<p>Define the function</p>

<p>$$
V(p, e(p, u)) - u = f(u) ~~~\mbox{for}~~~ u_\min \leq u \leq u_\max
$$</p>

<p>Your problem is then reduced to finding a point $u^* \in [u_\min, u_\max]$ such that $f(u^*)=0$. First, this problem may have not solution at all, even if the function is continuos. That being said, if the function is not continuos, you can have a situation like the one below
<a href="https://i.sstatic.net/hF9do.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/hF9do.png" alt="enter image description here"></a></p>

<p>Clearly, this function is not continuos at $u=u_0$, moreover, it is not possible to find a point $u^*$ such that $f(u^*) = 0$.</p>

CC BY-SA 3.0
2017-09-28T05:45:55.257Z
In your answer the function $e(p,u)$ itself is not defined at $u^*$. As long as that is allowed there is no need to bother with such a complicated example. Let $e(p,u)$ be undefined everywhere. Now $V(p,e(p,u))$ is also undefined everywhere, hence $V(p,e(p,u)) \neq u$ for any $u$.
Giskard
CC BY-SA 3.0
2017-09-29T20:23:34.080Z
@denesp Why is $e(p,u)$ necessarily undefined and not $V(\cdot)$ alone?
luchonacho
CC BY-SA 3.0
2017-09-29T20:53:13.873Z
@luchonacho What expenditure is requred to reach utility level $u$ at price $p$ if utility level $u$ can never be reached? (Because the utility function does not give value $u$ to anything.)
Giskard
CC BY-SA 3.0
2017-09-29T21:02:42.070Z
@denesp Elementary, my dear Watson...
luchonacho
CC BY-SA 3.0
2017-10-01T09:55:01.350Z
@denesp My point is exactly that: if the function is not defined at a point, the problem $f(u) = 0$ may not have a solution. That's the reason for the graph, to show it more explicitly
caverac
CC BY-SA 3.0
2017-10-01T12:33:21.747Z
But you are cheating with the definition. It is not that the function $V$ cannot take the value $u^*$ at $e(p,u^*)$, it is that the function $e(p,\cdot)$ is not defined at $u^*$. Again if this is allowed you can just say let the function $V$ be not defined anywhere. See, now it does not take any values.
Giskard
CC BY-SA 3.0
2017-10-01T12:36:14.327Z
(+1) Hard to know, perhaps this indeed what the OP wanted, but I feel like the more likely question is whether the equation holds if $e(p,u)$ is defined.
Giskard
CC BY-SA 3.0
2017-09-28T02:06:07.103Z
2017-09-28T02:06:07.103Z
caverac
<p>Most utility functions we work with are already continuous, so I thought it was understood that it must continuous. But how would I show that if the utility function is discontinuous the identity V(p, e(p,u))=u might not hold?</p>

<microeconomics><utility>
Utility function must be continuous for $V(p, e(p,u) )$ to hold?
Lucy Z
# Utility function must be continuous for $V(p, e(p,u) )$ to hold?

(Asked by: username_0 on 2017-09-27)

Most utility functions we work with are already continuous, so I thought it was understood that it must continuous. But how would I show that if the utility function is discontinuous the identity V(p, e(p,u))=u might not hold?

**Tags:** microeconomics, utility

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## 1 answers

### Answer by username_1 on 2017-09-28. Score: 1

Define the function

$$
V(p, e(p, u)) - u = f(u) ~~~\mbox{for}~~~ u\_\min \leq u \leq u\_\max
$$

Your problem is then reduced to finding a point $u^\* \in [u\_\min, u\_\max]$ such that $f(u^\*)=0$. First, this problem may have not solution at all, even if the function is continuos. That being said, if the function is not continuos, you can have a situation like the one below
[![enter image description here](https://i.sstatic.net/hF9do.png)](https://i.sstatic.net/hF9do.png)

Clearly, this function is not continuos at $u=u\_0$, moreover, it is not possible to find a point $u^\*$ such that $f(u^\*) = 0$.

> **Comments:**
> **username_2**: In your answer the function $e(p,u)$ itself is not defined at $u^*$. As long as that is allowed there is no need to bother with such a complicated example. Let $e(p,u)$ be undefined everywhere. Now $V(p,e(p,u))$ is also undefined everywhere, hence $V(p,e(p,u)) \neq u$ for any $u$.  
> **username_3**: @denesp Why is $e(p,u)$ necessarily undefined and not $V(\cdot)$ alone?  
> **username_2**: @luchonacho What expenditure is requred to reach utility level $u$ at price $p$ if utility level $u$ can never be reached? (Because the utility function does not give value $u$ to anything.)  
> **username_3**: @denesp Elementary, my dear Watson...  
> **username_1**: @denesp My point is exactly that: if the function is not defined at a point, the problem $f(u) = 0$ may not have a solution. That's the reason for the graph, to show it more explicitly  
> **username_2**: But you are cheating with the definition. It is not that the function $V$ cannot take the value $u^*$ at $e(p,u^*)$, it is that the function $e(p,\cdot)$ is not defined at $u^*$. Again if this is allowed you can just say let the function $V$ be not defined anywhere. See, now it does not take any values.  
> **username_2**: (+1) Hard to know, perhaps this indeed what the OP wanted, but I feel like the more likely question is whether the equation holds if $e(p,u)$ is defined.  

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