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Let function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{1}{2}(x+\sqrt{x^2+1})$.

I need to show that for each compact $K\subset\mathbb{R},\exists\alpha_K<1$ such that $|f(x)-f(y)|\leq\alpha_K|x-y|,\forall x,y \in K$.

As $K$ is compact, $\exists M_K>0$ such that $|x|\leq M_K, \forall x \in K$.

We have $|f(x)-f(y)|=\frac{1}{2}|(x-y)+(\sqrt{x^2+1}-\sqrt{y^2+1})|=\frac{1}{2}|(x-y)+\frac{x^2+1-(y^2+1)}{\sqrt{x^2+1}+\sqrt{y^2+1}}|=\frac{1}{2}|(x-y)+(x-y)\frac{x+y}{\sqrt{x^2+1}+\sqrt{y^2+1}}|=\frac{1}{2}|x-y||1+\frac{x+y}{\sqrt{x^2+1}+\sqrt{y^2+1}}|$.

Now, how to manipulate $\frac{1}{2}|1+\frac{x+y}{\sqrt{x^2+1}+\sqrt{y^2+1}}|$ and use the constant $M_K$ to find the upper bound $\alpha_C<1$?

user145306
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1 Answers1

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The derivative is $${1\over 2}\left(1+\frac{x}{\sqrt{1+x^2}}\right)$$ We always have $x<\sqrt{1+x^2}$ so $f'(x)<(1/2)\cdot (1+1)=1$ for all $x\in\mathbb{R}$. More generally, it is clear that $|f'(x)|<1$.

Over a compact region, $f'$ will have and attain its bounds. The bounds must be strictly less than $1$ in absolute value, otherwise $f'$ would attain an absolute value of $1$ contradicting the computation above. Therefore we can say that over any compact interval $K$, $$\exists \alpha_K<1\;:\;|f'(x)|<\alpha_K\,(\forall x\in K)$$ where of course $\alpha_K$ is positive. Kewl.

Now use the Mean Value Theorem. For every $x,y\in K$ there is a $c\in (x,y)\subset K$ - and hence a $c\in K$ - such that $$\frac{f(x)-f(y)}{x-y}=f'(c)$$ holds. But this implies $$|f(x)-f(y)|=|f'(c)|\cdot|x-y|<\alpha_K |x-y|$$ as was to be shown.