if $f $ is continuous then is $\sqrt f $continuous? I just want to know if $f $ is continuous on a compact interval, then does it follow that $\sqrt f$ is also continuous? 
 A: Suppose you have to prove that $\sqrt{f}$ is a continuous function (where $f$ is continuous). We go by the definition of continuity. Take any $\epsilon>0$. Then,
$$
\left|\sqrt{f(x)}-\sqrt{f(y)}\right|^2=f(x)+f(y)-2\sqrt{f(x)f(y)}.
$$
Since $f$ should be non-negative,
$$
\begin{align}
\left|\sqrt{f(x)}-\sqrt{f(y)}\right|^2&<f(x)+f(y)=\left|f(x)+f(y)\right|\\
\implies\quad\left|\sqrt{f(x)}-\sqrt{f(y)}\right|&<\sqrt{\left|f(x)+f(y)\right|}.
\end{align}
$$ 
Since $f$ is continuous we know that there exist a $\delta > 0$ such that,
$$
\left|f(x)+f(y)\right|<\epsilon^2\mbox{ whenever }|x-y|<\delta.
$$
Hence,
$$\left|\sqrt{f(x)}-\sqrt{f(y)}\right|<\sqrt{\left|f(x)+f(y)\right|}<\epsilon\mbox{ whenever }|x-y|<\delta$$
and $\sqrt{f}$ is also continuous.
A: Well, if $f\colon K\rightarrow \mathbb{R}_{+}$ is continuous, given that $\sqrt{\cdot} \ \colon \mathbb{R}_{+} \rightarrow \mathbb{R}_{+}$ is also continuous, then so is their composition: $\sqrt{f} \ \colon K \rightarrow \mathbb{R}_{+}$ (without any assumptions about K)
