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Homework Help: Complex uniform continuity

  1. Sep 16, 2017 #1
    1. The problem statement, all variables and given/known data
    Snap2.jpg

    2. Relevant equations
    $$a^2-b^2=(a-b)(a+b)$$

    3. 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:
    $$\left| \sqrt{y_2} - \sqrt{y_1} \right| \leq \sqrt{\sqrt{1-x_2^2} - \sqrt{1-x_1^2} }$$
     
  2. jcsd
  3. Sep 16, 2017 #2

    FactChecker

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    What are the conditions in Problem 74?
     
  4. Sep 16, 2017 #3
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  5. Sep 16, 2017 #4

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    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.
     
  6. Sep 17, 2017 #5
    $$\left| \sqrt{y_2}-\sqrt{y_1} \right| \leq \frac{1}{\sqrt{y_2-y_1}}\cdot \left| \sqrt{y_2}-\sqrt{y_1} \right| = \frac{\sqrt{y_2-y_1}}{y_2-y_1}\cdot \left| \sqrt{y_2}-\sqrt{y_1} \right| = \sqrt{y_2-y_1} \cdot \frac{\left| \sqrt{y_2}-\sqrt{y_1} \right| }{y_2-y_1}=\sqrt{y_2-y_1} \cdot \frac{\left| \sqrt{y_2}-\sqrt{y_1} \right| \cdot \left| \sqrt{y_2}+\sqrt{y_1} \right| }{(y_2-y_1)\cdot \left| \sqrt{y_2}+\sqrt{y_1} \right|} \leq \sqrt{y_2-y_1} \cdot 1$$
    $$\left| \sqrt{1-x_2^2}-\sqrt{1-x_1^2} \right| \leq \sqrt{1-x_2^2-1+x_1^2}=\sqrt{x_1^2-x_2^2}=\sqrt{(x_1-x_2)(x_1+x_2)} \leq \sqrt{2}\sqrt{x_1-x_2}$$
    $$\rightarrow~C=\sqrt{2},~m=\frac{1}{2}$$
    But in ##~\sqrt{x_1^2-x_2^2}~,~x_1^2-x_2^2 \leq 0##
     
  7. Sep 17, 2017 #6

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    Hi Karol, :oldsmile:
    How did you get this?
    I don't see it.
    For the record, this is the correct result.
    Can we first generalize to ##|\sqrt y_2 - \sqrt y_1| \le \sqrt{|y_2 - y_1|}## before substituting?
     
  8. Sep 17, 2017 #7
    Because the domain is [-1,1] ##~\displaystyle \frac{1}{\sqrt{y_2-y_1}} \geq 1##
     
    Last edited: Sep 17, 2017
  9. Sep 18, 2017 #8

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    Ah yes. Good!
    Can we prove ##|\sqrt y_2 - \sqrt y_1|\le \sqrt{|y_2-y_1|}## in the same way?
     
  10. Sep 18, 2017 #9
    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
     
  11. Sep 18, 2017 #10

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    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|##?
     
  12. Sep 18, 2017 #11

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    In the original statement, y2 > y1.
     
  13. Sep 18, 2017 #12

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    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.
     
  14. Sep 18, 2017 #13
    $$\left| \sqrt{y_2}-\sqrt{y_1} \right| \leq \frac{1}{\left| \sqrt{y_2}-\sqrt{y_1} \right|}\cdot \left| \sqrt{y_2}-\sqrt{y_1} \right| = \frac{\left| \sqrt{y_2}+\sqrt{y_1} \right|}{\left| \sqrt{y_2}+\sqrt{y_1} \right|} \frac{1}{\left| \sqrt{y_2}-\sqrt{y_1} \right|}\cdot \left| \sqrt{y_2}-\sqrt{y_1} \right| =$$
    $$=\left| \sqrt{y_2}-\sqrt{y_1} \right| \cdot \frac{\left| \sqrt{y_2}+\sqrt{y_1} \right| }{\left| y_2-y_1\right|} \leq \left| \sqrt{y_2}-\sqrt{y_1} \right| \cdot \frac{\left| \sqrt{y_2}+\sqrt{y_1} \right| }{\sqrt{\left| y_2-y_1\right|}}=$$
    $$=\frac{\left| y_2-y_1\right| }{\sqrt{ \left| y_2-y_1\right| }}=\sqrt{ \left| y_2-y_1\right| }$$
     
  15. Sep 19, 2017 #14

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    Oh, it's different now... and not correct any more... :oldeek:
    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.

    This step is incorrect.
    Consider for instance ##y_2-y_1=\frac 14##.
    Then:
    $$\frac{1}{|y_2-y_1|} = 4 \not\le 2 = \frac{1}{\sqrt{|y_2-y1|}}$$
     
  16. Sep 19, 2017 #15
    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:
    $$\frac{1}{\left| \sqrt{y_2}+\sqrt{y_1} \right|}=\frac{1}{\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}}=\frac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}$$
    $$\frac{\sqrt{9\cdot 4 }}{\sqrt{9}+\sqrt{4}}=\frac{6}{5}=1.2$$
    $$\frac{\sqrt{2\cdot 5 }}{\sqrt{2}+\sqrt{5}}=\frac{3.16}{3.64}=0.9$$
     
  17. Sep 20, 2017 #16

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    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?
     
  18. Sep 20, 2017 #17
    $$|\sqrt y_2 - \sqrt y_1| \overset ?\le \sqrt{y_2-y_1}~\rightarrow~y^2-2\sqrt{y_2}\sqrt{y_1}+y_1 \overset ?\le y_2-y_1$$
    $$\rightarrow~y_1-2\sqrt{y_2}\sqrt{y_1} \overset ?\le -y_1$$
    $$y_1-2\sqrt{y_2}\sqrt{y_1} \le y_1-2\sqrt{y_1}\sqrt{y_1} = y_1-2y_1=-y_1$$
     
  19. Sep 21, 2017 #18

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    That last line does not seem to lead anywhere does it?
    How about we make it:
    $$\rightarrow~2y_1 \overset ?\le 2\sqrt{y_2}\sqrt{y_1} \le 2\sqrt{y_2}\sqrt{y_2} = 2y_2$$
    This is always true isn't it (with ##y_2 > y_1 \ge 0##)?
     
  20. Sep 21, 2017 #19
    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:
    $$y_1-2\sqrt{y_2}\sqrt{y_1} \le y_1-2\sqrt{y_1}\sqrt{y_1} = y_1-2y_1=-y_1$$
    $$\rightarrow y_1-2\sqrt{y_2}\sqrt{y_1} \le...= -y_1$$
     
  21. Sep 21, 2017 #20

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    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.
     
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