Does no unstable manifold imply Lyapunov stability?

Let f:Rn→Rnf : \mathbb{R}^n \to \mathbb{R}^n be continuous and suppose 00 is a fixed point of ff (that is, f(0)=0f(0) = 0).
Does Wu(0)=∅W^u(0) = \varnothing imply that 00 is a Lyapunov stable fixed point of ff?
My guess is yes, but how can it be proved?
If not, what is a counterexample?
This seems rather fundamental, but I cannot find a solution online.

Definitions: the unstable manifold of 00, denoted Wu(0)W^u(0), is the set of all x≠0x \ne 0 for which there exists a backwards orbit with fi(x)→0f^i(x) \to 0 as i→−∞i \to -\infty.
The fixed point 00 is said to be Lyapunov stable if for every neighbourhood UU of 00 there exists a (smaller) neighbourhood VV of 00 such that fi(x)∈Uf^i(x) \in U for every x∈Vx \in V and every i≥0i \ge 0.

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Let’s imagine a system with a stable manifold Ws(0)W^s(0) and a center manifold Wc(0)W^c(0) through 00. Let Bϵ(0)B_{\epsilon}(0) be an ϵ\epsilon-ball around 00. Is there necessarily a δ\delta-ball Bδ(0)B_{\delta}(0) such that the points in Wc(0)∩Bδ(0)W^c(0)\cap B_{\delta}(0) stay within Bϵ(0)B_{\epsilon}(0) for all t≥0t \geq 0?
– Michael Lee
2 days ago

  

 

It depends on the dynamics on Wc(0)W^c(0). Wc(0)W^c(0) can contain parts of Ws(0)W^s(0) and Wu(0)W^u(0).
– djws
2 days ago

  

 

This all depends on how you define WsW^s and WuW^u. If you define them only as manifolds which are tangent to EsE^s and EuE^u respectively, then @MichaelLee gave an answer already. However, this requires ff to be at least C1C^1 or so. And there is another definition, which is written in your post, based purely on which trajectories converge to point in forward/backward time.
– Evgeny
yesterday

  

 

Yes I mean Ws(0)W^s(0) and Wu(0)W^u(0) to be points that converge to 00 under forwards and backwards iteration, respectively. If ff is differentiable and 00 is hyperbolic, then Ws(0)W^s(0) and Wu(0)W^u(0) are tangent to Es(0)E^s(0) and Eu(0)E^u(0), respectively.
– djws
yesterday

  

 

It’s not necessary for 00 to be hyperbolic, it’s enough to be differentiable. If it’s differentiable, then you have no WuW^u, but definitely have either WsW^s or WcW^c (or both). And if you have WcW^c, as @MichaelLee said, you can make point Lyapunov unstable — it won’t violate having no WuW^u.
– Evgeny
yesterday

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