Problem: If $g(x)=\sqrt{x}$ for $x\in[0,1]$,show that there does not exist a constant K such that $|g(x)|\leq K|x|$ for all $x\in[0,1]$. Conclude that the uniformly continuous $g$ is not a Lipschitz function.
I am stuck on this problem but I tried using contradiction to do this problem. Suppose that there exists a $K>0$ such that $|g(x)|\leq K|x|= \sqrt{x}\leq Kx\implies x\leq K^2x^2\implies 1\leq K^2x$ where $x\in(0,1]$. Then $\frac{1}{\sqrt{x}}\leq K$. From here I can't seem to find a contradiction. A hint would help thanks.a
Best Answer
Since your K must be uniform, your last line says that K has to be infinite, which is absurd