I have given two functions $f(x)=\frac{1}{2}x$ and $g(x)=\sqrt{1+x^2}$, $x\in \Bbb{R}$, I want to show that for some $K\geq 0$, $$|f(x)-f(y)|+|g(x)-g(y)|\leq K|x-y|$$ for all $x,y\in \Bbb{R}$. And furthermore I want to show that $$|f(x)|+|g(x)|\leq K(1+|x|)$$
Clearly $$\begin{align}|f(x)-f(y)|+|g(x)-g(y)|&=\frac{1}{2}|x-y|+\left|\sqrt{1+x^2}-\sqrt{1+y^2}\right|\end{align}$$ but I don't see how I can bound $\left|\sqrt{1+x^2}-\sqrt{1+y^2}\right|$ to get something of the form $C|x-y|$.
Also in the second part I'm stuck because $$\begin{align}|f(x)|+|g(x)|&=\frac{1}{2}|x|+|\sqrt{1+x^2}|\\&\leq\frac{1}{2}|x|+1+|x|\\&=1+\frac{3}{2}|x|\end{align}$$ but also here I don't find this $K$ as we wanted. Could someone help me further?
Thanks a lot!
Let me give a way which does not involves derivatives. Let me use the notation $\langle x\rangle = \sqrt{1+|x|^2}$ and notice that $|x| = \sqrt{|x|^2} \leq \langle x\rangle$, then $$ |\langle x\rangle - \langle y\rangle|^2 - |x-y|^2 = 2 \left(1+ x\cdot\,y - \langle x\rangle\,\langle y\rangle\right). $$ Since $|1+x\cdot y|^2 \leq 1 + |x|^2\,|y|^2 + 2\,|x|\,|y| \leq \langle x\rangle^2\,\langle y\rangle^2$, one deduces that $$ 1+ x\cdot\,y - \langle x\rangle\,\langle y\rangle \leq |1+ x\cdot\,y| - \langle x\rangle\,\langle y\rangle \leq 0, $$ and so from the first equation $$ |\langle x\rangle - \langle y\rangle| \leq |x-y|. $$
For the second part, since $1<3/2$, you get $1 + \frac{3}{2}\,|x| \leq \frac{3}{2} (1+|x|)$ ...
Of course one can also treat the first part with derivatives, notice that $\nabla \langle x\rangle = \frac{x}{\langle x\rangle} =: u(x)$ with $|u|\leq 1$ and so the fundamental theorem of calculus yields $$ \langle x\rangle - \langle y\rangle = (x-y)\cdot\int_0^1 u((1-t)x+ty)\,\mathrm d t $$ which implies $$ |\langle x\rangle - \langle y\rangle| \leq |x-y|\int_0^1 1\,\mathrm d t = |x-y|. $$