# Set of consumption over all period is convex in $\mathbb{R}^T_+$?

Today in class, the professor said that the set of all consumption $$c(S)$$ is non-empty, compact and convex subset of $$\mathbb{R}^T_+$$. i.e. we know $$\sum \limits_{t=1} ^T c_t = S$$ where $$c_t$$ is consumption in period t and $$S$$ is total wealth. The set $$c(S)$$ is the set of all T-period consumption plan.

I could understand why $$c(S)$$ is compact, but I have no clue why it has to be convex and non-empty.

Following the same question, the professor also mentioned that $$W(c)$$ the summation of all the T-period utility functions denoted by $$W(c)=\sum \limits_{t=1} ^T U(c_t)$$ is continuous on $$c(S)$$. I do not understand why $$W(c)$$ has to be continuous on $$c(S)$$

I think this has something to with Berge's Maximum theorem, but I am unable to link it properly. Any help will be very appreciated.

• How is $c(S)$ defined? Whether $W$ is continuous will depend on $U(.)$ is the sum of continuous functions itself a continuous function? Without more information about the context it is impossible to answer the question, so I think the best option is to consult the book/notes for the course or ask the professor. He is paid to teach after all. – Jesper Hybel Dec 12 '20 at 7:43
• Hey @JesperHybel, the professor just mentioned C(S) be the set of all T–period consumption plans, which is not empty. With regards to the utilty functions, he mentioned that $U(.)$ is twice differentiable concave functions.. Also, I asked the Professor, he said it is for me to figure it out! So...not much help either – Doubts Dec 12 '20 at 8:30
• Well I'm sorry to hear that. Still, you must have an optimization problem specifying some initial conditions as well as a transitionsmechanism. Without this there is nothing to suggest that $c(S)$ is even compact. Why could $c(S)=\{(1,1),(2,2)\}$ with $T=2$? Nothing in your question contradicts this. Short of someone guessing the details of the problem you are trying to grasp I am afraid you will not get an answer. So if you have anymore information pls. provide it. – Jesper Hybel Dec 12 '20 at 9:43

Assuming the following:

• $$S > 0$$, $$T > 0$$,
• $$c(S) = \left\{c\in\Bbb{R}^T_{+} : \sum_{t=1}^T{c_t} = S \right\}$$, and
• $$U(\cdot)$$ continuous.

Let's start with convexity. To demonstrate convexity, we need to show that for any two points $$c^1$$ and $$c^2$$ in $$c(S)$$, any linear combination of the two is also an element of $$c(S)$$.

Let $$\lambda \in [0,1]$$, $$c^1 \in c(S)$$, $$c^2 \in c(S)$$. Then, we have

$$\sum_{t=1}^T{\lambda c^1_t + (1-\lambda) c^2_t}$$ $$= \sum_{t=1}^T{\lambda c^1_t} + \sum_{t=1}^T{(1-\lambda) c^2_t}$$ $$= \lambda \sum_{t=1}^T{c^1_t} + (1-\lambda) \sum_{t=1}^T{c^2_t}$$ $$= \lambda S + (1-\lambda)S$$ $$= S$$ $$\Rightarrow \lambda c^1 + (1-\lambda) c^2 \in c(S)$$

Since any linear combination of any two points in c(S) is itself in c(S), we have c(S) convex.

For non-emptiness, we just need to show that a single point exists in c(S): $$\sum_{t=1}^T{\frac{S}{T}} = S$$ $$\Rightarrow \left\{ \frac{S}{T} \right\}^T \in c(S)$$

If $$U(\cdot)$$ is continuous on $$c(S)$$, then continuity of $$W(c)$$ follows from the triangle inequality (see e.g., here or here). If we don't know more about $$U(\cdot)$$, then there's not much we can say about $$W(c)$$. Berge's Theorem of the Maximum has to do with the continuity of an optimized function, not the sum of functions.