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I am currently examining whether the function defined by $$ f(x) = -\sum_{i=1}^3 x_i \ln x_i $$ for $x = (x_1, x_2, x_3) \in \mathbb{R}_{++}^3$ subject to the constraint $x_1 + x_2 + x_3 = 1$ possesses a maximum. Initially, I contemplated applying Lagrange multipliers to solve this constrained optimization problem, but uncertainties regarding the existence of a maximum remained.

The function $f$ appears to be concave and continuous. However, the domain $\mathbb{R}_{++}^3$ is not compact, precluding a direct application of the Weierstrass Theorem to guarantee the existence of a maximum.

Could someone provide guidance or suggest a method to verify whether a maximum exists for this function under the given constraints?

bruno
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1 Answers1

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You can extend the function to $\mathbb{R}^3_+$ by continuity ($-x \ln(x) \rightarrow 0$ when $x \rightarrow 0$). Then, the intersection of $\mathbb{R}^3_+$ with the plane $\{x_1+x_2+x_3=1\}$ is compact, so you can apply the Weierstrass theorem.
(sorry for this short answer, i dont have enough rep to comment)

quantum_unicorn
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  • But the function is not defined at $0$, your argument shows not being closed of the set? – bruno May 02 '24 at 17:21
  • You can extend the function to $0$ by continuity ($−x \ln(x)→0$ when $x→0$) (thanks to the person who upvoted me, I can now comment!) – quantum_unicorn May 02 '24 at 17:23
  • But this is the same as you have written before. I could not get how to extend a function to the point at which it is undefined. – bruno May 02 '24 at 17:26
  • Calculate the limit and see that it converges to $0$, and thus can be extended by continuity. – quantum_unicorn May 02 '24 at 17:29
  • Then the theorem's compactness part becomes redundant because as soon as we have a continuous function, we can extend it. Am I wrong? – bruno May 02 '24 at 17:37
  • No because you can't extend every continuous function, it is case by case – quantum_unicorn May 02 '24 at 17:39