# Thread: Is \sqrt{1-x^2} uniformly continuous?

1. ## Is \sqrt{1-x^2} uniformly continuous?

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

Here is my proposed solution.

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

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

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

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

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

Is that correct?

2. Originally Posted by redsoxfan325
Is $f(x)=\sqrt{1-x^2}$ uniformly continuous?

Here is my proposed solution.

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

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

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

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

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

Is that correct?
why did you divide the interval? $f(x)=\sqrt{1-x^2}$ is continuous over $[-1,1],$ which is a compact set. thus $f$ is uniformly continuous over $[-1,1].$

3. That's a good point. Man, do I feel like an idiot...