)$ and $B=\frac{1}{1-\beta}$ is a solution of the Bellman equation. I also know that it is the value function of the cake problem but the classic verification principle, i.e. $\lim_{t \to +\infty} \beta^tv(x_t) = 0$ is not met.

How should i proceed to actually show that $v(x)$ is the value function of the problem?

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  • $\begingroup$ Perhaps you did not complete the expression for the verification principle? To what should the limit evaluate? And I guess $0<\beta <1$? $\endgroup$ – Alecos Papadopoulos Jan 7 '16 at 20:19
  • $\begingroup$ Sorry for the delay. The esxpression above is one of the verification principle, which does not work in this case. I know there are others. $\beta \in (0,1)$, you get it. $\endgroup$ – PhDing Jan 9 '16 at 6:18
  • $\begingroup$ I cannot understand the expression, because it just writes down a limit. It does not write what the value of the limit should be. As is, it is not a verification principle, it is only an incomplete mathematical expression. $\endgroup$ – Alecos Papadopoulos Jan 9 '16 at 15:38
  • $\begingroup$ The value of the limit is not significant. It is enough to know that it is not equal to zero, i.e. the verification principle I wrote is not satisfied. Are there other ways to show that $v(x)$ is the value function of the RF problem above? $\endgroup$ – PhDing Jan 9 '16 at 23:15
  • $\begingroup$ So it appears that you should write in your question "$\lim_{t \to +\infty} \beta^tv(x_t) =0$ is not met". Why do you presume that readers will know that the verification principle implies that this limit should be zero? $\endgroup$ – Alecos Papadopoulos Jan 9 '16 at 23:25

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