Question

Hi, I need help proving the function f:x->x+sin(x) is lipschitz-continuous...

Answer 1

Definition of a lipschitzian function:
\[k \in \mathbb{R}^{+}, \forall (a,b) \in \mathbb{R}^{2}, |f(a)-f(b)|\le k|a-b|\]

Now so far I only have that
\[|f(a)-f(b)|=|\sin(a)-\sin(b)+a-b|=|2\sin(\frac{a-b}{2})\cos(\frac{a+b}{2})+a-b|\]

And I'm not sure about how to use a correct triangular inequality to prove it...

Answer 2

You can try using the mean value theorem.

Answer 3

Is it correct to use the Fundamental Theorem of calculus?

Answer 4

Well, I'm open to any type of solutions, but I can't see how you would use those theorems in this case

Answer 5

Im still thinking about it, but basically, you want to show that the possible slopes of all the secant lines are bounded.\[|f(a)-f(b)|\leq k|a-b| \iff \left|\frac{f(a)-f(b)}{a-b}\right|\le k\]So if we can show that, we are done.

Answer 6

You can use the derivative to come to the conclusion that a function is Lipschitz, i dont know if that constitutes as a valid proof though.

Answer 7

someone1348, doesn't the Lipschitz condition imply that\[|f(a)-f(b)|\leq k|a-b|\implies\left|\frac{f(a)-f(b)}{a-b}\right|\leq k\implies\left|\frac{f(a+h)-f(a)}{h}\right|\leq k\]by putting \(b=a+h\)? This, in turn, would imply\[|f'(x)|\leq k,\]which means all we have do to is determine whether the derivative of \(f\) is bounded or not.

Answer 8

Yes, that's a geometrical aspect of lipschitz continuity, but I don't know if I can use differentiation, since a and b are not necessarily in a close neighborhood....
And I'd like a general solution that doesn't require that f can be differentiated (my math teacher is a pain, and specifically wanted us to do it without differentiating)

Answer 9

by using that you only imply that if its lipschitzian then its derivative is bounded, but not the other way around

Answer 10

you can use the triangle inequality

Answer 11

That's what I was thinking to use in my first post, but I can't see how to use it right and to set the inequality on the right side :S

Answer 12

I thought of using, since
\[|f(a)-f(b)|=|a-b+2*\sin(\frac{a-b}{2})\cos(\frac{a+b}{2})|\]
\[ \leq |a-b|+2|\sin(\frac{a-b}{2})\cos(\frac{a+b}{2})|\iff 2|\sin(\frac{a-b}{2})\cos(\frac{a+b}{2})|\leq (k-1)|a-b|\]

But then.... I'm not sure, obviously the left term is bounded, so that means I'm done? Or are there any mistakes?

Answer 13

hmm, maybe you can prove that f(x) = x and g(x) = sin(x) are individually Lipschitz, then prove that the sum of two Lipschitz functions is also Lipschitz?

Let f(x) and g(x) be Lipschitz, and h(x) = f(x)+g(x). Is h(x) also Lipschitz?

\[|h(x)-h(y)| = |f(x)+g(x)-f(y)-g(y)|\]\[=|f(x)-f(y)+g(x)-g(y)|\le |f(x)-f(y)|+|g(x)-g(y)|\]because f and g are Lipschitz, we obtain:\[k_1|x-y|+k_2|x-y|=(k_1+k_2)|x-y|\]so we have shown that:\[|h(x)-h(y)|\le K|x-y|\]

Answer 14

So the sum of two Lipschitz functions is also Lipschitz. Show that f(x) = x and g(x) = sin(x) are Lipschitz.

Answer 15

This is easy using the triangle inequality:
\[ |f(a) - f(b) = | \sin a + a - \sin b - b | \]
\[ \leq | \sin a - \sin b| + |a-b| \ \hbox{, by the triangle inequality } \]
\[ \leq | a-b| + |a-b| \ \hbox{, by the Mean Value Theorem } \]
\[ = 2 |a-b| \]

Answer 16

....or that lol.

Answer 17

Thanks!
By the way, if
\[A=\left\{ k \in \mathbb{R}^{+}, f k-lipschitzian \right\} \]
Then
\[\inf(A)=1\]
Because, by calling B and C the sets of lipschitz constants for sin and x->x, inf(A)=inf(B)+inf(C)=1, no?

Answer 18

If you're asking whether the inf of the set of Lipschitz constants for f(x) = sin x + x is 1, I'd say no; it's more likely to be 2. Just look at
\[ | f(δ)−f(−δ) | \]

for arbitrarily small δ

Answer 19

true... I think, in fact, sin is 1-lipschitzian?
Let's see...
\[|\sin(a)-\sin(b)|=2|\sin(\frac{a-b}{2})\cos(\frac{a+b}{2})|\]
and then I'm stuck =/

Answer 20

For sure the best bound k such that
\[ | \sin x - \sin y | \leq k |x-y| \]
is k = 1; just look at it around the origin.

To formally prove it: suppose there exists a Lipschitz constant k where k < 1. Then for all x,y
\[ \left| \frac{ \sin x - \sin y }{x -y} \right| \leq k < 1 \]
in which case the maximum value of the derivative of sin would be k < 1. But this is false. Hence k must be at least 1. Now we can prove with the MVT that it can be 1. Hence the inf(Lip constants of f) = min(Lip constants of f) = 1.

Answer 21

But to prove it's k=1, you have to first admit that sin is lipschitzian?

Answer 22

No. use the MVT. This is a very useful result (the result of applying the MVT to the function sin) to figure out and learn if you don't know it.

Answer 23

If you mean we need to show first sin is Lipschitz, yes.

Answer 24

...and the way to do it is with the MVT.

Answer 25

Is there any way to do it without it?

Answer 26

Why? There's no reason not to use a result from first year calculus in a course in analysis, which is where Lipschitz continuity belongs.

Answer 27

Forgot to state my problem asks first to prove f is lipschitzian, then to calculate inf(a) without using the mean value theorem... I forgot to say it because the problem is in french, and I didn't know that the theorem I knew was MVT in english... :S
And yeah... but she specifically prohibited us from using it in this problem, which is why I think more and more that french math is twisted...

Answer 28

you can use the unit circle and a simple geometric argument to show that

\[\sin(\theta)\le \theta\]

Answer 29

or better \[|\sin(\theta)|\le |\theta|\]

Answer 30

Hmm true, especially since before they had me prove that \[|\sin(x)|\leq x\]

Answer 31

Yep, I have that, it was a particular pain to write the proof formally, but then afterwards what do you suggest I do with it?

Answer 32

use it on \[|\sin(a)-\sin(b)|=2|\sin(\frac{a-b}{2})\cos(\frac{a+b}{2})|\]

Answer 33

Then use the identities you've been playing with above:
\[ | \sin x - \sin y | = 2 | \sin((x-y)/2) . \cos((x+y)/2) | \]
\[ \leq ... what? \]

Answer 34

You know you want |x-y| on the right-hand side. See how to get there.

Answer 35

\[\leq|x-y|\cos(\frac{x+y}{2})\] ?

Answer 36

now bound the cosine function

Answer 37

yes and someone doing analysis should know how to bound cos

Answer 38

...yep, duh. crap. I feel extremely bad now hahaha

Answer 39

Thanks to you both.