# Show that $\sqrt{x+2}$ is a contraction

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

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

Since

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

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

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

hint

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

By MVT

$$f(x)-f(y)=(x-y)f'(c)$$

where $$0\le x

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

• Thanks! Basically to show that $f$ is a contraction it is enough to show that $\mid f'(x) \mid \leq \beta$, $\forall x \in X$ – Alessandro Oct 13 '18 at 19:18
• @Alessandro It is sufficient to prove that the derivative is bounded. $| f'(x) | \ le \beta$. – hamam_Abdallah Oct 13 '18 at 19:20
• @Salahamam_Fatima You mean that $|f'(x)|<1$, not just generally bounded. – mathematics2x2life Oct 13 '18 at 19:31
• @mathematics2x2life To be a contraction, you need $|f'(x)|\le \beta<1$. – 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 $$[0,\infty)$$.

• 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 – Alessandro Oct 13 '18 at 22:34
• this question can clarify the issue, I hope: math.stackexchange.com/questions/419392/… – Alessandro Oct 13 '18 at 22:41
• @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.. – mathematics2x2life Oct 15 '18 at 15:34
• 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) – Alessandro Oct 16 '18 at 15:51