Prove $\cos{\sqrt{x}}$ is uniformly continuous, $x \in [0,+\infty)$

Question

I try to prove $f(x)=\cos\sqrt{x},x\in[0,\infty)$ is uniformly continuous use definition: $\exists \delta>0,\forall x_1,x_2\in[0,+\infty) ,|x_1-x_2|<\delta,|\cos \sqrt{x_1} -\cos \sqrt{x_2}|< \varepsilon$

Here is my attempt $|\cos \sqrt{x_1} -\cos \sqrt{x_2}|=|-\frac{1}{2}\sin(\frac{\sqrt{x_1}+\sqrt{x_2}}{2})\sin(\frac{\sqrt{x_1}-\sqrt{x_2}}{2})|\leq|\frac{1}{2} \cdots\cdots|???$

I know $|\sin x|\leq 1$, how to select $\delta$?

Answer 1

Let $x,y\in [0,+\infty)$ be arbitrary and assume WLOG $x>y$, so by the mean value theorem there exists $\xi_{x,y} \in (y,x)$ such that: $$|\cos\sqrt{x}-\cos\sqrt{y}|=\left|-\frac{\sin \sqrt{\xi_{x,y}}}{2\sqrt{\xi_{x,y}}}\right|\cdot|x-y|=\frac{|\sin \sqrt{\xi_{x,y}}|}{2\sqrt{\xi_{x,y}}}|x-y|$$ Using that $|\sin t| \le |t|$ for each $t\in\mathbb{R}$, we have: $$\frac{|\sin \sqrt{\xi_{x,y}}|}{2\sqrt{\xi_{x,y}}}|x-y| \le \frac{|x-y|}{2}$$ Let $\epsilon>0$ be arbitrary and let $x,y \in [0,+\infty)$ be arbitrary. Set $\delta_{\epsilon}:=2\epsilon$, so $|x-y|<\delta_\epsilon$ implies: $$|\cos\sqrt{x}-\cos\sqrt{y}|\le \frac{|x-y|}{2}<\epsilon$$ Notice that this holds for arbitrary $x,y \in[0,+\infty)$ because of the WLOG argument.

Alternatively, we can save your first attempt avoiding the mean value theorem. Notice that, using the same inequality on sine mentioned above, we have: $$|\cos \sqrt{x}-\cos\sqrt{y}|=2\left|\sin\frac{\sqrt{x}+\sqrt{y}}{2}\right|\cdot\left|\sin\frac{\sqrt{x}-\sqrt{y}}{2}\right|$$ $$\le2\frac{|\sqrt{x}+\sqrt{y}|}{2}\cdot\frac{|\sqrt{x}-\sqrt{y}|}{2}=\frac{|x-y|}{2}$$ So, again, you can set $\delta_\epsilon:=2\epsilon$.

Answer 2

A real valued function $f:X\subseteq\mathbb{R}\to\mathbb{R}$ is uniformly continuous iff it maps equivalent sequences onto equivalent sequences. Having said that, let us consider two equivalent sequences $(a_{n})_{n\in\mathbb{N}}$ and $(b_{n})_{n\in\mathbb{N}}$ such that its terms belong to $[0,+\infty)$. Based on your attempt, one has that: \begin{align*} \lim_{n\to\infty}|\cos(\sqrt{a_{n}}) - \cos(\sqrt{b_{n}})| & = \lim_{n\to\infty}\left|2\sin\left(\frac{\sqrt{a_{n}} - \sqrt{b_{n}}}{2}\right)\sin\left(\frac{\sqrt{a_{n}} + \sqrt{b_{n}}}{2}\right)\right|\\\\ & \leq \lim_{n\to\infty}\left|2\left(\frac{\sqrt{a_{n}} - \sqrt{b_{n}}}{2}\right)\left(\frac{\sqrt{a_{n}} + \sqrt{b_{n}}}{2}\right)\right|\\\\ & = \lim_{n\to\infty}\frac{1}{2}|a_{n} - b_{n}| = 0. \end{align*}

Hopefully this helps!