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I have a following problem.

Let $X = R$, $d(x,y) = |x-y|$, $T(x) = \sqrt{x^2 + 1}$ but $T$ does not have a fixed point. Does this contradict Banach's Fixed Point Theorem?

I know that if $X$ was compact space then it would imply the existence and uniqueness of a fixed point, but not sure what to do in this case ($R$ is not compact). I would appreciate any help.

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    $\begingroup$ Are the premises of Banach's fixed point theorem satisfied? $\endgroup$ – Daniel Fischer Dec 22 '14 at 23:13
  • $\begingroup$ Perhaps you're thinking of the Brouwer fixed point theorem? That's one where compactness is part of the hypotheses. The Banach fixed point theorem does not mention compactness. $\endgroup$ – Robert Israel Dec 23 '14 at 0:34
  • $\begingroup$ No, it is a Banach fixed point theorem. Compactness is just my assumption, it does not have to be correct. $\endgroup$ – Jim Dec 23 '14 at 12:10
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No, $T(x)$ doesn't have the contraction property, consider $x_n=n$ and $y=0$, then $|x_n-0|=n$, but

$$|T(x_n)-T(0)|=|\sqrt{n^2+1}-1|={n^2+1-1\over \sqrt{n^2+1}+1}$$

$$={n^2\over \sqrt{n^2+1}+1}$$

With $\epsilon>0$ arbitrary and fixed we see for $n>\!>0$ we have

$$n(1+\epsilon)^{-1}={n^2\over\sqrt{(1+\epsilon)^2n^2}}< |T(x_n)-T(0)|$$

But then by taking $\epsilon$ sufficiently small, we see that there can be no $q\in [0,1)$ such that

$$|T(x)-T(y)|<q|x-y|$$

in fact we see that clearly $q\ne 0$, so if a candidate $0<q<1$ is given then $\epsilon =\min\{{1\over 2},q^{-1}-1\}$ works.

for all $x,y\in\Bbb R$, because for some $n$ we have $x=x_n, y=0$ is a counterexample, hence $T$ is not a contraction.

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  • $\begingroup$ Thanks! Could you just explain what your 'No' means - is it that my statement does not contradict Banach Fixed Point theorem? I don't understand why would that be since T is not a contraction and the theorem requires T to be a contraction. In that case it has to be a contradiction, right? $\endgroup$ – Jim Dec 23 '14 at 12:08
  • $\begingroup$ @Jim Well, it doesn't contradict the theorem, because a contradiction to a theorem is a counterexample, i.e. an example which satisfies the hypotheses, but fails to have the conclusion. In the map you give it is not one that satisfies the hypotheses of the theorem, hence does not contradict it. $\endgroup$ – Adam Hughes Dec 23 '14 at 16:42
  • $\begingroup$ Thanks you have been very helpful! $\endgroup$ – Jim Dec 23 '14 at 18:43
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To solve such a problem the first and foremost thing is to recall the result in question. So, what does the Banach Fixed Point Theorem say:

Let $(X, d)$ be a non-empty complete metric space with a contraction mapping $T : X \to X$. Then $T$ admits a unique fixed-point $x^\ast$ in $X$.

Now $\mathbb{R}$ with $d(x,y)= |x-y|$ is a complete metric space. And, your $T$ is a map from $\mathbb{R}$ to itself.

Yet, is it a contraction? What even is a contraction? Let us check the same Wikipedia page to find:

A map $T : X \to X$ is called a contraction mapping on $X$ if there exists $q \in [0, 1)$ such that $d(T(x),T(y)) \le q d(x,y)$ for all $x, y$ in $X$.

Is this true for your map $T$? (It can be important to pay attention to the fact that just $d(T(x),T(y)) < d(x,y)$ for all $x, y$ in $X$ is not sufficient but really the slightly stronger condition is needed.)

(Note: the answer ought to be "no," because 'clearly' there should not be an example that contradicts the well-known theorem, and your map has no fix point and this condition is the last we would need to check to get one.)

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  • $\begingroup$ Thanks! I have the same question as in my comment above, if one of you could clarify. $\endgroup$ – Jim Dec 23 '14 at 12:09
  • $\begingroup$ The map $T$ is not a contraction, as implied here, and worked out in the other answer. This means that the premises of the Banach FPT are not satisfied (see the first comment on your question). That its conclusion is also not satisfied is then no problem. The abstract form is "If [something], then [this]." The [this] is having a fixed point, and the [something] are the various conditions. The situations is that the [something] is not true, and this you cannot derive the conclusion. $\endgroup$ – quid Dec 23 '14 at 12:26
  • $\begingroup$ There is no contradiction to the BFPT, you only cannot apply it in this case. Contradiction to it would only be the case if "[something] is true" yet still "[this] is false." $\endgroup$ – quid Dec 23 '14 at 12:26
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    $\begingroup$ Great, thank you very much! $\endgroup$ – Jim Dec 23 '14 at 12:56

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