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I am currently doing some revision for an exam next week when I came across this question from 3 years previous. I am a bit confused on how to tackle this question and lack a thorough understanding of the Lipschitz continuity. Can anybody help?? :) enter image description here

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  • $\begingroup$ Well, which bits are you stuck on? Part a is the most straight-forward, how are you doing there? $\endgroup$ – Simon S Jun 13 '15 at 23:57
  • $\begingroup$ My understanding of Lipschitz continuity is a bit vague so I'm unsure how to go about it. Do you have any advice on Lipschitz?? $\endgroup$ – Chloe Jun 14 '15 at 0:05
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Hint: For (a), think about triangle inequality, $|f+g|\leq |f|+|g|$. So what would happen when both $f$ and $g$ are Lipschitz?

For (b), suppose by contradiction that $f(x)=x^2$ is indeed Lipschitz. Then there must a positive constant $L<\infty$ such that $|x^2-y^2|\leq L|x-y|$. Take $y:=0$, and $x>0$. What can you say about the behavior of $L$?

For (c), use (b) to get a counter-example.

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For the first one, just use the triangular inequality :

$$|(f+g)(x) - (f+g)(y)| = |( f(x) - f(y) ) + (g(x)-g(y)) | \leq |f(x) - f(y) | + |g(x)-g(y) |$$

$$\leq L|x-y|+L'|x-y| = (L+L')|x-y|$$

For the second one, you can write :

$$|x^2 - y^2| = |x+y|\cdot|x-y|$$

Now, for every constant $M$, there exist $(x,y) \in \mathbb{R}^2$ such that $|x+y| > M$, so $x\mapsto x^2$ cannot be Lipschitz on R

For the last question, you can take $f(x)=g(x)=x$ and use the previous question

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Hints:

a) Use the triangle inequality and choose the max of the two Lipschitz constants $L_f$ and $L_g$

b) Argue by contradiction and use the Mean Value Theorem

c) Use the example of part b). The function $f(x) = x$ is Lipschitz with $L = 1$ !

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  • $\begingroup$ How do I use b for c? I have completed b and got a contradiction obviously as |2L +1| is not smaller than L which is smaller than -1. But I'm not sure how to use this for c? $\endgroup$ – Chloe Jun 14 '15 at 0:20
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Loosely, what it means for a function to be $L$-Lipschitz continuous, is that for any two pair of points on the graph of $f$, the absolute value of the slope of the line connecting them is not larger than $L$. This puts a restriction on how "fast" the function can move.

$1)$ If $f$ and $g$ are (respectively) $L$ and $L'$ Lipschitz continuous, then:

$$|f(x) - f(y)| \le L|x - y|, \ \forall \ x,y$$

$$|g(x) - g(y)| \le L' |x - y|, \ \forall \ x,y$$

Then, $\forall$ $x,y$:

$$|(f+g)(x) - (f+g)(y)| = |f(x) + g(x) - f(y) - g(y)| = |f(x) - f(y) + g(x) - g(y)| \le |f(x) - f(y)| + |g(x) - g(y)| \le (L + L')|x - y|$$

Which shows that $f+g$ is $L+L'$-Lipschitz.

$2)$ If it is, then there would be $L>0$ for which $|x^2 - y^2| \le L|x - y|$ for all $x,y \in \mathbb R$. Take $x := L + 1$ and $y:= L$. Find a contradiction.

$3)$ The functions $f(x) = g(x) = x$ are $1$-Lipschitz on $\mathbb R$, though $(fg)(x) = x^2$ is not.

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  • $\begingroup$ with two, if I plug in x=L+1 and y=L I get |2L+1| I think if I did it correctly? Where do I go from here? $\endgroup$ – Chloe Jun 14 '15 at 0:10
  • $\begingroup$ On one hand $|x^2 - y^2| = 2L +1$. On the other, $L|x - y| = L$. So we get $2L + 1 \le L$, so $L \le -1$ which is contradictory. $\endgroup$ – user230734 Jun 14 '15 at 0:13
  • $\begingroup$ Okay good, I did that part right! Thankyou for all your help, going to give 3 a go now :) $\endgroup$ – Chloe Jun 14 '15 at 0:15

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