Is uniform continuity related to finding a bound on a complex function?

[Karol]

Homework Statement

Homework Equations

$$a^2-b^2=(a-b)(a+b)$$

The Attempt at a Solution

$$a^2=\sqrt{1-x_2^2}\,\,\, ,\ \ b^2=\sqrt{1-x_1^2}$$
$$|a^2-b^2|=\left| \sqrt{1-x_2^2}-\sqrt{1-x_1^2} \right|=\left| \sqrt[4]{1-x_2^2} - \sqrt[4]{1-x_1^2} \right|\cdot\left| \sqrt[4]{1-x_2^2} + \sqrt[4]{1-x_1^2} \right|$$
I have to reach:
$$\left| \sqrt{y_2} - \sqrt{y_1} \right| \leq \sqrt{\sqrt{1-x_2^2} - \sqrt{1-x_1^2} }$$

 

[FactChecker]

What are the conditions in Problem 74?

 

[Karol]

What are the conditions in Problem 74?

 

[FactChecker]

Thanks. I think you are supposed to prove the hint with generic y₂ > y₁ > 0 before you look at the specific case involving x₁, x₂, and your specific equation for f(x). You made it more complicated by using the equation for f(x) immediately.

 

[Karol]

$$\left| \sqrt{y_2}-\sqrt{y_1} \right| \leq \frac{1}{\sqrt{y_2-y_1}}\cdot \left| \sqrt{y_2}-\sqrt{y_1} \right| = \frac{\sqrt{y_2-y_1}}{y_2-y_1}\cdot \left| \sqrt{y_2}-\sqrt{y_1} \right| = \sqrt{y_2-y_1} \cdot \frac{\left| \sqrt{y_2}-\sqrt{y_1} \right| }{y_2-y_1}=\sqrt{y_2-y_1} \cdot \frac{\left| \sqrt{y_2}-\sqrt{y_1} \right| \cdot \left| \sqrt{y_2}+\sqrt{y_1} \right| }{(y_2-y_1)\cdot \left| \sqrt{y_2}+\sqrt{y_1} \right|} \leq \sqrt{y_2-y_1} \cdot 1$$
$$\left| \sqrt{1-x_2^2}-\sqrt{1-x_1^2} \right| \leq \sqrt{1-x_2^2-1+x_1^2}=\sqrt{x_1^2-x_2^2}=\sqrt{(x_1-x_2)(x_1+x_2)} \leq \sqrt{2}\sqrt{x_1-x_2}$$
$$\rightarrow~C=\sqrt{2},~m=\frac{1}{2}$$
But in $~\sqrt{x_1^2-x_2^2}~,~x_1^2-x_2^2 \leq 0$

 

[I like Serena]

Hi Karol,

$$\left| \sqrt{y_2}-\sqrt{y_1} \right| \leq \frac{1}{\sqrt{y_2-y_1}}\cdot \left| \sqrt{y_2}-\sqrt{y_1} \right|$$

How did you get this?
I don't see it.

$$\rightarrow~C=\sqrt{2},~m=\frac{1}{2}$$

For the record, this is the correct result.

But in $~\sqrt{x_1^2-x_2^2}~,~x_1^2-x_2^2 \leq 0$

Can we first generalize to $|\sqrt y_2 - \sqrt y_1| \le \sqrt{|y_2 - y_1|}$ before substituting?

 

[Karol]

Because the domain is [-1,1] $~\displaystyle \frac{1}{\sqrt{y_2-y_1}} \geq 1$

 

[I like Serena]

Because the domain is [-1,1] $~\displaystyle \frac{1}{\sqrt{y_2-y_1}} \geq 1$

Ah yes. Good!
Can we prove $|\sqrt y_2 - \sqrt y_1|\le \sqrt{|y_2-y_1|}$ in the same way?

 

[Karol]

I repeat in more detail what i did in post #5:
$$\left| \sqrt{y_2}-\sqrt{y_1} \right| \leq \frac{1}{\sqrt{y_2-y_1}}\cdot \left| \sqrt{y_2}-\sqrt{y_1} \right| = \frac{\sqrt{y_2-y_1}}{y_2-y_1}\cdot \left| \sqrt{y_2}-\sqrt{y_1} \right| = \sqrt{y_2-y_1} \cdot \frac{\left| \sqrt{y_2}-\sqrt{y_1} \right| }{y_2-y_1}=$$
$$=\sqrt{y_2-y_1} \cdot \frac{\left| \sqrt{y_2}-\sqrt{y_1} \right| \cdot \left| \sqrt{y_2}+\sqrt{y_1} \right| }{(y_2-y_1)\cdot \left| \sqrt{y_2}+\sqrt{y_1} \right|} =
\sqrt{y_2-y_1} \cdot \frac{ y_2-y_1 }{(y_2-y_1)\cdot \left| \sqrt{y_2}+\sqrt{y_1} \right|} = \sqrt{y_2-y_1} \cdot \frac{1}{\left| \sqrt{y_2}+\sqrt{y_1} \right|}
\leq \sqrt{y_2-y_1} \cdot 1$$
But the last action is wrong, i have to think it over

 

[I like Serena]

Currently the first inequality already has a problem, since $(y_2-y_1)$ can be negative.
How about replacing every occurrence of $(y_2-y_1)$ by $|y_2-y_1|$?

 

[FactChecker]

Currently the first inequality already has a problem, since $(y_2-y_1)$ can be negative.
How about replacing every occurrence of $(y_2-y_1)$ by $|y_2-y_1|$?

In the original statement, y₂ > y₁.

 

[I like Serena]

In the original statement, y₂ > y₁.

Indeed, but we also need to cover the situation that it's the other way around when addressing the actual problem.
Adding absolute value markers seems to be the simplest way to me.

 

[Karol]

$$\left| \sqrt{y_2}-\sqrt{y_1} \right| \leq \frac{1}{\left| \sqrt{y_2}-\sqrt{y_1} \right|}\cdot \left| \sqrt{y_2}-\sqrt{y_1} \right| = \frac{\left| \sqrt{y_2}+\sqrt{y_1} \right|}{\left| \sqrt{y_2}+\sqrt{y_1} \right|} \frac{1}{\left| \sqrt{y_2}-\sqrt{y_1} \right|}\cdot \left| \sqrt{y_2}-\sqrt{y_1} \right| =$$
$$=\left| \sqrt{y_2}-\sqrt{y_1} \right| \cdot \frac{\left| \sqrt{y_2}+\sqrt{y_1} \right| }{\left| y_2-y_1\right|} \leq \left| \sqrt{y_2}-\sqrt{y_1} \right| \cdot \frac{\left| \sqrt{y_2}+\sqrt{y_1} \right| }{\sqrt{\left| y_2-y_1\right|}}=$$
$$=\frac{\left| y_2-y_1\right| }{\sqrt{ \left| y_2-y_1\right| }}=\sqrt{ \left| y_2-y_1\right| }$$

 

[I like Serena]

$$\left| \sqrt{y_2}-\sqrt{y_1} \right| \leq \frac{1}{\left| \sqrt{y_2}-\sqrt{y_1} \right|}\cdot \left| \sqrt{y_2}-\sqrt{y_1} \right|$$

Oh, it's different now... and not correct any more...
This first step is actually correct, except that the right hand side evaluates to 1.
Consequently, we will not be able to get the right hand side to be less than $\sqrt{|y_2-y_1|}$ any more.

$$\left| \sqrt{y_2}-\sqrt{y_1} \right| \cdot \frac{\left| \sqrt{y_2}+\sqrt{y_1} \right| }{\left| y_2-y_1\right|} \leq \left| \sqrt{y_2}-\sqrt{y_1} \right| \cdot \frac{\left| \sqrt{y_2}+\sqrt{y_1} \right| }{\sqrt{\left| y_2-y_1\right|}}$$

This step is incorrect.
Consider for instance $y_2-y_1=\frac 14$.
Then:
$$\frac{1}{|y_2-y_1|} = 4 \not\le 2 = \frac{1}{\sqrt{|y_2-y1|}}$$

 

[Karol]

I got:
$$\left| \sqrt{y_2}-\sqrt{y_1} \right| \leq \frac{1}{\sqrt{\left| y_2-y_1 \right| }}\cdot \left| \sqrt{y_2}-\sqrt{y_1} \right| = ... = \sqrt{ \left| y_2-y_1 \right| } \cdot \frac{1}{\left| \sqrt{y_2}+\sqrt{y_1} \right|}$$
I have to prove $~\displaystyle \frac{1}{\left| \sqrt{y_2}+\sqrt{y_1} \right|} \leq 1$
But it can have any value. i can write any y<1 as $~\frac{1}{a}~,~a>1$ so:
$$\frac{1}{\left| \sqrt{y_2}+\sqrt{y_1} \right|}=\frac{1}{\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}}=\frac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}$$
$$\frac{\sqrt{9\cdot 4 }}{\sqrt{9}+\sqrt{4}}=\frac{6}{5}=1.2$$
$$\frac{\sqrt{2\cdot 5 }}{\sqrt{2}+\sqrt{5}}=\frac{3.16}{3.64}=0.9$$

 

[I like Serena]

Something is indeed still wrong. The last step in your original proof was not valid after all.
How about we go about it differently.
We want to prove that:
$$|\sqrt y_2 - \sqrt y_1| \overset ?\le \sqrt{y_2-y_1}$$
Suppose we square both sides and see where that leads us?

 

[Karol]

$$|\sqrt y_2 - \sqrt y_1| \overset ?\le \sqrt{y_2-y_1}~\rightarrow~y^2-2\sqrt{y_2}\sqrt{y_1}+y_1 \overset ?\le y_2-y_1$$
$$\rightarrow~y_1-2\sqrt{y_2}\sqrt{y_1} \overset ?\le -y_1$$
$$y_1-2\sqrt{y_2}\sqrt{y_1} \le y_1-2\sqrt{y_1}\sqrt{y_1} = y_1-2y_1=-y_1$$

 

[I like Serena]

That last line does not seem to lead anywhere does it?
How about we make it:

$$\rightarrow~y_1-2\sqrt{y_2}\sqrt{y_1} \overset ?\le -y_1$$

$$\rightarrow~2y_1 \overset ?\le 2\sqrt{y_2}\sqrt{y_1} \le 2\sqrt{y_2}\sqrt{y_2} = 2y_2$$
This is always true isn't it (with $y_2 > y_1 \ge 0$)?

 

[Karol]

yes it's true, but my post i think it's also good. I had to prove $~\displaystyle y_1-2\sqrt{y_2}\sqrt{y_1} \le -y_1~$ and i did in:
$$y_1-2\sqrt{y_2}\sqrt{y_1} \le y_1-2\sqrt{y_1}\sqrt{y_1} = y_1-2y_1=-y_1$$
$$\rightarrow y_1-2\sqrt{y_2}\sqrt{y_1} \le...= -y_1$$

 

[I like Serena]

yes it's true, but my post i think it's also good. I had to prove $~\displaystyle y_1-2\sqrt{y_2}\sqrt{y_1} \le -y_1~$ and i did in:
$$y_1-2\sqrt{y_2}\sqrt{y_1} \le y_1-2\sqrt{y_1}\sqrt{y_1} = y_1-2y_1=-y_1$$
$$\rightarrow y_1-2\sqrt{y_2}\sqrt{y_1} \le...= -y_1$$

Ah okay.

Anyway, we now have a chain of logic full with question marks, and implications that are going in the wrong direction.
So we have to start with the last step, and redo the reasoning in reverse, to prove the hint.

 

[Karol]

I will settle with that. we proved the hint and found C and m.
Thank you FactChecker and Serena