How to maximize the utility function?

Question

How to maximize $\sum_{i =0 }^{\infty} f(c_i)$ where $f \in C^2 $ increasing ($f'>0 $) and strictly concave ($f''<0 $) function with $f(0) = 0 $ subject to $\sum_{i =0 }^{\infty} c_i \leq 1$, $c_i \geq 0$?
The sum $\sum_{i =0 }^{\infty} f(c_i)$ is also known as utility function, so a consumer wants to maximize his/her utility by choosing a specific bundle of goods or in this exact problem choose how to split the cake over days ($i$). I think that it does not have solutions as every day the consumer can split the cake say into two parts infinitely (this means that $f(\frac{c}{2}) > \frac{f(c)}{2}$). However, if it's true, how it may be proven?

Answer 1

I think you are on the right track.

Let's assume there was an optimal choice of ${c_{opt}}_i$ that maximizes the utility function. From $f(0)=0$ and $f'>0$ follows $f(1)>0$, that means setting $c_0=1$ and $c_i=0$ for $i=1,2,\ldots$ gives a utility of $f(1) > 0$. So the maximal utility is bigger than $0$, so the ${c_{opt}}_i$ can't all be $0$ (because that has utility $0$, so not maximal). That's rather intuitive, I think, not eating any part of the cake at all isn't best.

So we know there is an $n$ with ${c_{opt}}_n > 0$. Now it's easy to propose a choice that's actually better then our assumed optimum. Set

$$c_i= \begin{cases} {c_{opt}}_i, & \text{ if } i < n; \\ \frac{{c_{opt}}_n}2, & \text{ if } i = n \text{ or }n+1; \\ {c_{opt}}_{i-1} & \text{ if } i > n+1; \\ \end{cases} $$ Basically we keep our assumed optimal ${c_{opt}}_i$, except we split ${c_{opt}}_n$ in $2$ equal halves and thus have to shift the following ${c_{opt}}_i$ one day back.

We have $\sum_{i=0}^{\infty}c_i = \sum_{i=0}^{\infty}{c_{opt}}_i$, and $\forall i: c_i \ge 0$, so the new choice is feasable. We also have

$$\sum_{i=0}^{\infty}f(c_i) - \sum_{i=0}^{\infty}f({c_{opt}}_i)=2f(\frac{{c_{opt}}_n}2)- f({c_{opt}}_n) > 0.$$

the latter inequality due to strict concavity.

So no optimal choice exists.

As an additional note, since $\forall x \in [0,1]: f(x) \le f'(0)x$ (again due to concavity), we have $\sum_{i=0}^{\infty}f(c_i) \le \sum_{i=0}^{\infty}f'(0)c_i \le f'(0)$, so $f'(0)$ is an upper bound on the utility that can be reached. This also justifies doing differences above of infinite series, which might have been infinite.

By chosing each $c_i$ positive but small enough that $|\frac {f(c_i)}{c_i}-f'(0)| < \epsilon$ holds, a utility that gets above $f'(0)-\epsilon$ can be reached.

So $f'(0)$ is the supremem of the possible utilities, though unattainable.