# Example of $f: \mathbb{S}\rightarrow \mathbb{S}\subset \mathbb{R},$ $\mathbb{S}$ closed with no fixed point

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)) 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 $$\mathbb{S}$$)"

Question?

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.

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 $$x, so $$e^{-x}>e^{-y}$$. Also note that by the choice of $$S$$, $$f$$ is increasing on $$S$$, so $$x+e^{-x}. 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 $$x \ge 1$$?

• The "weak" refers to the fact that the RHS is just $d(x,y)$ and not $\lambda d(x,y)$ for some constant $\lambda<1$.
– kccu
Sep 23, 2019 at 2:59
• 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? Sep 23, 2019 at 2:59
• @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. Sep 23, 2019 at 3:00
• @LucasMation: No, at $x=1$ we have $f'(x)=1-\frac 1e$ Sep 23, 2019 at 3:11