Let $f:X \rightarrow X$ where $X=[0,\infty)$ be defined as $f(x)=\sqrt{x+2}$. I have to show that this mapping is a contraction and find its unique fixed point. The second part is easy: by the CMT, it has a unique fixed point in $X$ and it is $x^\ast = 2$.

For $f$ being a contraction I wts the following: $ \exists \beta \in [0,1)$ such that

$\mid\sqrt{x+2}-\sqrt{y+2}\mid \leq \beta \mid x-y \mid, \ (\forall x,y\geq0)$


$\mid\sqrt{x+2}-\sqrt{y+2}\mid = \dfrac{\mid x-y \mid}{\sqrt{x+2}+\sqrt{y+2}} $

I'm tempted to set $ \beta = \dfrac{1}{\sqrt{x+2}+\sqrt{y+2}}$ but $\beta$ cannot depend on $x$ or $y$.... Any ideas about how to proceed? Thanks!


Note that for all $x,y\geqslant 0$ $$\frac{1}{\sqrt{x+2}+\sqrt{y+2}}\leqslant \frac{1}{\sqrt{2}+\sqrt{2}}=\frac{\sqrt{2}}{4}$$

  • $\begingroup$ Great! I had just to find an upper bound smaller than one! $\endgroup$ – Alessandro Oct 13 '18 at 18:59


Let $x,y\in [0,+\infty)$.



where $$0\le x<c<y.$$

$$f'(c)=\frac{1}{2\sqrt{c+2}}\le \frac{1}{2\sqrt{2}}$$

  • $\begingroup$ Thanks! Basically to show that $f$ is a contraction it is enough to show that $ \mid f'(x) \mid \leq \beta$, $\forall x \in X$ $\endgroup$ – Alessandro Oct 13 '18 at 19:18
  • $\begingroup$ @Alessandro It is sufficient to prove that the derivative is bounded. $| f'(x) | \ le \beta$. $\endgroup$ – hamam_Abdallah Oct 13 '18 at 19:20
  • $\begingroup$ @Salahamam_Fatima You mean that $|f'(x)|<1$, not just generally bounded. $\endgroup$ – mathematics2x2life Oct 13 '18 at 19:31
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    $\begingroup$ @mathematics2x2life To be a contraction, you need $|f'(x)|\le \beta<1$. $\endgroup$ – hamam_Abdallah Oct 13 '18 at 19:48

As per Salahamam's solution, to show a function is a contraction, it is sufficient to show that its derivative has $|f'(x)|<1$.

Prop. If $f(x)$ is a differentiable function function with $|f'(x)|<1$ for all $x$, then $f(x)$ is a contraction.

Proof. Let $x,y \in \mathbb{R}$. By the Mean Value Theorem, we have $|f(x)-f(y)|= |f'(c)(x-y)|= |f'(c)||x-y|$ for some $c$ between $x$ and $y$. But as $|f'(c)|<1$ by assumption, we must have $$|f(x)-f(y)|=|f'(c)||x-y| < 1 \cdot |x-y|=|x-y|.$$ Therefore, $f$ is a contraction.

Note that the converse is false as contractions need not be differentiable.

So in your case, you only need show that $\sqrt{x+2}$ has bounded derivative. But $\dfrac{d}{dx} \; \sqrt{x+1}= \dfrac{1}{2\sqrt{x+2}}$ which is at most $\frac{1}{2\sqrt{2}}$ on the interval $[0,\infty)$.

  • $\begingroup$ To show that $ f $ is a contraction you need to show that the derivative is uniformly bounded by a number less than one. Proving that $f'(x)<1$ is not enough. That's what I understood $\endgroup$ – Alessandro Oct 13 '18 at 22:34
  • $\begingroup$ this question can clarify the issue, I hope: math.stackexchange.com/questions/419392/… $\endgroup$ – Alessandro Oct 13 '18 at 22:41
  • $\begingroup$ @Alessandro That was exactly what I said. You have a function which is differentiable on the intervals which you are considering, so it suffices to show that $|f'(x)|<1$. You do not need your derivative to be uniformly bounded. A strong contraction is a function for which there exists $|f(x)-f(y)| \leq c<1$. But of course, this depends on what one calls a contraction. But most mean $|f(x)-f(y)|<|x-y|$ when they say 'contraction' and reserve the former when speaking of something stronger like strong contraction, Lipschitz continuity, etc.. $\endgroup$ – mathematics2x2life Oct 15 '18 at 15:34
  • $\begingroup$ ok but to apply the contraction mapping theorem I need what you call "a strong contraction". Furthermore the standard definition of contraction requires $ \mid f(y)-f(x) \mid \leq \beta \mid x-y \mid $ (see en.wikipedia.org/wiki/Contraction_mapping) $\endgroup$ – Alessandro Oct 16 '18 at 15:51

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