# Prove that budget constraint is Lower Hemi Continuos (LHC)

I need to prove that the following constraint is LHC.

$$B=\{x \in R^n : px\leqslant pw)$$

But Im not capable of finding and sequence $$\{x_n\}$$ such that $$x_n \in B(p_n,w_n) \forall n$$ and that $$x_n\longrightarrow x$$.

I tried with setting $$x_n=\frac{x}{1+\beta^n}$$, $$p_n=\frac{p}{1+\beta^n}$$ , $$w_n=\frac{w}{1+\beta^n}$$ because in that case $$x_n\longrightarrow x$$. and $$x_n \in B(p_n,w_n) \forall n$$ but I thing I'm not covering the $$\forall p_n, w_n$$ part.

I don't believe it is lower semicontinous.

Let $$w = (0,\dots,0)$$, $$p \in \mathbb{R}^n_+$$ be any vector such that $$p_1 = 0$$ (the first coordinate being 0).

The allocation $$x=(1,0,\dots,0) \in B(p,w)$$.

Define the sequence $$p_n = p + (\frac{1}{n},0,\dots,0)$$ and $$w_n = (\frac{1}{n},0,\dots,0)$$. $$w_n \rightarrow w$$ and $$p_n \rightarrow p$$.

For any $$x^n \in B(p_n,w_n)$$, $$p_n x^n_1 \leq w_np_n$$, so $$x_1^n \leq \frac{1}{n}$$.

Hence for any sequence such that $$x^n \in B(p_n,w_n)$$, $$x^n \not \rightarrow x$$.

• Actually, it is. It is proved in De la Fuente's book problem 2.2 Chap 8.I didn't understand what you have done up there, but DLF prove is magical in the sense that I wouldn't have came with that answer. Good name BTW – Martin Mendina Jun 21 '20 at 19:26
• @MartinMendina De la Fuente proves that the Budget correspondence is lhc at points where all prices are positive and the endowment is not zero; so there is no contradiction here. By the way: de la Fuente takes the budget set to include only points in the nonnegative orthant, which is different from your definition. – Michael Greinecker Jun 22 '20 at 9:56
• Yes you're right Michael, I was not precise enough. – Martin Mendina Jun 22 '20 at 12:09

One approach could be the following. For a $$(p_n,w_n)$$ in the sequence and $$x \in B(p,w)$$ define: $$\alpha_n = 1 \text{ if } p_n x \le w_n$$ and $$\alpha_n = \frac{w_n}{p_n x} \text{ if } p_n x > w_n$$ Then define: $$x_n = \alpha_n x$$ Here $$x_n$$ equals $$x$$ if $$x$$ is in the budget $$B(p_n,w_n)$$. If not, then $$x_n$$ is the radial projection of $$x$$ onto the budget line.

Notice that $$p_n x_n = p_n x \le w_n \text{ if } p_n x \le w_n$$ and $$p_n x_n = p_n \frac{w_n}{p_n x} x = w_n \text{ if } p_n x > w_n$$ which shows that $$x_n \in B(p_n, w_n)$$.

As such, the only thing left to show is that $$x_n \to x$$ or equivalently, $$\alpha_n \to 1$$.

If $$p_n \to p \gg 0$$ and $$w_n \to w > 0$$. Then for $$n$$ big enough one can show that $$\alpha_n = \min\left\{\frac{w_n}{p_n x}, 1\right\}.$$ As the min function is continuous, it follows that $$\lim_n \alpha_n = \lim_n \left(\min \left\{\frac{w_n}{p_n x}, 1\right\}\right) = \min\left\{\frac{w}{p x},1\right\} = 1.$$