<p><span class="math-container">$$\bigg(\frac{1}{2}3\bigg)^2+\bigg(\frac{1}{2}3\bigg)^2=\bigg(\frac{3}{2}\bigg)^2+\bigg(\frac{3}{2}\bigg)^2=\frac{3^2}{2^2}+\frac{3^2}{2^2}=\frac{9}{4}+\frac{9}{4}=\frac{9}{2}&lt;9.$$</span></p>

CC BY-SA 4.0
2023-03-14T17:38:48.357Z
2023-03-14T17:38:48.357Z
Michael Greinecker
<p>I am given a utility function <span class="math-container">$u(x)=x_1^2+x_2^2$</span> and I am asked to see whether this function satisfies strict convexity. The answer is saying this:</p>
<p>We see that <span class="math-container">$u(3,0) = 9$</span>, <span class="math-container">$u(0,3) = 9$</span>, <span class="math-container">$u(1.5,1.5) = 4.5$</span>. Here, <span class="math-container">$x = (3,0)$</span>, <span class="math-container">$y = (0,3)$</span>, and <span class="math-container">$z = (3,0)$</span>, we have <span class="math-container">$x ≳ z$</span> and <span class="math-container">$y ≳ z$</span>, but <span class="math-container">$tx + (1 − t)y ≺ z$</span> for <span class="math-container">$t=1/2$</span>.</p>
<p>I am lost at this part here <span class="math-container">$tx + (1 − t)y ≺ z$</span> for <span class="math-container">$t=1/2$</span>; isn't the utility still 9, which is the same as the utility of every bundle there?</p>

<consumer-theory>
Proving that strict convexity is violated
kyrie3
# Proving that strict convexity is violated

(Asked by: username_0 on 2023-03-14)

I am given a utility function $u(x)=x\_1^2+x\_2^2$ and I am asked to see whether this function satisfies strict convexity. The answer is saying this:

We see that $u(3,0) = 9$, $u(0,3) = 9$, $u(1.5,1.5) = 4.5$. Here, $x = (3,0)$, $y = (0,3)$, and $z = (3,0)$, we have $x ≳ z$ and $y ≳ z$, but $tx + (1 − t)y ≺ z$ for $t=1/2$.

I am lost at this part here $tx + (1 − t)y ≺ z$ for $t=1/2$; isn't the utility still 9, which is the same as the utility of every bundle there?

**Tags:** consumer-theory

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## 1 answers

### Answer by username_1 on 2023-03-14. Score: 2 | Accepted answer

$$\bigg(\frac{1}{2}3\bigg)^2+\bigg(\frac{1}{2}3\bigg)^2=\bigg(\frac{3}{2}\bigg)^2+\bigg(\frac{3}{2}\bigg)^2=\frac{3^2}{2^2}+\frac{3^2}{2^2}=\frac{9}{4}+\frac{9}{4}=\frac{9}{2}<9.$$


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