<p>A counterexample:
<span class="math-container">$$
f(x,y)= \ln ((x-5)(y-5))
$$</span>
and the line is
<span class="math-container">$$
x+y=1.
$$</span>
All level curves of <span class="math-container">$f$</span> will stay above the point (5,5), not touching the line at all.
<span class="math-container">$f$</span> is differentiable and has full range (takes all <span class="math-container">$t\in\mathbb{R}$</span> values) as specified.</p>

CC BY-SA 4.0
2022-03-15T05:15:48.527Z
What about $(x,y) = (0.5,0.5)$ and $f = 2\ln(4.5)$? I think an easier counter would be $\frac{1}{x} = yc$ and line equation $x+y=0$. But the goal of the question is not to turn it into choosing the right range where the function exists and counter it. So reconsider the problem:  the line equation lies between the ICs with the minimum and the maximum utilities (if max and min exist), otherwise, the line equation cuts at least one level curve.
not tdm's twin
CC BY-SA 4.0
2022-03-15T08:09:19.397Z
If you wish to reconsider the problem make a *final* edit; do not invent new details later on. It is on the OP to include all assumptions in their question.
Giskard
CC BY-SA 4.0
2022-03-15T17:16:56.930Z
I think the current version is final. As I write these questions myself, I tend to miss out certain edge cases. But here's what I can do then -- I will create a new post from now on rather than modifying it.
not tdm's twin
CC BY-SA 4.0
2022-03-14T23:03:47.083Z
2022-03-14T23:03:47.083Z
Giskard
<p>Suppose we are given a differentiable function <span class="math-container">$f(x,y)$</span> such that <span class="math-container">$\forall$</span> <span class="math-container">$t \in \mathbb{R}$</span>, <span class="math-container">$f = t$</span> yields strictly convex level curves. If we are given a line equation <span class="math-container">$y = mx + c$</span> such that it always intersects two points of a level curve or is tangent to it or does not intersect at all, will it mean that a level curve tangent to <span class="math-container">$y = mx + c$</span> always exists (given the line cuts at least one level curve)?</p>
<p>I am not sure if this holds, but intuitively it feels like that, although I don't have a solid intuition to solve this. Using the Mean Value Theorem, one can say that a tangent with slope <span class="math-container">$m$</span> exists that will intersect a given level curve. But I can't extend this to prove that the line <span class="math-container">$y = mx + c$</span> itself will be tangent to some level curve.</p>

CC BY-SA 4.0
2022-03-14T20:22:18.877Z
Not sure I completely understand you but I would assume that quasi-linear preferences with a corner solution are a counterexample.
Jesper Hybel
CC BY-SA 4.0
2022-03-14T20:26:43.027Z
@JesperHybel It doesn't work I suppose. Because the budget line has to intersect the curves $U=c$ either at two points, or at a point of tangency, or nowhere.
not tdm's twin
CC BY-SA 4.0
2022-03-14T20:45:55.873Z
1. You should define what precisely you mean by tangency.
Giskard
CC BY-SA 4.0
2022-03-14T20:47:06.140Z
2. Are you familiar with the [hyperplane separation theorem](https://en.wikipedia.org/wiki/Hyperplane_separation_theorem)? You can apply it here directly.
Giskard
CC BY-SA 4.0
2022-03-14T20:51:05.393Z
@Giskard (1) If $U(x,y) = c$ touches a line $y = rx + s$ at exactly one point where $\frac{\partial U}{\partial x} = r$, then it's a point of tangency. (2) No.
not tdm's twin
<microeconomics><mathematical-economics>
Line equation tangent to convex level curve
not tdm's twin
# Line equation tangent to convex level curve

(Asked by: username_0 on 2022-03-14)

Suppose we are given a differentiable function $f(x,y)$ such that $\forall$ $t \in \mathbb{R}$, $f = t$ yields strictly convex level curves. If we are given a line equation $y = mx + c$ such that it always intersects two points of a level curve or is tangent to it or does not intersect at all, will it mean that a level curve tangent to $y = mx + c$ always exists (given the line cuts at least one level curve)?

I am not sure if this holds, but intuitively it feels like that, although I don't have a solid intuition to solve this. Using the Mean Value Theorem, one can say that a tangent with slope $m$ exists that will intersect a given level curve. But I can't extend this to prove that the line $y = mx + c$ itself will be tangent to some level curve.

**Tags:** microeconomics, mathematical-economics

> **Comments:**
> **username_1**: Not sure I completely understand you but I would assume that quasi-linear preferences with a corner solution are a counterexample.  
> **username_0**: @JesperHybel It doesn't work I suppose. Because the budget line has to intersect the curves $U=c$ either at two points, or at a point of tangency, or nowhere.  
> **username_2**: 1. You should define what precisely you mean by tangency.  
> **username_2**: 2. Are you familiar with the [hyperplane separation theorem](https://en.wikipedia.org/wiki/Hyperplane_separation_theorem)? You can apply it here directly.  
> **username_0**: @username_2 (1) If $U(x,y) = c$ touches a line $y = rx + s$ at exactly one point where $\frac{\partial U}{\partial x} = r$, then it's a point of tangency. (2) No.  

---

## 1 answers

### Answer by username_2 on 2022-03-14. Score: 0

A counterexample:
$$
f(x,y)= \ln ((x-5)(y-5))
$$
and the line is
$$
x+y=1.
$$
All level curves of $f$ will stay above the point (5,5), not touching the line at all.
$f$ is differentiable and has full range (takes all $t\in\mathbb{R}$ values) as specified.

> **Comments:**
> **username_0**: What about $(x,y) = (0.5,0.5)$ and $f = 2\ln(4.5)$? I think an easier counter would be $\frac{1}{x} = yc$ and line equation $x+y=0$. But the goal of the question is not to turn it into choosing the right range where the function exists and counter it. So reconsider the problem:  the line equation lies between the ICs with the minimum and the maximum utilities (if max and min exist), otherwise, the line equation cuts at least one level curve.  
> **username_2**: If you wish to reconsider the problem make a *final* edit; do not invent new details later on. It is on the OP to include all assumptions in their question.  
> **username_0**: I think the current version is final. As I write these questions myself, I tend to miss out certain edge cases. But here's what I can do then -- I will create a new post from now on rather than modifying it.  

---

