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I just want to know if $f $ is continuous on a compact interval, then does it follow that $\sqrt f$ is also continuous?

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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)

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  • $\begingroup$ Well you should at least assume $K$ is a topological space if you want the first statement to make sense. $\endgroup$ – JSchlather Dec 13 '13 at 14:08
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    $\begingroup$ True ... I was thinking of $K$ as a subset of the real line, and therefore, no special property (such as compactedness) was needed, but you're right. $\endgroup$ – aaaaa Dec 13 '13 at 14:45
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Suppose you have to prove that $\sqrt{f}$ is a continuous function (where ). 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,

$\left|\sqrt{f(x)}-\sqrt{f(y)}\right|^2<f(x)+f(y)=\left|f(x)+f(y)\right|$

$\left|\sqrt{f(x)}-\sqrt{f(y)}\right|<\sqrt{\left|f(x)+f(y)\right|}$

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,

$\left|\sqrt{f(x)}-\sqrt{f(y)}\right|<\sqrt{\left|f(x)+f(y)\right|}<\epsilon\mbox{ whenever }|x-y|<\delta$

and $\sqrt{f}$ is also continuous.

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  • $\begingroup$ why does $\delta$ have to be from $\mathbb{N}$ and not $\mathbb{R}$ ? $\endgroup$ – Viktor Glombik Apr 7 '18 at 9:26
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Fix $\epsilon>0$. There exists $\delta>0$ such that if $x\in I(x_0;\delta)\cap D$, then $|f(x)-f(x_0)|<\epsilon$.

Note that $|f(x)-f(x_0)|=|(\sqrt{f(x)}-\sqrt{f(x_0)})(\sqrt{f(x)}+\sqrt{f(x_0)})|=|\sqrt{f(x)}-\sqrt{f(x_0)}||\sqrt{f(x)}+\sqrt{f(x_0)}|$ Hence, $|\sqrt{f(x)}-\sqrt{f(x_0)}|=\frac{|f(x)-f(x_0)|}{|\sqrt{f(x)}+\sqrt{f(x_0)}|}\leq |f(x)-f(x_0)|<\epsilon$. Therefore, $\sqrt{f(x)}$ is also continuous at $x_0$.

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