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I am working on the following exercise:

Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a continuously differentiable and let $x_0 \in \mathbb{R}^n$. We generate the sequence $\{x_k\}_k$ in the following way:

  • For each $k > 0$ take a $y_{k-1} \in \mathbb{R}^n$ such that a $t_k \in \mathbb{R}$ exists with $$f(x_{k-1}+t \cdot y_{k-1}) < f(x_{k-1})$$ for all $t \in [0,t_{k-1}]$. Then $x_k = x_{k-1}+t_{k-1} \cdot y_{k-1}$.

Suppose the generated sequence $\{x_k\}_k$ is not finite. Can an accumulation point $x^*$ of $\{x_k\}_k$ be a local maximum of $f$? What if $\{x_k\}_k$ is finite?

I do not see how I could solve this exercise. Could you give me a hint?

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  • $\begingroup$ does the sequence need to converge? or it includes a partial limit? $\endgroup$
    – BinyaminR
    Sep 5, 2020 at 19:53
  • $\begingroup$ @BinyaminR: No, we may only assume that $x^*$ is an accumulation point. $\endgroup$
    – 3nondatur
    Sep 5, 2020 at 19:55

1 Answer 1

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let's prove that x* can't be a local maximum.

Let $B_{\epsilon}\left(x^{\star}\right)$ be the ball with radius epsilon for some $\epsilon>0$. Since x* is an accumulation point, we know that there exists some sub sequence $\left(x_{n_{l}}\right)_{l=1}^{\infty}$ that converges to x*. Let $x_{n_{k}}$ be an element of the subsequence such that $x_{n_{k}}\in B_{\dfrac{\epsilon}{3}}\left(x^{\star}\right)$.

All we need in order to finish the proof is to show that $f\left(x_{n_{k}}\right)>f\left(x^{\star}\right)$. But we know that the sequence $\left(f\left(x_{n_{l}}\right)\right)_{l=1}^{\infty}$ is strictly downward monotonic, and so it's limit is strictly smaller then any of its elements. We also know since f is continuous that: $$lim_{l\to\infty}\left(f\left(x_{n_{l}}\right)\right)=f\left(lim_{l\to\infty}\left(x_{n_{l}}\right)\right)=f\left(x^{\star}\right)$$ which finishes the proof

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