# 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.

• $g(x) = \sqrt{x}$ is continuous. Composition of continuous functions is continuous. – Noé AC Aug 4 '17 at 21:42
• $f(x)=1$ is continuous and $|f(x)+f(y)|=2, \forall x,y\in \mathbb{R}.$ – mfl Aug 4 '17 at 21:45
• – jimjim Aug 4 '17 at 21:48

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 $$D$$

• Why do we need to say "$x_0$ was an arbitrary point"? How is arbitrary point related to continuity? – user13985 Sep 13 '20 at 16:06
• @user13985 I mean arbitrary point in $D$. – Marios Gretsas Sep 13 '20 at 18:27
• Would you mind write this in $\epsilon - \delta$ proof? I'm trying to write it let this, don't know if I'm doing it right: "Let $(x_n)$ be a sequence in D that converges to $x_0$..." – user13985 Sep 13 '20 at 20:44
• A separate question, in: "If f(x_0) then from continuity of $\sqrt(x)...$, how do you show $sqrt(x)$ is continuous, instead of assuming it? – user13985 Sep 14 '20 at 14:50
• @user13985 you can do what i did in the last lines of my proof using the epsilon-delta definition – Marios Gretsas Sep 14 '20 at 18:59

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.