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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?

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.

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    $\begingroup$ 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? $\endgroup$
    – Michael Greinecker
    Commented Jun 24, 2022 at 22:21
  • $\begingroup$ @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). $\endgroup$
    – xyz123
    Commented Jun 25, 2022 at 0:32
  • $\begingroup$ @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. $\endgroup$
    – xyz123
    Commented Jun 25, 2022 at 0:35
  • $\begingroup$ @MichaelGreinecker The point is, I want a proof of precisely what you described but without assumptions that are not necessary. $\endgroup$
    – xyz123
    Commented Jun 25, 2022 at 0:38
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    $\begingroup$ @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. $\endgroup$
    – Amit
    Commented Jun 25, 2022 at 9:28

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

From Wikipedia:

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.

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  • $\begingroup$ (+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. $\endgroup$
    – xyz123
    Commented Jun 25, 2022 at 14:38
  • $\begingroup$ @xyz123 $\max(x,y)$ is not a convex utility function, so it is not surprising that this proof does not cover it. $\endgroup$
    – Giskard
    Commented Jun 25, 2022 at 14:43
  • $\begingroup$ 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. $\endgroup$
    – Giskard
    Commented Jun 25, 2022 at 14:44
  • $\begingroup$ 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 $x$? $\endgroup$
    – xyz123
    Commented Jun 25, 2022 at 14:58
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    $\begingroup$ @xyz123 I would like to see (if you can provide some reference) where these generalisations are quoted without the proof. $\endgroup$
    – Amit
    Commented Jun 26, 2022 at 16:49

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