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Let $f:\mathbb{R}^n\rightarrow ]-\infty,+\infty]$ be proper and strictly convex. Show that $f$ has at most one minimizer.

I am hoping someone can give me some feedback on the proof for this. I feel that it may not be rigorous enough.

proof: Let $\mu=\inf_{x\in\mathbb{R}^n} f(x)$ (i.e. $\mu$ is a global infimum) and choose a sequence $(x_n)_{n=1}^{\infty}$ in $\mathbb{R}^n$ such that $f(x_n)\rightarrow\mu.$ Then $(x_n)$ is a bounded sequence

[otherwise $(x_n)\rightarrow+\infty$, so $f(x_n)\rightarrow+\infty$. Therefore, $\mu\rightarrow+\infty$, which means that $dom(f)=\emptyset$, which negates our assumption that $f$ is proper.]

Moreover, by the Bolzano-Wierstrass theorem there exists a convergent subsequence $(x_{n_{k}})_{k=1}^{\infty}$, which means $(x_{n_{k}})\rightarrow x\in\mathbb{R}^{n}$. Thus, we see that \begin{align} \mu&\leq f(x)\;\;\;\;\;\;\;\;\;\;\;\;\;\qquad\text{as $f$ is lsc}\\ &\leq\liminf_{k \to \infty} \;(f(x_{n_k}))\\ &=\mu. \end{align}

Thus, $\mu=f(x)$ is our minimizer, and since $f$ is strictly convex we get that $\mu$ is unique.$\qquad\;\;\;\;\Box$

gerw
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Jeremy
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1 Answers1

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There are several problems with your proof:

  • You do not prove the assertion. Indeed, if your proof would be correct, you would have shown that $f$ has a at least a minimizer.
  • I do not get the point of your last sentence. Indeed, $\mu$ is always unique, by definition, independently of the properties of $f$.
  • Your proof is wrong:
    • Indeed, your minimizing sequence $\{x_n\}$ might not be bounded. Consider the counterexample $f(x) = \mathrm{e}^x$.
    • $f$ might not be lower semicontinuous. Consider $$f(x) = \begin{cases} x^2 & x > 0 \\ +\infty & x \le 0\end{cases}$$ and $x_n = 1/n$.

Here is a nice start for a correct proof: Let $x_1$, $x_2$ be global minimizers of $f$. Set $x_3 = \frac{x_1 + x_2}{2}$. Then,....

gerw
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