<p>We can make life easier taking a monotonic transformation of <span class="math-container">$$U=log (x_1)^2+log (x_2)^2.$$</span></p>
<p>This is possible as monotonic transformations of a utility function represent the same preferences.</p>
<p>We can see that the utility function above can be reconducted to a Cobb-Douglas utility function   <span class="math-container">$U(x_1,x_2)= x_1^ax_2^b$</span>, <span class="math-container">$a,b&gt;0$</span>.</p>
<p>Take</p>
<p><span class="math-container">$U_1(x_1,x_2)=e^{log (x_1)^2+log (x_2)^2}= x_1^2 x_2^2,   $</span></p>
<p>which can be furthermore transformed (as the square is a monotonic transformation for positive values) as:</p>
<p><span class="math-container">$$U_2(x_1,x_2)=x_1 x_2.$$</span></p>
<p>The indifference curves are given by:</p>
<p><span class="math-container">$$U_2(x_1,x_2)=x_1 x_2=c.$$</span> <span class="math-container">$c$</span> constant,</p>
<p>that is:</p>
<p><span class="math-container">$$x_1=\frac{c}{x_2}.$$</span></p>
<p>The indifference curves are hyperbolas, that are convex, the preferences are convex.</p>

CC BY-SA 4.0
2022-11-22T14:50:22.837Z
Isn't the question whether the utility function is convex? (As opposed to whether it generates convex indifference curves, or rationalises convex preferences.)
afreelunch
CC BY-SA 4.0
2022-11-22T15:06:47.163Z
They are connected questions. A  preference can defined as ' convex if the upper contour set is convex". This definition is equivalent to the definition with $\lambda$ and $[0,1]$  and so on. To see these definitions you can refer to Mas Colell, *Microeconomic Theory*.
BakerStreet
CC BY-SA 4.0
2022-11-22T15:08:16.547Z
I don't think the connection is clear at all. For example, Cobb Douglas preferences can be represented using a concave utility function, but generate convex indifference curves. In contrast, there *is* a connection between convexity of indifference curves and *quasi*-concavity of the utility function.
afreelunch
CC BY-SA 4.0
2022-11-22T15:14:24.343Z
Yes, you are right, but here we are talking about the convexity of preferences and indifference curves, the question of the OP is about convexity of preferences,  that can be defined  the two way I mentioned above.  We are not talking  about concavity of utility functions and convexity of indifference curves, this is another question..
BakerStreet
CC BY-SA 4.0
2022-11-22T15:21:38.443Z
@BakerStreet If you read OP's question, it seems to be about the convexity of the utility function. Like you, I suspected that this was not the intended question.
afreelunch
CC BY-SA 4.0
2022-11-22T15:23:01.153Z
@afreelunch the title explicitly says "if the preference represented by the utility function is convex"
reindeer
CC BY-SA 4.0
2022-11-22T15:24:36.277Z
@reindeer My apologies, I seem to have missed that! I was reading the text of the question, where you wrote down the utility function and then asked if it was convex.
afreelunch
CC BY-SA 4.0
2022-11-22T15:27:43.210Z
@ freelunch your observations are interesting , and it is not a very easy topic, you can go in depht using a microeconomic textbook.  But the question is clearly about the convexity of preferences, as the OP confirms.
BakerStreet
CC BY-SA 4.0
2022-11-22T15:31:25.823Z
@BakerStreet So, if the preferences are convex, in terms of the utility function: does it mean, the utility function evaluated at convex combination should yield a higher utility?
reindeer
CC BY-SA 4.0
2022-11-22T15:38:31.153Z
Think about the graph of the usual indifference curves, convex as in Cobb-Douglas, and take the segment of convex combinations: you will see immediatly the answer to your question.
BakerStreet
CC BY-SA 4.0
2022-11-20T10:56:19.663Z
2022-11-22T12:42:01.010Z
2022-11-22T12:42:01.010Z
BakerStreet
<p>A proof that used a calculator:</p>
<p>Let <span class="math-container">$y = (\frac{3}{4},\frac{1}{4}), z = (\frac{1}{4},\frac{3}{4}), x = (e,1)$</span></p>
<p>We have that</p>
<p><span class="math-container">$U(x) = (\log(e))^2 + (\log(1))^2 = 1^2 + 0^2 = 1 + 0 = 1$</span>,</p>
<p><span class="math-container">$U(y) = (\log{\frac{3}{4}})^2 + (\log{\frac{1}{4}})^2 = (\log(3) - \log(4))^2 + (-\log(4))^2 = (\log(4) - \log(3))^2 + (\log(4))^2 = (\log(4))^2 - 2 \log(3)\log(4) + (\log(4))^2 = 2 (\log(4))^2 - 2 \log(3) \log(4)$</span></p>
<p><span class="math-container">$= 2 (2 \log(2))^2 -4 \log(3) \log(2) = 8 \log(2) - 4 \log(3) \log(2) = (8-4\log(3)) \log(2) \approx 2$</span></p>
<p>Note that <span class="math-container">$U(x_1,x_2)$</span> is symmetric in <span class="math-container">$x_1, x_2$</span>.</p>
<p>From symmetry, <span class="math-container">$U(z) = (8-4\log(3)) \log(2) \approx 2$</span>.</p>
<p>This implies that <span class="math-container">$y \geq x$</span> and <span class="math-container">$z \geq x$</span>.</p>
<p>But for <span class="math-container">$\lambda = \frac{1}{2} \in [0,1]$</span>,</p>
<p><span class="math-container">$\lambda y + (1-\lambda)z = \frac{1}{2} y + \frac{1}{2} z = (\frac{1}{2},\frac{1}{2})$</span>,</p>
<p>and</p>
<p><span class="math-container">$U(\frac{1}{2},\frac{1}{2}) = (\log(\frac{1}{2}))^2 + (\log(\frac{1}{2}))^2 = 2 (\log(\frac{1}{2}))^2 = 2 (-\log(2))^2 = 2 (\log(2))^2 \approx 0.96 &lt; 1 = U(x)$</span></p>
<p>Therefore, <span class="math-container">$\lambda y + (1-\lambda)z &lt; x$</span>.</p>
<p>From this, we can conclude that the preference relation represented by the utility function is not convex.</p>

CC BY-SA 4.0
2022-11-22T15:07:45.127Z
the other answer from @BakerStreet contradicts this??
reindeer
CC BY-SA 4.0
2022-11-22T16:19:18.223Z
My answer if for (ln(x_i))^2 and @BakerStreet ‘s is for (ln(x_i^2)). They’re not the same function. I thought you meant the former, hence I answered like that.
Nicolas Torres
CC BY-SA 4.0
2022-11-19T21:47:16.820Z
2022-11-19T21:47:16.820Z
Nicolas Torres
<p>In one line: the utility function you posted is monotonic and continous. Any monotonic function is quasi-concave. Quasi-concavity implies convex preferences.</p>
<p>Quickest possible way to see that this function is strictly concave is to simply notice (no algebra) that the Hessian is negative semi-definite, as the cross derivatives are zero.</p>

CC BY-SA 4.0
2022-11-22T22:13:06.677Z
Maybe be you refer to functions from $\mathbb{R}$  to $\mathbb{R}$. A function defined on an interval of  $\mathbb{R}$  is quasi-concave if it is monotonic, that’ s true.

But the function in the question is a function of two variables. You can’t extend this statement  to function from $\mathbb{R^2}$ to $\mathbb{R}$, because in $\mathbb{R^2}$ , in mathematical analysis,   we don’t speak of ‘monotonic function’.
A definition of monotonicity requires an order. In $\mathbb{R}$  there is an order, in $\mathbb{R^2}$  an order  isn’t defined
BakerStreet
CC BY-SA 4.0
2022-11-22T18:47:40.207Z
2022-11-22T19:16:44.767Z
2022-11-22T19:16:44.767Z
Matteo Bulgarelli
<img src="https://latex.codecogs.com/svg.image?u(x_1,x_2)=ln(x_1)%5E2&plus;ln(x_2)%5E2" title="https://latex.codecogs.com/svg.image?u(x_1,x_2)=ln(x_1)^2+ln(x_2)^2" />
<p>How do I prove that it is convex? How to prove it with the typical t in [0,1] definition? I am having a bad time typing equations here, sorry. please help me out</p>

CC BY-SA 4.0
2022-11-19T13:03:33.833Z
Do you insist on proving it with the [0,1] definition, or would you consider using derivatives?
Giskard
CC BY-SA 4.0
2022-11-19T13:24:12.910Z
@Giskard using derivative? If i have to use the Hessian definition, it should be positive semi-definite? But I don't know what is positive semi-definite though? Also, here is there any difference in proving the preference is convex vs the function is convex?
reindeer
CC BY-SA 4.0
2022-11-20T08:35:00.313Z
Is it $\ln[(x_1)^2]$ or $[\ln(x_1)]^2$?
Richard Hardy
CC BY-SA 4.0
2022-11-22T12:35:31.557Z
I think it is enough clear that it is $ln [(x_1)^2]$ The formula in the question is correct, maybe there are superfluous brackets, but with brackets maybe is better: without brackets, a square, as there is a subscript near the $x_i$, could be confused with a superscript.
BakerStreet
CC BY-SA 4.0
2022-11-22T14:51:22.467Z
@reindeer Just to clarify, you are asking whether this utility function is convex? (As opposed to whether it represents convex preferences, or generates convex indifferences curves.)
afreelunch
CC BY-SA 4.0
2022-11-22T15:04:57.583Z
I think it means to tell if it represents convex preferences, that is if the utility at the convex bundle would yield me a higher utility than the convex combination?
reindeer
CC BY-SA 4.0
2022-11-22T15:10:24.083Z
@reindeer I see; I should emphasise that this is *not* the same question as whether the utility function itself is convex (which one could check using the [0, 1] definition, or examining the associated Hessian matrix).
afreelunch
<microeconomics>
Is the preference represented by the utility function below convex?
reindeer
# Is the preference represented by the utility function below convex?

(Asked by: username_0 on 2022-11-19)

![](https://latex.codecogs.com/svg.image?u(x_1,x_2)=ln(x_1)%5E2+ln(x_2)%5E2 "https://latex.codecogs.com/svg.image?u(x_1,x_2)=ln(x_1)^2+ln(x_2)^2")

How do I prove that it is convex? How to prove it with the typical t in [0,1] definition? I am having a bad time typing equations here, sorry. please help me out

**Tags:** microeconomics
> **Comments:**
> **username_1**: Do you insist on proving it with the [0,1] definition, or would you consider using derivatives?  
> **username_0**: @Giskard using derivative? If i have to use the Hessian definition, it should be positive semi-definite? But I don't know what is positive semi-definite though? Also, here is there any difference in proving the preference is convex vs the function is convex?  
> **username_2**: Is it $\ln[(x_1)^2]$ or $[\ln(x_1)]^2$?  
> **username_3**: I think it is enough clear that it is $ln [(x_1)^2]$ The formula in the question is correct, maybe there are superfluous brackets, but with brackets maybe is better: without brackets, a square, as there is a subscript near the $x_i$, could be confused with a superscript.  
> **username_4**: @reindeer Just to clarify, you are asking whether this utility function is convex? (As opposed to whether it represents convex preferences, or generates convex indifferences curves.)  
> **username_0**: I think it means to tell if it represents convex preferences, that is if the utility at the convex bundle would yield me a higher utility than the convex combination?  
> **username_4**: @reindeer I see; I should emphasise that this is *not* the same question as whether the utility function itself is convex (which one could check using the [0, 1] definition, or examining the associated Hessian matrix).  

---

## 3 answers

### Answer by username_3 on 2022-11-20. Score: 2

We can make life easier taking a monotonic transformation of $$U=log (x\_1)^2+log (x\_2)^2.$$

This is possible as monotonic transformations of a utility function represent the same preferences.

We can see that the utility function above can be reconducted to a Cobb-Douglas utility function $U(x\_1,x\_2)= x\_1^ax\_2^b$, $a,b>0$.

Take

$U\_1(x\_1,x\_2)=e^{log (x\_1)^2+log (x\_2)^2}= x\_1^2 x\_2^2, $

which can be furthermore transformed (as the square is a monotonic transformation for positive values) as:

$$U\_2(x\_1,x\_2)=x\_1 x\_2.$$

The indifference curves are given by:

$$U\_2(x\_1,x\_2)=x\_1 x\_2=c.$$ $c$ constant,

that is:

$$x\_1=\frac{c}{x\_2}.$$

The indifference curves are hyperbolas, that are convex, the preferences are convex.

> **Comments:**
> **username_4**: Isn't the question whether the utility function is convex? (As opposed to whether it generates convex indifference curves, or rationalises convex preferences.)  
> **username_3**: They are connected questions. A  preference can defined as ' convex if the upper contour set is convex". This definition is equivalent to the definition with $\lambda$ and $[0,1]$  and so on. To see these definitions you can refer to Mas Colell, *Microeconomic Theory*.  
> **username_4**: I don't think the connection is clear at all. For example, Cobb Douglas preferences can be represented using a concave utility function, but generate convex indifference curves. In contrast, there *is* a connection between convexity of indifference curves and *quasi*-concavity of the utility function.  
> **username_3**: Yes, you are right, but here we are talking about the convexity of preferences and indifference curves, the question of the OP is about convexity of preferences,  that can be defined  the two way I mentioned above.  We are not talking  about concavity of utility functions and convexity of indifference curves, this is another question..  
> **username_4**: @BakerStreet If you read OP's question, it seems to be about the convexity of the utility function. Like you, I suspected that this was not the intended question.  
> **username_0**: @afreelunch the title explicitly says "if the preference represented by the utility function is convex"  
> **username_4**: @reindeer My apologies, I seem to have missed that! I was reading the text of the question, where you wrote down the utility function and then asked if it was convex.  
> **username_3**: @ freelunch your observations are interesting , and it is not a very easy topic, you can go in depht using a microeconomic textbook.  But the question is clearly about the convexity of preferences, as the OP confirms.  
> **username_0**: @BakerStreet So, if the preferences are convex, in terms of the utility function: does it mean, the utility function evaluated at convex combination should yield a higher utility?  
> **username_3**: Think about the graph of the usual indifference curves, convex as in Cobb-Douglas, and take the segment of convex combinations: you will see immediatly the answer to your question.  

---

### Answer by username_5 on 2022-11-19. Score: 1

A proof that used a calculator:

Let $y = (\frac{3}{4},\frac{1}{4}), z = (\frac{1}{4},\frac{3}{4}), x = (e,1)$

We have that

$U(x) = (\log(e))^2 + (\log(1))^2 = 1^2 + 0^2 = 1 + 0 = 1$,

$U(y) = (\log{\frac{3}{4}})^2 + (\log{\frac{1}{4}})^2 = (\log(3) - \log(4))^2 + (-\log(4))^2 = (\log(4) - \log(3))^2 + (\log(4))^2 = (\log(4))^2 - 2 \log(3)\log(4) + (\log(4))^2 = 2 (\log(4))^2 - 2 \log(3) \log(4)$

$= 2 (2 \log(2))^2 -4 \log(3) \log(2) = 8 \log(2) - 4 \log(3) \log(2) = (8-4\log(3)) \log(2) \approx 2$

Note that $U(x\_1,x\_2)$ is symmetric in $x\_1, x\_2$.

From symmetry, $U(z) = (8-4\log(3)) \log(2) \approx 2$.

This implies that $y \geq x$ and $z \geq x$.

But for $\lambda = \frac{1}{2} \in [0,1]$,

$\lambda y + (1-\lambda)z = \frac{1}{2} y + \frac{1}{2} z = (\frac{1}{2},\frac{1}{2})$,

and

$U(\frac{1}{2},\frac{1}{2}) = (\log(\frac{1}{2}))^2 + (\log(\frac{1}{2}))^2 = 2 (\log(\frac{1}{2}))^2 = 2 (-\log(2))^2 = 2 (\log(2))^2 \approx 0.96 < 1 = U(x)$

Therefore, $\lambda y + (1-\lambda)z < x$.

From this, we can conclude that the preference relation represented by the utility function is not convex.

> **Comments:**
> **username_0**: the other answer from @BakerStreet contradicts this??  
> **username_5**: My answer if for (ln(x_i))^2 and @BakerStreet ‘s is for (ln(x_i^2)). They’re not the same function. I thought you meant the former, hence I answered like that.  

---

### Answer by username_6 on 2022-11-22. Score: 1

In one line: the utility function you posted is monotonic and continous. Any monotonic function is quasi-concave. Quasi-concavity implies convex preferences.

Quickest possible way to see that this function is strictly concave is to simply notice (no algebra) that the Hessian is negative semi-definite, as the cross derivatives are zero.

> **Comments:**
> **username_3**: Maybe be you refer to functions from $\mathbb{R}$  to $\mathbb{R}$. A function defined on an interval of  $\mathbb{R}$  is quasi-concave if it is monotonic, that’ s true.  But the function in the question is a function of two variables. You can’t extend this statement  to function from $\mathbb{R^2}$ to $\mathbb{R}$, because in $\mathbb{R^2}$ , in mathematical analysis,   we don’t speak of ‘monotonic function’. A definition of monotonicity requires an order. In $\mathbb{R}$  there is an order, in $\mathbb{R^2}$  an order  isn’t defined  

---

