<p>Assuming the following:</p>
<ul>
<li><span class="math-container">$ S &gt; 0 $</span>, <span class="math-container">$T &gt; 0 $</span>,</li>
<li><span class="math-container">$ c(S) = \left\{c\in\Bbb{R}^T_{+} : \sum_{t=1}^T{c_t} = S \right\} $</span>, and</li>
<li><span class="math-container">$ U(\cdot) $</span> continuous.</li>
</ul>
<p>Let's start with convexity. To demonstrate convexity, we need to show that for any two points <span class="math-container">$c^1$</span> and <span class="math-container">$c^2$</span> in <span class="math-container">$c(S)$</span>, any linear combination of the two is also an element of <span class="math-container">$c(S)$</span>.</p>
<p>Let <span class="math-container">$\lambda \in [0,1]$</span>, <span class="math-container">$c^1 \in c(S)$</span>, <span class="math-container">$c^2 \in c(S)$</span>. Then, we have</p>
<p><span class="math-container">$$ \sum_{t=1}^T{\lambda c^1_t + (1-\lambda) c^2_t} $$</span>
<span class="math-container">$$ = \sum_{t=1}^T{\lambda c^1_t} + \sum_{t=1}^T{(1-\lambda) c^2_t} $$</span>
<span class="math-container">$$ = \lambda \sum_{t=1}^T{c^1_t} + (1-\lambda) \sum_{t=1}^T{c^2_t} $$</span>
<span class="math-container">$$ = \lambda S + (1-\lambda)S $$</span>
<span class="math-container">$$ = S $$</span>
<span class="math-container">$$ \Rightarrow \lambda c^1 + (1-\lambda) c^2 \in c(S) $$</span></p>
<p>Since any linear combination of any two points in c(S) is itself in c(S), we have c(S) convex.</p>
<p>For non-emptiness, we just need to show that a single point exists in c(S):
<span class="math-container">$$ \sum_{t=1}^T{\frac{S}{T}} = S $$</span>
<span class="math-container">$$ \Rightarrow \left\{ \frac{S}{T} \right\}^T \in c(S) $$</span></p>
<p>If <span class="math-container">$U(\cdot)$</span> is continuous on <span class="math-container">$c(S)$</span>, then continuity of <span class="math-container">$W(c)$</span> follows from the triangle inequality (see e.g., <a href="https://math.stackexchange.com/questions/744896/continuity-of-the-sum-of-continuous-functions">here</a> or <a href="https://math.stackexchange.com/questions/2750160/prove-that-the-sum-of-two-continuous-functions-is-continuous">here</a>). If we don't know more about <span class="math-container">$U(\cdot)$</span>, then there's not much we can say about <span class="math-container">$W(c)$</span>. Berge's Theorem of the Maximum has to do with the continuity of an optimized function, not the sum of functions.</p>

CC BY-SA 4.0
2020-12-28T08:59:45.893Z
2020-12-28T08:59:45.893Z
Amaan M
<p>Today in class, the professor said that the set of all consumption <span class="math-container">$c(S)$</span> is non-empty, compact and convex subset of <span class="math-container">$\mathbb{R}^T_+$</span>. i.e. we know <span class="math-container">$\sum \limits_{t=1} ^T c_t = S$</span> where <span class="math-container">$c_t$</span> is consumption in period t and <span class="math-container">$S$</span> is total wealth. The set  <span class="math-container">$c(S)$</span> is the set of all T-period consumption plan.</p>
<p>I could understand why <span class="math-container">$c(S)$</span> is compact, but I have no clue why it has to be convex and non-empty.</p>
<p>Following the same question, the professor also mentioned that <span class="math-container">$W(c)$</span> the summation of all the T-period utility functions denoted by <span class="math-container">$W(c)=\sum \limits_{t=1} ^T U(c_t)$</span> is continuous on <span class="math-container">$c(S)$</span>. I do not understand why <span class="math-container">$W(c)$</span> has to be  continuous on <span class="math-container">$c(S)$</span></p>
<p>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.</p>

CC BY-SA 4.0
2020-12-12T07:43:59.433Z
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
CC BY-SA 4.0
2020-12-12T08:30:28.100Z
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
CC BY-SA 4.0
2020-12-12T09:43:51.757Z
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
<microeconomics><mathematical-economics><utility>
Set of consumption over all period is convex in $\mathbb{R}^T_+$?
Doubts
# Set of consumption over all period is convex in $\mathbb{R}^T_+$?

(Asked by: username_0 on 2020-12-12)

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.

**Tags:** microeconomics, mathematical-economics, utility
> **Comments:**
> **username_1**: 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.  
> **username_0**: 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  
> **username_1**: 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.  

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## 1 answers

### Answer by username_2 on 2020-12-28. Score: 1

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](https://math.stackexchange.com/questions/744896/continuity-of-the-sum-of-continuous-functions) or [here](https://math.stackexchange.com/questions/2750160/prove-that-the-sum-of-two-continuous-functions-is-continuous)). 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.


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