# When does $x_k$ converge if we know that $f(x_k)$ converges

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 $$\lim_{k\rightarrow \infty} x_{k} = x^{*}$$?

• 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. – nicomezi Jan 22 at 14:42

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.

• 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? – Olivier Roche Jan 22 at 16:16
• 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 – Arararararagi-kun Jan 22 at 16:27
• Yes, thanks. :) – Olivier Roche Jan 22 at 19:31
• @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. – User257 Jan 23 at 7:09
• 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. – Arararararagi-kun Jan 23 at 13:50