This is a textbook question.

"A function $f: \mathbb{S}\rightarrow \mathbb{S}\subset \mathbb{R}$ with $\mathbb{S}$ closed is called weakly contractive if \begin{align*} d(f(x), f(y))<d(x, y)\qquad \forall x,y\in \mathbb{S} \end{align*} Given an example of a weakly contractive mapping with no fixed point. Show that if $\mathbb{S}$ is compact, then $f$ has a fixed point. (Hint: Define $g(x)=|f(x)-x|$ and check its extreme value property on $\mathbb{S}$)"


Is the definition of weakly contractive correct/standard or should it be a $\leq$ symbol instead of $<$? (textbook, although good, does have some math typos)

IF it was $\leq$ then if $\mathbb{S}=\mathbb{R}_+$ and $f(x)=x+c, c>0$ could be a solution.

If it is indeed $<$ then I have no idea how to proceed. Any suggestions?

EDIT: perhaps $f(x)=-1/x$ could do it even with the strict inequality? (the idea here being exploiting $\mathbb{S}$ closed but not compact) Is that right? EDIT2: Just realized this does not work. $f(x)=-1/x$ is not defined at $x=0$. Making $\mathbb{S}$ not closed. EDIT3: There was a hint for the question, just added now.


2 Answers 2


No, the definition is correct. The point is that for the contraction mapping theorem to hold, it's not enough to say "points get closer when you map them under $f$". You need them to get closer by some factor $\lambda<1$ that does not depend on the points you plug in.

By the contraction mapping theorem, $f$ will have a fixed point if there exists $\lambda<1$ such that $d(f(x),f(y))<\lambda d(x,y)$ for all $x,y \in S$. So this tells you that no such $\lambda$ can exist for your $f$. In other words, for any $\lambda<1$, there exist $x,y \in S$ such that $d(f(x),f(y)) \geq \lambda d(x,y)$.

I think you'll find that Ross Millikan's function does the trick, since as $x,y \to \infty$, $d(f(x),f(y)) \approx d(x,y)$. To show that the inequality holds, suppose $x<y$, so $e^{-x}>e^{-y}$. Also note that by the choice of $S$, $f$ is increasing on $S$, so $x+e^{-x}<y+e^{-y}$. Then: $$d(f(x),f(y)) = (y+e^{-y})-(x+e^{-x}) = (y-x)+ (e^{-y}-e^{-x})<(y-x)=d(x,y).$$


The adjective weakly would suggest that $\le$ is appropriate. I do not know what is standard. Your answer would be a good one in that case. Without the $\le$ you need to push things off to infinity like your $+c$ does, but still have a contraction. How about $x \to x+e^{-x}$ on $x \ge 1$?

  • 1
    $\begingroup$ The "weak" refers to the fact that the RHS is just $d(x,y)$ and not $\lambda d(x,y)$ for some constant $\lambda<1$. $\endgroup$
    – kccu
    Sep 23, 2019 at 2:59
  • $\begingroup$ I was thinking about some sort of asymptotic function approaching f(x)=x from above. But that invalidates the inequality for small values of x, no? $\endgroup$ Sep 23, 2019 at 2:59
  • $\begingroup$ @LucasMation: Not necessarily. That is why I chose $x \ge 1$ for $S$, so the derivative would be strictly less than $1$ in absolute value. $\endgroup$ Sep 23, 2019 at 3:00
  • $\begingroup$ @LucasMation: No, at $x=1$ we have $f'(x)=1-\frac 1e$ $\endgroup$ Sep 23, 2019 at 3:11

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