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$$x\in \Bbb R^2$$ $$4x_1^2 + 4x_2^2 \le 2x_1x_2 - x_1 + 2$$

I don't know how to prove that this set is convex, I can't find anything understandable either.

The only thing I found is:

$f(\theta x + (1 - \theta)y) = \theta f(x) + (1 - \theta)f(x)$
$\theta \in [0,1]$

But I don't know how to use this.

Is there some simple way...or well, quite understandable, to solve this? How to do this?

khernik
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2 Answers2

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Or without much machinery: Your inequalities describe a closed ellipse.

mrf
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The workflow:

If a function $f\in\mathcal C^2$ has a positive definite Hessian (matrix of second derivatives), then it's a convex function.

The Lebesgue sets (i.e. $\{x\in \Bbb R^2: f(x)\le a\}$, $a$ is a constant) of a convex function are convex.

TZakrevskiy
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  • What's the function here? It's an inequality. $f(x)=4x_1^2 + 4x_2^2 - 2x_1x_2 + x_1 - 2$ ? What if the function would not be differentiable? – khernik Dec 15 '13 at 00:00
  • You are studying the set ${f(x)=4x_1^2 + 4x_2^2 - 2x_1x_2 + x_1 - 2\le0}$. Now it's sufficient to prove that your $f$ is convex. If the function is not differentiable, then you have to use other methods. The criterion with Hessian is only sufficient, not necessary. – TZakrevskiy Dec 15 '13 at 00:03
  • Ok, thanks. And btw. positive definite Hessian means it has to have positive $a_{11}$ element and positive determinant? (being 3x3 matrix) – khernik Dec 15 '13 at 00:06
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    In the case $2\times 2$ it's equivalent to say that $a_{11}$ is positive and the determinant is positive. In the case $3\times 3$ it's equivalent ot say that all principal minors are positive. In general case it's equivalent to say that all eigenvalues are positive. – TZakrevskiy Dec 15 '13 at 00:11