<p>The proof below is not very rigorous - I do not discuss discontinuities - but I think it captures the logic well.</p>
<hr />
<p>From <a href="https://en.wikipedia.org/wiki/Convex_function#Functions_of_one_variable" rel="nofollow noreferrer">Wikipedia</a>:</p>
<blockquote>
<p>Suppose <span class="math-container">$f$</span> is a function of one real variable defined on an interval, and let
<span class="math-container">$$ R(x_{1},x_{2})={\frac {f(x_{2})-f(x_{1})}{x_{2}-x_{1}}} $$</span>
[...]
<span class="math-container">$f$</span> is convex if and only if <span class="math-container">$R(x_{1},x_{2})$</span> is monotonically non-decreasing in <span class="math-container">$x_{1}$</span> for every fixed <span class="math-container">$x_{2}$</span> (or vice versa).</p>
</blockquote>
<p>Let us denote the point of tangency by <span class="math-container">$(x^*,y^*)$</span>, and let us denote the function describing the indifference curve at level <span class="math-container">$U(x^*,y^*)$</span> by <span class="math-container">$f$</span>.</p>
<p>Since <span class="math-container">$U$</span> is differentiable at <span class="math-container">$(x^*,y^*)$</span>, we have
<span class="math-container">$$
MRS(x^*,y^*) = f'(x^*).
$$</span></p>
<p>By the definition of derivatives, we have,
<span class="math-container">$$
f'(x^*) = \lim_{h \to 0} \frac{f(x^* + h)-f(x^*)}{x^* + h - x^*}.
$$</span></p>
<p>Using the Wikipedia lemma, we have for any <span class="math-container">$x &lt; x^*$</span>
<span class="math-container">$$
\frac{f(x)-f(x^*)}{x - x^*} \leq f'(x^*)
$$</span>
or
<span class="math-container">$$
f(x) \geq f(x^*) + (x - x^*)f'(x^*).
$$</span>
Thus any basket <span class="math-container">$(x,y)$</span> on the indifference curve defined by <span class="math-container">$f$</span> that is to the left <span class="math-container">$x^*$</span> is above or on the tangent line. Since the tangent line coincides with the budget constraint, these baskets are unattainable or just attainable. Assuming positive prices, no baskets of higher utility are attainable, as that would violate monotonicity.</p>
<p>One can similarly show that
<span class="math-container">$$
f(x) \geq f(x^*) + (x - x^*)f'(x^*)
$$</span>
also hold for all <span class="math-container">$x &gt; x^*$</span>, thus baskets on the indifference curve to the right of <span class="math-container">$x$</span> will also be unattainable or just attainable.</p>

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2022-06-25T14:38:32.303Z
(+1) Very interesting, I like the idea! I think it does not work for all continuous case either, for example when $U(x,y) = \max(x,y)$. We can't define $f(x)$ for its ICs. I want to give this some time before accepting it in case @MichaelGreinecker wants to chime in later or if you would like to capture some more cases.
xyz123
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2022-06-25T14:43:15.003Z
@xyz123 $\max(x,y)$ is not a convex utility function, so it is not surprising that this proof does not cover it.
Giskard
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2022-06-25T14:44:11.810Z
In case you meant $\min(x,y)$, you wrote "*not necessarily differentiable except for a region of the corresponding IC around the tangency point*", but $\min(x,y)$ is not differentiable around the tangency point.
Giskard
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2022-06-25T14:58:51.793Z
That's true, I missed it. Does that mean we can always say that for convex and monotone preferences, the indifference curves can be written as a function of $x$?
xyz123
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2022-06-25T16:16:39.767Z
If you don't insist on full range (non-weak) monotonicity is enough.
Giskard
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2022-06-26T03:45:54.677Z
With non-weak monotonicity, it is certainly enough as a point lying on top of another needs to have more utility (by definition of strong monotonicity). I am not assuming strong monotonicity in my case. When the ICs are straight lines moving toward the right, then it's an example where convex and monotone preferences won't have a point of tangency. Do we need strict convexity or strict monotonicity (or both) to ensure the tangency condition holds?
xyz123
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2022-06-26T08:35:05.257Z
We seem to be going back and forth over the same details. In your question you are *assuming* local differentiabality and tangency, and asking under what conditions can we assume this to be optimal. In your last comment you seem to do away with this assumption.
Giskard
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2022-06-26T08:39:39.543Z
In case you would have something like Leontief preferences with the kink smoothed out, you would have tangency and weak monotonicity. This is one of those special cases which the proof would have to discuss seperately. It would be fine though, as the proof works for cases where the slope is 0.
Giskard
CC BY-SA 4.0
2022-06-26T14:06:25.073Z
Ignore the example please; I was tired and sleepy. There could be several such special (or even pathological) cases which can not distinctively be turned into a *case*. The proof works where the slope is $0$, but not in the vertical line portions of the Leontief ICs. (I'm accepting this answer as I think generalizing it would be difficult. Strangely, I see these generalizations being quoted often but never really with a proof.)
xyz123
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2022-06-26T14:23:11.857Z
"*The proof works where the slope is 0, but not in the vertical line portions of the Leontief ICs.*" It would though, as you can treat these cases seperately, and by switching the axes you can turn the 'infinite' slope into 0, making the proof directly applicable. IMO the proof lacks rigor, but I see no flaws.
Giskard
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2022-06-26T14:23:42.147Z
Seems like you are very quick with your conclusions and examples; perhaps take more time?
Giskard
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2022-06-26T16:49:43.380Z
@xyz123 I would like to see (if you can provide some reference) where these generalisations are quoted without the proof.
Amit
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2022-06-25T13:35:01.300Z
2022-06-25T13:35:01.300Z
Giskard
<p>When an indifference curve is tangent to the budget line such that the preferences are convex and monotone, why is the point of tangency an optimal for an <a href="https://en.wikipedia.org/wiki/Utility_maximization_problem" rel="nofollow noreferrer">UMP</a>?</p>
<p>Given the budget line <span class="math-container">$p_1 x + p_2 y = I$</span>, let MRS<span class="math-container">$_{xy} = \frac{p_1}{p_2}$</span> at some point <span class="math-container">$(a,b)$</span>. Define MRS as <span class="math-container">$\frac{dy(x)}{dx}$</span> where <span class="math-container">$y : (a-\epsilon, a+\epsilon) \to \mathbb{R}$</span> for an appropriately small <span class="math-container">$\epsilon &gt; 0$</span>. This is to say that the optimal IC is differentiable around the point of tangency, and any other point (in this or other IC or of <span class="math-container">$U$</span> may not be differentiable).</p>
<p>I have come across this often but I haven't seen a proof. If <span class="math-container">$U$</span> is differentiable everywhere, then the Kuhn-Tucker condition is satisfied and the result follows. The problem is when every point is not necessarily differentiable except for a region of the corresponding IC around the tangency point.</p>

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2022-06-24T22:21:35.773Z
You want a proof that if the indifference curve has the same slope as the budget line on a point on the budget line, then that point solves the UMP?
Michael Greinecker
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2022-06-25T00:32:55.213Z
@MichaelGreinecker We can define the tangency condition in a slightly different way: Suppose MRS$_{xy} = \frac{p_1}{p_2}$ at the point $(a,b)$. Define MRS as $\frac{dy(x)}{dx}$ where $y : (a-\epsilon, a+\epsilon) \to \mathbb{R}$ for an appropriately small $\epsilon > 0$. This is to say that only the IC around the point of tangency is differentiable. That IC, as an whole, or other ICs or even $U$ are not assumed to be differentiable (or partially differentiable).
xyz123
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2022-06-25T00:35:54.217Z
@MichaelGreinecker Take $U(x,y) = x+y+ \lfloor x+y \rfloor$ for example. It's not differentiable when $x+y \in \mathbb{Z}$, but the IC given by $U(x,y) = 10$ is tangent to the budget line $B : x+y=5$. Here, optimality and tangency condition hold despite the partial derivatives not being defined.
xyz123
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2022-06-25T00:38:14.740Z
@MichaelGreinecker The point is, I want a proof of precisely what you described but without assumptions that are not necessary.
xyz123
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2022-06-25T08:45:20.453Z
Can you please edit your question so that it contains all the information posted in the comments? Presently this is kind of hard to read. If Michael's comment helped, you can also use it to clarify your question.
Giskard
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2022-06-25T09:28:37.360Z
@xyz123 By the way, utility function $u(x, y) = x + y + \lfloor x +y \rfloor$ represents the same preference as the utility function $v(x, y) = x + y$ which is a differentiable and quasi-concave function.
Amit
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2022-06-25T10:30:19.323Z
@Amit The example was to justify my point that partial derivatives may not necessarily exist.
xyz123
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2022-06-25T10:31:37.583Z
@Giskard I added the comment to the post.
xyz123
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2022-06-25T18:14:45.387Z
@xyz123 I know that there exist utility functions that are non-differentiable so that partial derivatives may not exist. I was just trying to make a point that some preferences represented by non-differentiable utility functions might also have other utility representations that are differentiable so that they can be treated just like the way we treat differentiable utility functions.
Amit
<microeconomics><consumer-theory><choice-theory>
Proof of the tangency condition in UMP
xyz123
# Proof of the tangency condition in UMP

(Asked by: username_0 on 2022-06-24)

When an indifference curve is tangent to the budget line such that the preferences are convex and monotone, why is the point of tangency an optimal for an [UMP](https://en.wikipedia.org/wiki/Utility_maximization_problem)?

Given the budget line $p\_1 x + p\_2 y = I$, let MRS$\_{xy} = \frac{p\_1}{p\_2}$ at some point $(a,b)$. Define MRS as $\frac{dy(x)}{dx}$ where $y : (a-\epsilon, a+\epsilon) \to \mathbb{R}$ for an appropriately small $\epsilon > 0$. This is to say that the optimal IC is differentiable around the point of tangency, and any other point (in this or other IC or of $U$ may not be differentiable).

I have come across this often but I haven't seen a proof. If $U$ is differentiable everywhere, then the Kuhn-Tucker condition is satisfied and the result follows. The problem is when every point is not necessarily differentiable except for a region of the corresponding IC around the tangency point.

**Tags:** microeconomics, consumer-theory, choice-theory
> **Comments:**
> **username_1**: You want a proof that if the indifference curve has the same slope as the budget line on a point on the budget line, then that point solves the UMP?  
> **username_0**: @MichaelGreinecker We can define the tangency condition in a slightly different way: Suppose MRS$_{xy} = \frac{p_1}{p_2}$ at the point $(a,b)$. Define MRS as $\frac{dy(x)}{dx}$ where $y : (a-\epsilon, a+\epsilon) \to \mathbb{R}$ for an appropriately small $\epsilon > 0$. This is to say that only the IC around the point of tangency is differentiable. That IC, as an whole, or other ICs or even $U$ are not assumed to be differentiable (or partially differentiable).  
> **username_0**: @MichaelGreinecker Take $U(x,y) = x+y+ \lfloor x+y \rfloor$ for example. It's not differentiable when $x+y \in \mathbb{Z}$, but the IC given by $U(x,y) = 10$ is tangent to the budget line $B : x+y=5$. Here, optimality and tangency condition hold despite the partial derivatives not being defined.  
> **username_0**: @MichaelGreinecker The point is, I want a proof of precisely what you described but without assumptions that are not necessary.  
> **username_2**: Can you please edit your question so that it contains all the information posted in the comments? Presently this is kind of hard to read. If Michael's comment helped, you can also use it to clarify your question.  
> **username_3**: @xyz123 By the way, utility function $u(x, y) = x + y + \lfloor x +y \rfloor$ represents the same preference as the utility function $v(x, y) = x + y$ which is a differentiable and quasi-concave function.  
> **username_0**: @Amit The example was to justify my point that partial derivatives may not necessarily exist.  
> **username_0**: @Giskard I added the comment to the post.  
> **username_3**: @xyz123 I know that there exist utility functions that are non-differentiable so that partial derivatives may not exist. I was just trying to make a point that some preferences represented by non-differentiable utility functions might also have other utility representations that are differentiable so that they can be treated just like the way we treat differentiable utility functions.  

---

## 1 answers

### Answer by username_2 on 2022-06-25. Score: 3 | Accepted answer

The proof below is not very rigorous - I do not discuss discontinuities - but I think it captures the logic well.

---

From [Wikipedia](https://en.wikipedia.org/wiki/Convex_function#Functions_of_one_variable):

> Suppose $f$ is a function of one real variable defined on an interval, and let
> $$ R(x\_{1},x\_{2})={\frac {f(x\_{2})-f(x\_{1})}{x\_{2}-x\_{1}}} $$
> [...]
> $f$ is convex if and only if $R(x\_{1},x\_{2})$ is monotonically non-decreasing in $x\_{1}$ for every fixed $x\_{2}$ (or vice versa).

Let us denote the point of tangency by $(x^\*,y^\*)$, and let us denote the function describing the indifference curve at level $U(x^\*,y^\*)$ by $f$.

Since $U$ is differentiable at $(x^\*,y^\*)$, we have
$$
MRS(x^\*,y^\*) = f'(x^\*).
$$

By the definition of derivatives, we have,
$$
f'(x^\*) = \lim\_{h \to 0} \frac{f(x^\* + h)-f(x^\*)}{x^\* + h - x^\*}.
$$

Using the Wikipedia lemma, we have for any $x < x^\*$
$$
\frac{f(x)-f(x^\*)}{x - x^\*} \leq f'(x^\*)
$$
or
$$
f(x) \geq f(x^\*) + (x - x^\*)f'(x^\*).
$$
Thus any basket $(x,y)$ on the indifference curve defined by $f$ that is to the left $x^\*$ is above or on the tangent line. Since the tangent line coincides with the budget constraint, these baskets are unattainable or just attainable. Assuming positive prices, no baskets of higher utility are attainable, as that would violate monotonicity.

One can similarly show that
$$
f(x) \geq f(x^\*) + (x - x^\*)f'(x^\*)
$$
also hold for all $x > x^\*$, thus baskets on the indifference curve to the right of $x$ will also be unattainable or just attainable.

> **Comments:**
> **username_0**: (+1) Very interesting, I like the idea! I think it does not work for all continuous case either, for example when $U(x,y) = \max(x,y)$. We can't define $f(x)$ for its ICs. I want to give this some time before accepting it in case @MichaelGreinecker wants to chime in later or if you would like to capture some more cases.  
> **username_2**: @xyz123 $\max(x,y)$ is not a convex utility function, so it is not surprising that this proof does not cover it.  
> **username_2**: In case you meant $\min(x,y)$, you wrote "*not necessarily differentiable except for a region of the corresponding IC around the tangency point*", but $\min(x,y)$ is not differentiable around the tangency point.  
> **username_0**: That's true, I missed it. Does that mean we can always say that for convex and monotone preferences, the indifference curves can be written as a function of $x$?  
> **username_2**: If you don't insist on full range (non-weak) monotonicity is enough.  
> **username_0**: With non-weak monotonicity, it is certainly enough as a point lying on top of another needs to have more utility (by definition of strong monotonicity). I am not assuming strong monotonicity in my case. When the ICs are straight lines moving toward the right, then it's an example where convex and monotone preferences won't have a point of tangency. Do we need strict convexity or strict monotonicity (or both) to ensure the tangency condition holds?  
> **username_2**: We seem to be going back and forth over the same details. In your question you are *assuming* local differentiabality and tangency, and asking under what conditions can we assume this to be optimal. In your last comment you seem to do away with this assumption.  
> **username_2**: In case you would have something like Leontief preferences with the kink smoothed out, you would have tangency and weak monotonicity. This is one of those special cases which the proof would have to discuss seperately. It would be fine though, as the proof works for cases where the slope is 0.  
> **username_0**: Ignore the example please; I was tired and sleepy. There could be several such special (or even pathological) cases which can not distinctively be turned into a *case*. The proof works where the slope is $0$, but not in the vertical line portions of the Leontief ICs. (I'm accepting this answer as I think generalizing it would be difficult. Strangely, I see these generalizations being quoted often but never really with a proof.)  
> **username_2**: "*The proof works where the slope is 0, but not in the vertical line portions of the Leontief ICs.*" It would though, as you can treat these cases seperately, and by switching the axes you can turn the 'infinite' slope into 0, making the proof directly applicable. IMO the proof lacks rigor, but I see no flaws.  
> **username_2**: Seems like you are very quick with your conclusions and examples; perhaps take more time?  
> **username_3**: @xyz123 I would like to see (if you can provide some reference) where these generalisations are quoted without the proof.  

---

