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.

Help please and thanks in advance.


2 Answers 2


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 $x^n \in B(p_n,w_n)$, $x^n \not \rightarrow x$.

  • $\begingroup$ 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 $\endgroup$
    – mmendina
    Commented Jun 21, 2020 at 19:26
  • 2
    $\begingroup$ @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. $\endgroup$ Commented Jun 22, 2020 at 9:56
  • $\begingroup$ Yes you're right Michael, I was not precise enough. $\endgroup$
    – mmendina
    Commented Jun 22, 2020 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 $$ \lim_n \alpha_n = \lim_n \left(\min \left\{\frac{w_n}{p_n x}, 1\right\}\right) = \min\left\{\frac{w}{p x},1\right\} = 1.$$


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