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Let $(f_n)_n$ be a sequence of functions $f_n: [0,1] \rightarrow [0,1]$ that (pointwise) converges to a function $f$. Suppose that all $f_n$ are contractions, prove that $f$ is a contraction as well.

Here's what I have:

$$ \begin{array}{rl} | f(x) - f(y) | &= | \lim_{n \rightarrow \infty}f_n(x) - \lim_{n \rightarrow \infty}f_n(y) |\\ &= | \lim_{n \rightarrow \infty}(f_n(x) - f_n(y)) |\\ &= \lim_{n \rightarrow \infty}|f_n(x) - f_n(y) |\\ &< \lim_{n\rightarrow \infty}|x-y| &= |x-y| \end{array} $$

That's a bunch of symbols. I fear that I might have waved my hands too much at the details. I can't point to any theorems that tell my that I'm alowed to proceed with the third equality.

Is this correct and why? Also, why does it matter that the $f_n$ are defined as $f_n: [0,1] \rightarrow [0,1]$?

(The part of the question is: "prove that the convergence is uniform")

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  • $\begingroup$ Haven't you missed to include a Lipschitz constant $k<1$ in the definition of contraction? In that case the constants can be different for each $f_n$ and might converge to $1$. Then the conclusion does not hold. Perhaps you have a different definition of contraction than I'm used to.The one I'm used to is the same as stated on Wikipedia: en.wikipedia.org/wiki/Contraction_mapping $\endgroup$ – dioid Jun 3 '15 at 13:47
  • $\begingroup$ I'm using this one: (It seems equivalent) A function $f$ is a contraction if this holds: $\forall x,y : |f(x)-f(y)| < |x-y|$ $\endgroup$ – Syd Kerckhove Jun 3 '15 at 13:49
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    $\begingroup$ They're not equivalent. According to the definition on Wikipedia the function $f(x) = x$ is not a contraction. According to your definition no function is a contraction (let $x=y$). If you add $x \neq y$ to your definition they are still not equivalent since the definition on Wikipedia says you have to stay away by some fixed distance from $|x-y|$ while your (modified) definition allows this distance to depend on $x, y$ and become arbitrarily small. $\endgroup$ – dioid Jun 3 '15 at 13:53
  • $\begingroup$ Oh, I made a mistake, the definition in our notes is $\forall x,y:\ |f(x)-f(y)| \le |x-y|$ (so $\le$ instead of $<$). Also the notes define a strict contraction as $\exists c\in [0,1[, \forall x,y:\ |f(x)-f(y)| \le c|x-y|$ $\endgroup$ – Syd Kerckhove Jun 3 '15 at 13:55
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    $\begingroup$ Ok, you're all good then. The terminology I'm used to is to call it non-expansion and contraction instead of contraction and strict contraction. $\endgroup$ – dioid Jun 3 '15 at 13:57
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You can swap $|\cdot|$ and $\lim_{n \to +\infty}$ because $|\cdot|$ is a continuous function. But when you take the limit, $<$ becomes $\leq$. It is not essential that the functions are from $[0,1]$ to itself here - however, fixed point theorems are stated with $[0,1]$ at first for simplicity. You can look at Brouwer's, Banach's fixed point theorems, etc.

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  • $\begingroup$ Wait a minute, are you saying the proof is correct? (If I add the necessary explanation?) $\endgroup$ – Syd Kerckhove Jun 3 '15 at 13:29
  • $\begingroup$ Yes, although I would have stated it like this: $$|f_n(x)-f_n(y)|< |x-y|,~\forall\,n \implies \lim_{n \to +\infty}|f_n(x)-f_n(y)| \leq |x-y| \implies |f(x)-f(y)| \leq |x-y|.$$ $\endgroup$ – Ivo Terek Jun 3 '15 at 13:31

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