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**redsoxfan325** Is $\displaystyle f(x)=\sqrt{1-x^2}$ uniformly continuous?

Here is my proposed solution.

Break $\displaystyle [-1,1]$ into three partitions: $\displaystyle [-1,-4/5]\cup(-4/5,4/5)\cup[4/5,1]$

On $\displaystyle [-1,-4/5]$, $\displaystyle f(x)$ is defined on a compact set, so $\displaystyle \forall~\epsilon>0, \exists~\delta_{-1}>0$ such that $\displaystyle |x-y|<\delta_{-1} \implies |f(x)-f(y)|<\epsilon$.

On $\displaystyle (-4/5,4/5)$, $\displaystyle f(x)$ has a bounded derivative: $\displaystyle |f'(x)| \leq \frac{4}{3}$. Let $\displaystyle \delta_0 = \frac{3}{4}\epsilon$. Thus, when $\displaystyle |x-y|<\delta_0$, by the MVT, $\displaystyle |f(x)-f(y)|=|x-y|f'(t) \leq M|x-y| < \frac{4}{3}\cdot\frac{3}{4}\epsilon = \epsilon$.

On $\displaystyle [4/5,1]$, $\displaystyle f(x)$ is defined on a compact set, so $\displaystyle \forall~\epsilon>0, \exists~\delta_1>0$ such that $\displaystyle |x-y|<\delta_1 \implies |f(x)-f(y)|<\epsilon$.

Let $\displaystyle \delta = \min\{\delta_{-1},\delta_0,\delta_1\}$ and it follows that $\displaystyle \forall~x\in[-1,1], \forall~y\in[-1,1], |f(x)-f(y)|<\epsilon$ when $\displaystyle |x-y|<\delta$, so $\displaystyle f(x)$ is uniformly continuous.

Is that correct?