Prove that $f(x)=\sqrt{x(1-x)}$ is continuous on $(0,1)$

Question

Attempt: By definition, it is to prove $ \forall a \in (0,1),\exists \delta>0, s.t.\lvert x-a \rvert < \delta$ implies $\lvert \sqrt{x(1-x)} - \sqrt{a(1-a)} \rvert < \epsilon$, which is equivalent to $ \frac{\lvert x-a \rvert \lvert 1-(x+a) \rvert}{\sqrt{x(1-x)} + \sqrt{a(1-a)}} < \epsilon $. But I don't know how to proceed. The textbook shows that I should choose $\eta = \min\{ a, 1-a\}$, then $ \frac{\lvert x-a \rvert \lvert 1-(x+a) \rvert}{\sqrt{x(1-x)} + \sqrt{a(1-a)}} < \frac{\lvert x-a \rvert}{\sqrt{a(1-a)}}$ holds. Then it's trivial to see if we set $\delta = min\{\eta, \sqrt{a(1-a)}\epsilon\} $, then $\lvert \sqrt{x(1-x)} - \sqrt{a(1-a)} \rvert < \frac{\lvert x-a \rvert}{\sqrt{a(1-a)}} < \epsilon $ . But the choice of $\eta$ is not intuitive, could anyone help explain that more clearly?

Answer 1

The elegant approach:

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Look at that numerator: $|x-a||1-(x+a)|$. On $(0,1)$, $|1-(x+a)|$ is at most $1$, so I can really just ignore that. That is, I am replacing $|1-(x+a)|$ with "$1$" and this keeps the $|\sqrt{x(1-x)}-\sqrt{a(1-a)}|<\cdots$ direction of the inequality. We now have an upper bound by: $$\frac{|x-a|}{\sqrt{x(1-x)}+\sqrt{a(1-a)}}$$

But also $\sqrt{x(1-x)}>0$ so I can simply ignore that too! We have an upper bound by: $$\frac{|x-a|}{\sqrt{a(1-a)}}$$Since the denominator has now been made smaller. This has nothing to do with "$\eta$". Then $k:=\sqrt{a(1-a)}$ is just some positive constant and $|x-a|<\delta$ is our future plan, so $\delta:=k\epsilon$ will suffice very nicely: if $0<x<1$ and $|x-a|<\delta$, we have $|\sqrt{x(1-x)}-\sqrt{a(1-a)}|<\epsilon$.

Now that you have elaborated on $\eta$... $\eta$ is chosen so that if $|x-a|<\delta\le\eta$ we must have $x\in(0,1)$. Because then $x-a<\eta<1-a$, so $x<1$, and also $a-x<\eta<a$ so that $x>0$.

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The inelegant, but much more realistic approach in general - use your theorems!:

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It is the composition of $\sqrt{\cdot}$ with the product of continuous functions $x\mapsto x,\,x\mapsto1-x$, this is continuous wherever it is defined.

We can unravel that into a proof. I know for any given $\upsilon>0$ and all $\epsilon>0$ there is some $\delta>0$ with $|\sqrt{x}-\sqrt{\upsilon}|<\epsilon$ whenever $\upsilon-\delta<x<\upsilon+\delta$. How can I find such $\delta$? Well, $\sqrt{x}-\sqrt{\upsilon}=\frac{x-\upsilon}{\sqrt{x}+\sqrt{\upsilon}}$ which can clearly be bounded in magnitude by $\frac{\delta}{2\sqrt{\upsilon-\delta}}$ (bound numerator from above, bound the denominator from below, note $\sqrt{.}$ is increasing). If $\delta<\upsilon/2$, then that bounds by $\frac{\delta}{\sqrt{2\upsilon}}$, and if I want that to be $<\epsilon$ I can easily set $\delta:=\min(\upsilon/2, \epsilon\sqrt{2\upsilon})$.

Now for continuity of $x\mapsto x(1-x)$. Well, fix any $x$ and any $\epsilon>0$. Note $x(1-x)-y(1-y)=x-y+y^2-x^2=(x-y)(1+y+x)$ so if $\delta<1$ and $x-\delta<y<x+\delta$ then $2x<1+y+x<2x+2$ and so the absolute value is bounded by $\max(|2x|,|2x+2|)$. Then for the specific $\delta:=\min(1,\frac{\epsilon}{2\max(|x|,|x+1|)})$ we have: $|y-x|<\delta\implies|x(1-x)-y(1-y)|<\epsilon$.

Then in composition, fix $0<x<1$ and $\epsilon>0$. We want $\delta'$ such that $|\sqrt{a}-\sqrt{b}|<\epsilon$ for $|a-b|<\delta'$, with $a:=x(1-x)$ and moreover we want $\delta>0$ with $|a-b|<\delta'$ whenever $|x-y|<\delta$, setting $b:=y(1-y)$.

Let's just use the above (with $\upsilon=x(1-x)$). $\delta':=\min(x(1-x)/2,\epsilon\sqrt{2x(1-x)})$ works and $\delta:=\min(1,\frac{\delta'}{2\max(|x|,|x+1|)})$ works. Inelegant, but it works.

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Visual "proof" that this all works:

Answer 2

If $a=0$, then $$ \begin{align} \left|\sqrt{x(1-x)}-\sqrt{a(1-a)}\,\right| &=\sqrt{x(1-x)}\tag{1a}\\ &\le\sqrt{x}\tag{1b}\\ &=\sqrt{x-a}\tag{1c} \end{align} $$ so let $\delta=\epsilon^2$.

If $a=1$, then $$ \begin{align} \left|\sqrt{x(1-x)}-\sqrt{a(1-a)}\,\right| &=\sqrt{x(1-x)}\tag{2a}\\ &\le\sqrt{1-x}\tag{2b}\\ &=\sqrt{a-x}\tag{2c} \end{align} $$ so let $\delta=\epsilon^2$.

If $0\lt a\lt1$, then $$ \begin{align} \left|\sqrt{x(1-x)}-\sqrt{a(1-a)}\,\right| &=\frac{|\,x(1-x)-a(1-a)\,|}{\sqrt{x(1-x)}+\sqrt{a(1-a)}}\tag{3a}\\ &=\frac{|(x-a)(1-x-a)|}{\sqrt{x(1-x)}+\sqrt{a(1-a)}}\tag{3b}\\ &\le\frac{|x-a|}{\sqrt{a(1-a)}}\tag{3c} \end{align} $$ so let $\delta=\epsilon\sqrt{a(1-a)}$.