Is $f(x)=\sqrt x$ locally Lipschitz continuous everywhere?
It is definitely not (globally) Lipschitz continuous. I wonder if it is locally Lipschitz continuous at $x=0$.
2026-03-25 23:56:40.1774483000
Is $\sqrt x$ locally Lipschitz continuous everywhere?
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Hint: Consider any interval $[0,\varepsilon)$ where $\varepsilon > 0$. For $x,y\in [0,\varepsilon)$, with $x < y$, we have $$\begin{align*} \frac{|f(x)- f(y)|}{|x-y|} &= \frac{\sqrt{y} - \sqrt{x}}{y - x}\\ &= \frac{1}{\sqrt{y}+\sqrt{x}}. \end{align*}$$ Can you show that this can be made arbitrarily large, and if so, what can you conclude?