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I want to prove that the function $f:[-1,1]\rightarrow [-1,1]$ given by \begin{equation*} f(x)=\frac{1}{2}x^2 \end{equation*} is not a Rakotch contraction. To give the definition of a Rakotch contraction I need to give the definition of a $\phi$- contraction first: Let $\phi$ be a function on $\mathbb{R}_{+}$. A function $f$ on $X$ is a $\phi$-contraction if \begin{equation} d(f(x),f(y))\leq \phi(d(x,y)), \quad \forall\ x,y\in X \nonumber. \end{equation} Let f be a $\phi$-contraction. If $\phi$ is such that $\frac{\phi(x)}{x}<1$ and $\frac{\phi(x)}{x}\leq\frac{\phi(y)}{y},\forall \ x>y$. Then, the function $f$ is a Rakotch contraction.

Aiswarya
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$f$ is in fact a Rakotch contraction given your definition. First, let me shorten $\frac{φ(δ)}{δ} ≕ Φ(δ)$, which we need to be decreasing and strictly less than 1 (but defined on positive numbers only).

It is equivalent data to your function $φ$. If $γ ≔\lim_{δ ⟶ 0} Φ(δ) < 1$, then $f$ is Lipschitz with Lipschitz constant $< 1$ and any such function is a Rakotch contraction with constant $Φ(δ)$. So, it is most interesting when $γ=1$.

In our example, $\frac{d(f(x), f(y))}{d(x, y)} = \frac 12 |x+y| \overset ! ≤ Φ(|x-y|)$. (using the third binomial formula)

Now, WLOG (flip signs or exchange x and y), $y = x - δ$ and $x ≥ \frac δ2$ for some $δ > 0$. The only requirement that we need to have is that $Φ(δ) ≥\frac 12 |x+y|= x - \frac δ 2$ for all such choices of $x$.

Thus, define $Φ(δ) = 1 - \frac δ 2$.

TheByteBreaker
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  • Thanks for the answer.. I was looking for a function $f$ such that $f$ is a contractive map in the sense that it strictly decreases the distance between points, but it is not a Rakotch contraction in a compact set. – Aiswarya Jun 18 '25 at 15:30
  • This is an interesting question. Such functions do not exist: We can define $Φ$ for any (continuous) function $f$: $Φ(δ) = \operatorname{sup}\left{\frac{d(f(x),f(y))}{d(x,y)} \middle | x, y \in X : d(x, y) ≥ δ\right}$. This is the smallest decreasing $Φ$-function, s.t. $f$ is a $φ$-contraction. Thus, $f$ is a Rakotch contraction iff this $Φ$ is $<1$ everywhere. The supremum is over a compact set (closed subset of $X×X$) thus the continuous function in it attains a maximum. By your assumption that $f$ strictly decreases distances, this maximum is $<1$. – TheByteBreaker Jun 25 '25 at 16:52
  • On non-compact spaces, one can probably find such functions: I'm thinking about a diagonal matrix with entries approaching 1 on an infinite dimensional vector space like $\mathbb R^{(\mathbb N)}$. I haven't thought about the question "is there a function $f:X ⟶ Y$ (X, Y metric spaces) that strictly decreases distances but there is a compact set restricted to which it is not a Rakotch contraction" which is a potential different interpretation of your (commented) question – TheByteBreaker Jun 25 '25 at 16:56