Show that f(x)=$\sqrt{x}$ is uniformly continuous, but not Lipschitz continuous.
I can prove that it's uniformly continuous. But why is it not Lipschits? How do I check the definition?
[Math] Lipschitz and uniform continuity
continuity
continuity
Show that f(x)=$\sqrt{x}$ is uniformly continuous, but not Lipschitz continuous.
I can prove that it's uniformly continuous. But why is it not Lipschits? How do I check the definition?
Best Answer
The basic idea is that around $0$, the change in $f(x)$ is not bounded by a constant times the amount that we change $x$ (note that the derivative becomes arbitrarily large). More formally, for any fixed $k>0$,
$$\sqrt{\frac{1}{(k+1)^2}} = \frac{1}{k+1} > k \frac{1}{(k+1)^2}$$
which violates the Lipschitz condition.