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Given the set

$V(q)=\{(x_1,x_2) \in \mathbb R^2:ax_1≥\log y \text{ and } bx_2≥\log y\}$

with $a$, $b$, and $y$ strictly positive, I have to show that $V$ is closed and convex.

My idea for convexity is to show that:

$z=t\cdot ax_1+(1-t)\cdot bx_2≥\log y$

with $t∈[0,1]$ (or $t∈(0,1)$?) so that $z∈V(q)$.

I do not have ideas at the moment on how to show that is closed and convex, could you help me?

EDIT: in general, given a vector of inputs $(x_1,...,x_{L-1})$, only one output $y$, $V$ can be defined as

$V(q)=(x∈ℝ_+^{L-1}:(y,-x)∈Y)$

where $Y$ is the set of all combinations of inputs and outputs. $V$ can take various mathematical forms, which one is that reported in the text of my question.

Levi
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Kolmogorovwannabe
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  • Is $y$ independent of $x_1,x_2$? If it were, it would be easy to show both since $V(q)$ (if I'm understanding your definition correctly -- please fix your typos and use the macro \log to produce $\log$) would simply be a closed rectangle that extends infinitely into the positive reals, i.e. $V(q)=[\frac{\log y}{a},,+\infty)\times[\frac{\log y}{b},,+\infty)$. – B. Núñez Nov 10 '19 at 09:01
  • No, $x_1$ and $x_2$ are inputs for $y$ (which is the output), so they are not independent. – Kolmogorovwannabe Nov 10 '19 at 09:10
  • @Kolmogorovwannabe Can you then rewrite the defintion of $V(q)$ which is not clear enough. What is $y$ versus $x_1,x_2$? – mathcounterexamples.net Nov 10 '19 at 09:34
  • I added an edit for a general understanding on what $V(q)$ is. To put it in simple words, the $x$ are inputs used to produce $y$. All the details provided in the question are sufficient to solve it. Intuitively I know how to do it, but I need maths to show it. – Kolmogorovwannabe Nov 10 '19 at 09:51

1 Answers1

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Recall that if $f: X \rightarrow Y$ is a continuous map between, say, metric spaces, then the pre-image $f^{-1}(C)$ of a closed set $C$ is closed. Applying this, the set $V$ is closed because it is the intersection of pre-images of the closed set $[\log y, \infty)$ under the continuous functions $(x_1, x_2) \mapsto ax_1$ and $(x_1,x_2) \mapsto bx_2$. As far as convexity goes, your approach is almost correct. Let $$t(x_1, x_2) + (1-t)(x_1', x_2') = (tx_1 + (1-t)x_1', tx_2 +(1-t)x_2)$$ be an element of the line between $(x_1, x_2)$ and $(x_1', x_2')$, both of which are assumed to lie in $V$. Compute for the first cooredinate, using $x_1,x_1' \geq \log y$, $$a (tx_1+(1-t)x_1') = tax_1 + (1-t)ax_1' \geq t\log y + (1-t) \log y = \log y.$$ Similarly, for the second coordinate, $$b(tx_2 + (1-t)x_2') \geq \log y.$$

Levi
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  • Wonderful. Only two questions and for me this is the best answer:
    1. Convexity is perfectly clear to me, but in $a (tx_1+(1-t)x_1') = tax_1 + (1-t)x_1' \geq t\log y + (1-t) \log y = \log y$, there is an $a$ missing before $(1-t)x_1'$ right?
    2. what do you exactly mean with "pre-images"? Could you argument in words the rationale behind your proof?
     – Kolmogorovwannabe Nov 10 '19 at 10:57
  • @Kolmogorovwannabe Thanks, found the mistake. About being closed, I added a first sentence. Does this help or what definition of "closed" would you like to have checked? – Levi Nov 10 '19 at 11:00