# Can such an accumulation point be local maximum?

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?

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

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