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Suppose we have a twice continuously differentiable function $f:\mathbb{R}^n \longrightarrow \mathbb{R}$ and $\lim_{k\rightarrow \infty} f(x_{k}) = 0$ for some series $(x_k)_k$.

Are there any conditions on $f$ (like unicity of a root, lipschitz continuity of the gradient, positive definiteness of jacobian matrix,...) that imply existence of the limit $\lim_{k\rightarrow \infty} x_{k}$.

For instance, suppose $x^*$ is the unique solution of $f(x)=0$, do we know $\lim_{k\rightarrow \infty} x_{k} = x^{*}$?

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    $\begingroup$ Root unicity is not enough. Consider $f(x) = x e^{-x}$ and $x_k =k$. Root unicity AND limit of $f(x)$ not being $0$ whenever $|x| \to \infty$ is enough. $\endgroup$ – nicomezi Jan 22 at 14:42
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Clearly we have to have an unique root, for if $a$ and $b$ are distinct roots the sequence $x_n$ equal to $a$ for odd $n$ and $b$ otherwise does not converge. Second, we must have $f(\pm\infty)\neq 0$, because we want the sequence $x_n$ to be bound. If we have both, then we shall obtain what you want: The sequence $x_n$ will be bound, thus will have an accumulation point, let's say $x=\lim x_{k_n}$. As we have $f(x)=\lim f(x_{k_n})=0$, by continuity, we conclude that $x=x*$. Thus we conclude that the accumulations points of the sequence are unique. As the sequence is bounded and has only one accumulation point it must converge to it.

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  • $\begingroup$ You claim "As the sequence is bounded and has only one accumulation point it must converge to it.". I don't think this holds in general. Can you give some more details please? $\endgroup$ – Olivier Roche Jan 22 at 16:16
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    $\begingroup$ If a sequence is bounded and has only one accumulation point then it must converge to it: take any open neighborhood of this point. The number of elements of the sequence outside this neighborhood must be finite, otherwise it would have a sub sequence in a compact set, which would have another accumulation point outside the neighborhood $\endgroup$ – Arararararagi-kun Jan 22 at 16:27
  • $\begingroup$ Yes, thanks. :) $\endgroup$ – Olivier Roche Jan 22 at 19:31
  • $\begingroup$ @Arararararagi-kun So if I understand correctly, for the multivariate case the condition $f(\pm\infty) \neq 0$ just has to be replaced with $\lim_{\|x\| \rightarrow \infty} f(x) \neq 0$ to make sure the sequence is bounded? The rest of the argument will just stay the same. $\endgroup$ – User257 Jan 23 at 7:09
  • $\begingroup$ Actually we must have a little stronger hypothesis, even in the dimension 1 case. We could have a sequence such that its image converge to zero without the function converge to zero. We must ensure that 0 is not an accumulation point of the image of any sequence that goes to infinite. $\endgroup$ – Arararararagi-kun Jan 23 at 13:50

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