Showing that $\sqrt{f}$ is continuous given $f$ is continuous If we know that $f$ is continuous, how can we show that $\sqrt{f}$ is continuous? (Assume that $f(x) \ge 0$ for all $x \in D$)
I find this answer if $f $ is continuous then is $\sqrt f $continuous? but I am not sure that I agree with his response.

Since $f$ is continuous we know that there exist a $\delta\in\mathbb{N}$ such that,
$\left|f(x)+f(y)\right|<\epsilon^2\mbox{ whenever }|x-y|<\delta$
Hence,
$\left|\sqrt{f(x)}-\sqrt{f(y)}\right|<\sqrt{\left|f(x)+f(y)\right|}<\epsilon\mbox{ whenever }|x-y|<\delta$

Do we actually know that there exists a $\delta$ such that $|f(x) +f(y)| \lt \epsilon^{2}$? I would agree that there is a $\delta$ such that $|f(x) -f(y)| \lt \epsilon^{2}$
Intuitively, I feel that $\delta \lt \epsilon^{2}$ would work, but I am not sure how to show this.
 A: Let $x_0 \in D$ and  $x_n \in D$ such that $x_n \rightarrow x_0$.
Then from the continuity of $f$ we have that $f(x_n) \rightarrow f(x_0) \geqslant 0$ .
If $f(x_0)=0$ then from continuity of $\sqrt{x}$ we have that $\sqrt{f(x_n)} \rightarrow 0= \sqrt{f(x_0)}$
If $f(x_0)>0$ then $$|\sqrt{f(x_n)}-\sqrt{f(x_0)}|=
\frac{|f(x_n)-f(x_0)|} {\sqrt{f(x_n)} + \sqrt{f(x_0)}}  \leqslant \frac{|f(x_n)-f(x_0)|}{\sqrt{f(x_0)}} \rightarrow 0 $$
$x_0$ was an arbitrary point in $D$ thus $f$ is continuous in $D$
A: Note that $\sqrt{a}+\sqrt{b-a} \ge \sqrt{b}$ for $0 \le a \le b$. As such,
$$\sqrt{f(x)} \le \sqrt{f(y)} + \sqrt{f(x)-f(y)},$$ when $f(x) \ge f(y)$. Similarly, $$\sqrt{f(y)} \le \sqrt{f(x)} + \sqrt{f(y)-f(x)},$$ when $f(y) \ge f(x)$. Consequently,
$$\left|\sqrt{f(x)}-\sqrt{f(y)}\right| \le \sqrt{|f(x)-f(y)|}.$$
Since $f(x)$ is continuous, there exists an $\epsilon > 0$ such that $|f(x)-f(y)| <\epsilon^2$ for $|x-y|<\delta$. As such, $|\sqrt{f(x)}-\sqrt{f(y)}| < \epsilon$ when $|x-y|<\delta$. Therefore, $\sqrt{f(x)}$ is also continuous.
