If $f,f_m$ are Lipschitz with $\lim_{m\to+\infty} d(f_m(x),f(x)) = 0 $, is it true that the fixed point of $f_m$ converges to the fixed point of $f$?

Question

I'm going through a proof in which at one point, the authors make the following statement :

Let $(X,d)$ be a complete metric space and $f:X\to X$ and $f_m:X\to X$ Lipschitz with $lip(f)<1$ and $lip(f_m)<1\; \forall m\in\mathbb{N}$. Let $x^*_m$ the fixed point of $f_m$ and $x^*$ the fixed point of $f$. If for each $x\in X$ we have that
$$\lim_{m\to+\infty} d(f_m(x),f(x)) = 0, $$ it follows that $$\lim_{m\to+\infty} d(x^*_m,x^*) = 0 .$$

I tried to figure out how they came up with this by an argument along the lines of \begin{align*}d(x^*_m,x^*) &= d(f_m(x^*_m),f(x^*)) \leq d(f_m(x^*_m),f(x_m^*))+ d(f(x_m^*),f(x^*))\\&\leq d(f_m(x^*_m),f(x_m^*))+lip(f)d(x_m^*,x^*). \end{align*}

Then $$0\leq (1-lip(f))d(f(x_m^*),f(x^*))\leq d(f_m(x^*_m),f(x_m^*)).$$ But now we can't apply the first limit because $x_m^*$ depends on $m$.

Also another thing I tried is by using the fact that $$\lim_{n \to+\infty}d(f^{\circ n}(x),x^*)=0 \; \forall x\in X$$ Where $f^{\circ n} = \underbrace{f\circ f\circ \dots \circ f}_{\text {n times}}.$

Then for a fixed $x\in X$ \begin{equation*}d(x^*_m,x^*)\leq d(x^*_m,f^{\circ n}_m(x)) + d(f^{\circ n}_m(x),f^{\circ n}(x))+ d(f^{\circ n}(x),x^*),\; \forall n\in\mathbb{N} \end{equation*}

Now let $\varepsilon > 0$ and set $N$ big enough so that $d(f^{\circ N}(x),x^*)<\varepsilon/3$ and $d(f_m^{\circ N}(x),x_m^*)< \varepsilon/3$.

Then \begin{align*} d(x^*_m,x^*) &\leq d(x^*_m,f^{\circ N}_m(x)) + d(f^{\circ N}_m(x),f^{\circ N}(x))+ d(f^{\circ N}(x),x^*)\\&<2\varepsilon/3 + d(f^{\circ N}_m(x),f^{\circ N}(x)). \end{align*}

Now it can be shown that $\lim\limits_{m\to+\infty} d(f^{\circ N}_m(x),f^{\circ N}(x)) =0$, so there is some $M$ such that $d(f^{\circ N}_m(x),f^{\circ N}(x)) <\varepsilon/3\; \forall m>M$, from which we get $$d(x^*_m,x^*)<\varepsilon.$$

But I feel like something is not right in my second attempt.

Is something missing from the statement?

Answer 1

If there exists $c\in(0,1)$ such that $lip(f_{n})\leq c$ for all $n$, then it is easy. For, observe that \begin{eqnarray*} d(x_{n}^{\ast},x^{\ast}) & = & d(f_{n}(x_{n}^{\ast}),f(x^{\ast}))\\ & \leq & d(f_{n}(x_{n}^{\ast}),f_{n}(x^{\ast}))+d(f_{n}(x^{\ast}),f(x^{\ast}))\\ & \leq & lip(f_{n})d(x_{n}^{\ast},x^{\ast})+d(f_{n}(x^{\ast}),f(x^{\ast}))\\ & \leq & c\cdot d(x_{n}^{\ast},x^{\ast})+d(f_{n}(x^{\ast}),f(x^{\ast})) \end{eqnarray*} Therefore, $d(x_{n}^{\ast},x^{\ast})\leq\frac{1}{1-c}\cdot d(f_{n}(x^{\ast}),f(x^{\ast})).$ Letting $n\rightarrow\infty$ and observe that $d(f_{n}(x^{\ast}),f(x^{\ast}))\rightarrow0,$ the result follows.

Answer 2

I have a strong feeling that one needs uniform convergence and not only pointwise convergence to conclude this. Let me give you a partial answer (I'll think about a counterexample) when we assume $$\lim_{m\to\infty} \sup_{x \in X} d(f_m(x),f(x)) = 0 $$

Indeed note that under this assumption we can write:

$$d(x^*_m,x^*) = d(f_m(x^*_m),f(x^*)) \leq \sup_{y \in X} d(f_m(y),f(y)) \xrightarrow{m \rightarrow \infty} 0$$

In your last attemp you cannot say "set $N$ big enough so that $d(f_m^{\circ N}(x),x_m^*)< \varepsilon/3$" since to write this passage you need uniform convergence (as in the comment is kindly pointed out - I slipped on the dependence of $N$ on $m$, as written in the comment below that does not allow to pass to the limit).

Answer 3

Your theorem is wrong. Next time, I would like you to check your statement carefully to save time of other MSE fellows who want to help you. You know, it is not always easy to tell whether a theorem is true or not.

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Counter example:
Choose $X= l^1(\mathbb{N})$, for all $a \in l^1$ and $m \in \mathbb{N}_*$, $f(a)=0$ and $$f_m(a)=(0,\dots,\underbrace{\frac{1}{m}+(1-\frac{1}{m})a_m}_{m-\text{th position}}, \dots )$$ where $a=(a_0,a_1,\dots,...)$.
So firstly, you can see that $X$ is complete with its canonical norm.
Secondly, $f_m$ converges pointwise to $f$ ( note that $X$ is $l^1$) And thirdly, the fixed points of $f_m$ are $e_m$, which doesn't converge to $0$- the only fixed point of $f$.