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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.

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user345
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2 Answers2

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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 $D$

Marios Gretsas
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  • 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
  • My attempt: Give $\epsilon>0$, we choose $\delta= \sqrt{f(x_0)} \epsilon > 0$. Then, whenever $|x_n - x_0| < \delta$, we have $|\sqrt{f(x_n)} - \sqrt{f(x_0)}| \le \frac{|f(x_n) - f(x_0)|}{\sqrt{f(x_0)}} <\epsilon .$ I need help with the last inequality. How do I know it's less than $\epsilon$? In the past, I'm always given a specific $f$. I just solve for the epsilon-delta relationship. – user13985 Sep 15 '20 at 15:06
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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.

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