# Proving duality of UMP and EMP arguing with continuity of utility

In Mas-Colell et al.'s Microeconomic Theory Proposition 3.E.1(ii) (p. 58) states that if $$\succsim$$ is a rational (i.e. complete and transitive), continuous, and locally nonsatiated preference relation on ($$X\equiv\mathbb{R}_{+}^L$$) represented by continuous utility function $$u(\cdot)$$, then $$x^*\in\text{arg}\min\{p\cdot x:u(x)\ge \underline{u}>u(0)\}$$ implies $$x^*\in\text{arg}\max\{u(x):p\cdot x\le p\cdot x^*\}$$ (in words, that under the assumptions the solution to the EMP also solves the UMP).

Now I have trouble deciphering a step of the proof (bolded). The proof goes as follows.

Proof. Suppose $$x^*\in\text{arg}\min\{p\cdot x:u(x)\ge u(x^*)>u(0)\}$$ and $$x^*\notin\text{arg}\max\{u(x):p\cdot x\le p\cdot x^*\}$$. First we note that $$x^*\ne0$$. Now, by our hypothesis we have that there exists $$x'\in X$$ with $$u(x')>u(x^*)$$ and $$p\cdot x'\le p\cdot x^*$$. Consider a bundle $$x''=\alpha x'$$ where $$\alpha\in(0,1)$$ ($$x''$$ is a scaled version of $$x'$$). By continuity of $$u(\cdot)$$, if $$\alpha$$ is close enough to $$1$$, then we will have $$u(x'')>u(x^*)$$ and $$p\cdot x''. And the conclusion clearly follows from the previous argument. $$\square$$

I don't understand the argument bolded (which I transcribed exactly as in the book), I suppose it has something to do with continuity of preferences that in the limit (i.e. "$$\alpha$$ close to $$1$$") $$x''\succ x'$$ but I don't see how, I think I should build some sort of sequence like $$(1-1/(n+1))x'$$ that in the limit has that property, but not sure how to see this argument.

I would be very grateful if anyone can point me the underlying argument of the reasoning mentioned.

Thanks.

By definition, a function is continuous if arbitrarily small changes in its value can be assured by choosing sufficiently small changes of its argument. Since $$u(x')>u(x^*)$$ and the utility function $$u$$ is continuous, a small enough change from $$x'$$ from $$x''$$ will ensure that $$u(x'')>u(x^*)$$ too.

Formally, let $$\varepsilon=\frac{1}{2}[u(x')-u(x^*)]$$. Then, by continuity of $$u$$, there exists a $$\delta>0$$ such that for all $$x\in X$$

$$|x-x'|<\delta \implies u(x')-\varepsilon

In particular, there exists a $$\delta>0$$ such that for all $$x\in X$$:

$$|x-x'|<\delta \implies u(x)>u(x')-\varepsilon=\frac{1}{2}[u(x')+u(x^*)]>u(x^*)$$

In particular, we know that if $$x''=\alpha x'$$ with: $$|\alpha x'-x'|<\delta \iff (1-\alpha)<\frac{\delta}{|x'|} \iff \alpha>1-\frac{\delta}{|x'|}$$

then $$u(x'')>u(x^*)$$.

• Thanks for your answer. How do you guarantee that $\alpha\in(0,1)$? Sep 8 at 15:30
• Since $\delta>0$, it must be that $1-\frac{\delta}{|x'|}<1$, so there certainly exist $\alpha\in(0,1)$ such that $\alpha>1-\frac{\delta}{|x'|}$.
– smcc
Sep 8 at 16:44