Be $X$ and $\Lambda$ metric spaces, with $X$ complete, and $f\in C(X\times\Lambda,X)$. Suppose that exists some $\alpha\in[0,1)$ and, for each $\lambda\in\Lambda$, some $q(\lambda)\in[0,\alpha]$ such that $$d(f(x,\lambda),f(y,\lambda))\le q(\lambda) d(x,y),\quad\forall x,y\in X$$ By the Banach fixed point theorem, for each $\lambda\in\Lambda$, $f(\cdot,\lambda)$ has a unique fixed point $x(\lambda)$. Prove that $[\lambda\mapsto x(\lambda)]\in C(\Lambda,X)$.

Im totally stuck with this exercise, I dont have a clue about what to do... I tried to show the continuity of $h$ defined as

$$h:\Lambda\to X,\quad \lambda\mapsto x(\lambda)$$

trough the $\epsilon-\delta$ definition of continuity of $f$ and the information of the problem but I cant do it. Geometrically is easy to see it veracity because the function $h$ is just the intersection of $f$ with the plane defined by the set $\{\langle x,y\rangle\in X\times X:x=y\}\times\lambda$.

My work: from the continuity of $f$ we have that for any fixed point $x:=f(x,\lambda_x)$ for any $\epsilon>0$ exists a $\delta>0$ such that

$$d(\langle y,\lambda\rangle,\langle x,\lambda_x\rangle)<\delta\implies d(f(y,\lambda),x)<\epsilon$$

If we set $\langle y,\lambda\rangle=\langle x_0,\lambda_x\rangle$ then for the sequence defined as


that converges to the fixed point $x$, we can rewrite the above as

$$d(\langle x_0,\lambda_x\rangle,\langle x,\lambda_x\rangle)<\delta\implies d(x_1,x)<\epsilon\tag{1}$$

and from the contraction of $g_\lambda:=f(x,\lambda)$ for fixed $\lambda\in\Lambda$ we knows that

$$d(f(x_0,\lambda_x),x)=d(x_1,x)\le q(\lambda_x) d(x_0,x)\tag{2}$$

for a fixed point $x$ (with any $0\le q(\lambda)<1$, hence the contraction). Moreover: we can suppose that the metric in $X\times\Lambda$ is the standard product metric, then:

$$d(\langle a,b\rangle,\langle c,d\rangle)=\max\{d(a,c),d(b,d)\}\tag{3}$$

Then applying $(3)$ in $(1)$ we get

$$d(x_0,x)<\delta\implies d(x_1,x)<\epsilon$$

But as I said Im stuck, I dont know how to show the desired continuity of $h$. Probably the last two (or three) identities are useless, I just take them to see if I can get something from there. Some help will be appreciated, thank you.


If $h$ were not continuous at some $\lambda \in \Lambda$, then it would be true

\begin{align}\tag{1}\label{eq1}\exists \epsilon_\lambda>0,\,\forall \delta >0,\, \exists \lambda_{\delta}\in\Lambda \text{ with } d_{\Lambda}(\lambda,\lambda_{\delta})<\delta,\, d_X(x_{\lambda},x_{\lambda_{\delta}})\geq \epsilon_\lambda\end{align}

In other words, we can find a sequence $\lambda_n \to \lambda$ such that $x_{\lambda_n}$ is always at least $\epsilon_\lambda$-away from $x_{\lambda}$.

Now, since $f$ is continuous at $(x_\lambda,\lambda)$, it is true that:

$$\forall \epsilon >0,\, \exists \delta > 0,\, \forall y\in X,\eta \in \Lambda \text{ with } d_{\Lambda}(\lambda,\eta) <\delta\text{ and }d_X(x_\lambda,y)<\delta,\,\\d_X(f_\lambda(x_\lambda),f_\eta(y))=d_X(x_\lambda,f_\eta(y))< \epsilon$$

Whatever $\delta$ is in the statement above, we may take $y=x_\lambda$ and $\eta=\lambda_n$ for some large enough $n$. More precisely, there is a subsequence $\lambda_{n_k}$ of $\lambda_n$ with $d_X(x_\lambda,f_{\lambda_{n_k}}(x_{\lambda}))\to 0$ as $k \to \infty$.

Now, $$d_X(x_\lambda,x_{\lambda_{n_k}})\leq d_X(x_\lambda,f_{\lambda_{n_k}}(x_{\lambda}))+d_X(f_{\lambda_{n_k}}(x_{\lambda}),\underbrace{f_{\lambda_{n_k}}(x_{\lambda_{n_k}})}_{x_{\lambda_{n_k}}})\\\leq d_X(x_\lambda,f_{\lambda_{n_k}}(x_{\lambda}))+\alpha\cdot d_X(x_\lambda,x_{\lambda_{n_k}})$$

So that

$$(1-\alpha)\cdot d_X(x_\lambda,x_{\lambda_{n_k}})\leq d_X(x_\lambda,f_{\lambda_{n_k}}(x_{\lambda}))$$

If $\eqref{eq1}$ did hold, then the LHS above would be bounded from below by $(1-\alpha)\cdot\epsilon_\lambda > 0$, but the RHS goes to $0$ as $k \to \infty$, a contradiction. It follows that $\eqref{eq1}$ may not hold, and $h$ is continuous.

  • $\begingroup$ Thank but I dont understand some things: why you define a subsequence $\lambda_{n,k}$ if we know that $\lambda_n\to\lambda$? Other question: how are you sure that such $x_{\lambda_{n,k}}$ exists? It seems circular, because if such $x_{\lambda_{n,k}}$ exists the statement is already proved before to end. $\endgroup$ – Masacroso Jan 20 '17 at 7:50
  • $\begingroup$ Take $\epsilon=1/k$ in the continuity statement for $f$. For each $k$, let $\delta_k$ be the corresponding $\delta$ in the statement. Since $\lambda_n\to\lambda$, for each $k$, there are infinitely many indices $n$ with $d_{\Lambda}(\lambda,\lambda_n)<\delta_k$. Just pick a subsequence $\lambda_{n_k}$ where each $n_k$ satisfies this and moreover $n_{k+1}>n_k$. This guarantees the existence of such a subsequence. We choose this subsequence because now we have that $d_X(x_\lambda,f_{\lambda_{n_k}}(x_\lambda))<1/k$, that is, $d_X(x_\lambda,f_{\lambda_{n_k}}(x_\lambda))\to0$ as $k\to\infty$. $\endgroup$ – Fimpellizieri Jan 20 '17 at 15:49
  • $\begingroup$ As a side note, if you're a more visual kind of person, drawing might help. Draw $\Lambda$ and $X$ separately. Draw some $\lambda \in \Lambda$ and a sequence $\lambda_n\to\lambda$. Draw $x_\lambda \in X$ and a circle of radius $\epsilon_\lambda$ centered on $x_\lambda$; the $x_{\lambda_n}$'s are never inside the circle. Now, $g_\Lambda(\cdot)=f(x_\lambda,\cdot)$ is continuous, so $g_\Lambda(\lambda_n)\to g_\Lambda(\lambda)=x_\lambda$. continues $\endgroup$ – Fimpellizieri Jan 20 '17 at 21:23
  • $\begingroup$ (continued) On the other hand, each $f(\cdot, \lambda_n)$ is a $q(\lambda_n)$-contraction. More precisely, with $x_{\lambda_n}$ being the unique fixed point of $f(\cdot, \lambda_n)$, we have that for all $y \in X$: $$d_X\Big(f(y, \lambda_n),x_{\lambda_n}\Big)= d_X\Big(f(y, \lambda_n),f(x_{\lambda_n},\lambda_n)\Big) \leq q(\lambda_n)\cdot d_X(y,x_{\lambda_n}) \leq \alpha\cdot d_X(y,x_{\lambda_n})$$ continues $\endgroup$ – Fimpellizieri Jan 20 '17 at 21:33
  • $\begingroup$ (continued) Now, consider $y=x_\lambda$ on the previous inequality. On the one hand, as $n\to \infty$, $f(\cdot,\lambda_n)$ takes $x_\lambda$ arbitrarily close to itself. On the other hand, it also shrinks the distance between everything and $x_{\lambda_n}$ by a factor of at least $\alpha$. Since $x_\lambda$ and $x_{\lambda_n}$ are always at least $\epsilon_\lambda$ at distant, these two happening at the same time would be a contradiction. I hope this more 'visual' approach helps. end $\endgroup$ – Fimpellizieri Jan 20 '17 at 21:38

I am going to prove that $\lambda\mapsto x(\lambda)$ is continuous at some $\lambda\in \Lambda$. For this, let $\epsilon>0$ be arbitrary and let $\delta>0$ be such that $d(f(x(\lambda),\lambda),f(x(\lambda),\eta))\leq \epsilon$ holds for any $\eta\in\Lambda$ with $d(\eta,\lambda)\leq \delta$. Fix such a $\eta\in\Lambda$.

I will use the following fact (see * below for a proof)

By the proof of Banach's fixed-point theorem for the mapping $f(\cdot,\eta)$ we have $$ d(x(\eta),x_0)\leq \frac{1}{1-q(\eta)}d(f(x_0,\eta),x_0) $$ for any $x_0\in X$

Applying this fact with $x_0:=x(\lambda)$ we get $$ d(x(\eta),x(\lambda))\leq \frac{1}{1-q(\eta)}d(f(x(\lambda),\eta),x(\lambda))\\ = \frac{1}{1-q(\eta)}d(f(x(\lambda),\eta),f(x(\lambda),\lambda))\\ \leq \frac{1}{1-\alpha}\epsilon. $$ This proves the claim.

Summary in words: For any $\lambda\in\Lambda$ the fixed point $x(\lambda)$ can be found by just applying $f_{\lambda}(\cdot):=f(\cdot,\lambda)$ repeatedly, with an arbitrary initial point. The uniform bound on $q(\lambda)$ shows that the limit of this iteration is at most a constant factor further away from the initial point than the first iteration. If we now start the iteration of $\eta\in\Lambda$ in the fixed point of $\lambda\in\Lambda$, then the first iteration only induces a small step, by continuity of $f$ in the second argument. It follows that the fixed point of $\eta$ is close to that of $\lambda$

Note: I only used continuity of $f$ in the second argument. Continuity in the first argument is needed for the existence of fixed points only.

* The proof of Banach's fixed point theorem proceeds by repeatedly applying the contraction mapping to an arbitrary $x_0$. The term $f(x_0,\eta)$ in the right-hand side above is the first iteration, $x_1$, and the term $\frac{1}{1-q(\eta)}$ comes from summing up the geometric series in $d(x(\eta),x_0)\leq \sum_{n=0}^{\infty}d(x_{n+1},x_{n})\leq \sum_{n=0}^{\infty}q^nd(x_{1},x_0)$

  • $\begingroup$ Sorry, I dont see the proof... observe that $\frac1{1-q(\eta)}\epsilon\ge \epsilon$. But anyway Im tired now. $\endgroup$ – Masacroso Jan 20 '17 at 8:46
  • $\begingroup$ Oh, sorry, after doing math for a while you get lazy and just stop when you arrive at such a point. To fix it, just chose the $\delta$ in the definition of continuity so that the distance is actually smaller than $\epsilon(1-\alpha)$. In other words: it is okay to stop a proof as soon as you could show that something is smaller than a constant (independent of $\epsilon$) times $\epsilon$ $\endgroup$ – Bananach Jan 20 '17 at 11:20
  • $\begingroup$ I personally find it actually better to write it the way I did (otherwise I wouldn't have done it, right?). The reason is that if I started with "Choose $\delta$ such that the distance is smaller than $\epsilon 1/(1-\alpha)$" you would immediately stop and wonder where and why $\alpha$ is coming from and it would just seem like a rabbit from the hat rather than the simple calculation that it is $\endgroup$ – Bananach Jan 20 '17 at 11:21
  • $\begingroup$ Just a comment: kid you need to care about your words when chatting with unknown people. Maybe you are wrong about some assumptions and interpretations. $\endgroup$ – Masacroso Jan 21 '17 at 2:54
  • $\begingroup$ @Masacroso Wait are you talking about the "after doing math for a while you get lazy"? Because with the "you" here I meant myself, trying to explain why I stopped the proof seemingly uncompleted. Anyway, I'm not sure if you noticed, but English is not my native language, so I apologize for possible misunderstandings. However, addressing me as kid seems more blatantly condescending than any of what I wrote $\endgroup$ – Bananach Jan 21 '17 at 7:50

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