<p>One approach could be the following. For a <span class="math-container">$(p_n,w_n)$</span> in the sequence and <span class="math-container">$x \in B(p,w)$</span> define:
<span class="math-container">$$ \alpha_n = 1 \text{ if } p_n x \le w_n$$</span>
and
<span class="math-container">$$ \alpha_n = \frac{w_n}{p_n x} \text{ if } p_n x &gt; w_n$$</span>
Then define:
<span class="math-container">$$ x_n = \alpha_n x$$</span>
Here <span class="math-container">$x_n$</span> equals <span class="math-container">$x$</span> if <span class="math-container">$x$</span> is in the budget <span class="math-container">$B(p_n,w_n)$</span>. If not, then <span class="math-container">$x_n$</span> is the radial projection of <span class="math-container">$x$</span> onto the budget line.</p>
<p>Notice that
<span class="math-container">$$p_n x_n = p_n x \le w_n \text{ if } p_n x \le w_n$$</span>
and
<span class="math-container">$$p_n x_n = p_n \frac{w_n}{p_n x} x = w_n \text{ if } p_n x &gt; w_n$$</span>
which shows that <span class="math-container">$x_n \in B(p_n, w_n)$</span>.</p>
<p>As such, the only thing left to show is that <span class="math-container">$x_n \to x$</span> or equivalently, <span class="math-container">$\alpha_n \to 1$</span>.</p>
<p>If <span class="math-container">$p_n \to p \gg 0$</span> and <span class="math-container">$w_n \to w &gt; 0$</span>. Then for <span class="math-container">$n$</span> big enough one can show that
<span class="math-container">$$ \alpha_n = \min\left\{\frac{w_n}{p_n x}, 1\right\}.$$</span>
As the min function is continuous, it follows that
<span class="math-container">$$ \lim_n \alpha_n = \lim_n \left(\min \left\{\frac{w_n}{p_n x}, 1\right\}\right) = \min\left\{\frac{w}{p x},1\right\} = 1.$$</span></p>

CC BY-SA 4.0
2020-06-22T06:20:32.803Z
2020-06-22T06:20:32.803Z
tdm
<p>I don't believe it is lower semicontinous.</p>
<p>Let <span class="math-container">$w = (0,\dots,0)$</span>, <span class="math-container">$p \in \mathbb{R}^n_+$</span> be any vector such that <span class="math-container">$p_1 = 0$</span> (the first coordinate being 0).</p>
<p>The allocation <span class="math-container">$x=(1,0,\dots,0) \in B(p,w)$</span>.</p>
<p>Define the sequence <span class="math-container">$p_n = p  + (\frac{1}{n},0,\dots,0)$</span> and <span class="math-container">$w_n = (\frac{1}{n},0,\dots,0)$</span>. <span class="math-container">$w_n \rightarrow w$</span> and <span class="math-container">$p_n \rightarrow p$</span>.</p>
<p>For any <span class="math-container">$x^n \in B(p_n,w_n)$</span>, <span class="math-container">$p_n x^n_1 \leq w_np_n$</span>, so <span class="math-container">$x_1^n \leq \frac{1}{n}$</span>.</p>
<p>Hence for any sequence such that <span class="math-container">$x^n \in B(p_n,w_n)$</span>, <span class="math-container">$x^n \not \rightarrow x$</span>.</p>

CC BY-SA 4.0
2020-06-21T19:26:55.137Z
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
mmendina
CC BY-SA 4.0
2020-06-22T09:56:41.767Z
@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
CC BY-SA 4.0
2020-06-22T12:09:08.047Z
Yes you're right Michael, I was not precise enough.
mmendina
CC BY-SA 4.0
2020-06-21T07:07:35.613Z
2020-06-21T07:07:35.613Z
Walrasian Auctioneer
<p>I need to prove that the following constraint is LHC.</p>
<p><span class="math-container">$B=\{x \in R^n : px\leqslant pw)$</span></p>
<p>But Im not capable of finding and sequence <span class="math-container">$\{x_n\}$</span> such that <span class="math-container">$x_n \in B(p_n,w_n) \forall n$</span> and that <span class="math-container">$x_n\longrightarrow x$</span>.</p>
<p>I tried with setting  <span class="math-container">$x_n=\frac{x}{1+\beta^n}$</span>, <span class="math-container">$p_n=\frac{p}{1+\beta^n}$</span> , <span class="math-container">$w_n=\frac{w}{1+\beta^n}$</span> because in that case <span class="math-container">$x_n\longrightarrow x$</span>. and <span class="math-container">$x_n \in B(p_n,w_n) \forall n$</span> but I thing I'm not covering the <span class="math-container">$\forall p_n, w_n$</span> part.</p>
<p>Help please and thanks in advance.</p>

<microeconomics><mathematical-economics><utility><consumer-theory><demand>
Prove that budget constraint is Lower Hemi Continuos (LHC)
mmendina
# Prove that budget constraint is Lower Hemi Continuos (LHC)

(Asked by: username_0 on 2020-06-21)

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.

**Tags:** microeconomics, mathematical-economics, utility, consumer-theory, demand


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## 2 answers

### Answer by username_2 on 2020-06-21. Score: 5

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 $x^n \in B(p\_n,w\_n)$, $x^n \not \rightarrow x$.

> **Comments:**
> **username_0**: 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  
> **username_3**: @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.  
> **username_0**: Yes you're right Michael, I was not precise enough.  

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### Answer by username_1 on 2020-06-22. Score: 4

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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