Thomas

If $X=Id$ (the identity matrix), the equation is true for any $A,B \in O(n)$.

Still (in 3), 4)) you have witten $\in$, and not $\notin$.

welcome

Finally ;-)

you found two different expressions for the inverse, and know that the inverse is unique. What does that imply?

no, the assumption is given

what does that tell you?

the axioms for groups say that the inverse of an element is unique. Now you know that a+b is an inverse of a+b and also that b+a is an inverse.

By assumption, the inverse of any $a\in G$ is $a$. So the inverse of $a+b$ is $a+b$. My calculation shows that also $b+a$ is inverse to $a+b$. What do you now need to finish? Look at my previous comment.

You know a bit more - first, you make an assumption about the inverse in your question. Then you have just, correctly, written "the" inverse. (As opposed to "some" inverse).

@Lisa The calculation shows that $b+a$ is the inverse of $a+b$. What do you know about the inverse?

@peek-a-boo is referring to the $L^p$ norm, wich is defined through an integral. And the answer to your question is no, this is not correct. The correct answer depends on $n$.

Fine. Good luck.

I'll delete my comments.

Sorry for the misunderstanding.

So it's wrongly put. But the obviously intended behaviour is to conclude something about the behvior at $a$, from an assumption made about the behavior at $a$

But the behavior in $b$ is not what they are actually intending to look at.

The authors made the mistake of writing $C^1([a,b])$

I think I can now see your point.

Ok. last try. The question is about the behaviour in $a$ and the relevant claim is about the behaviour in $a$, which is the open end of the interval. Sorry, but I really don't see how to explain this to you if you are constantly ignoring this.

Please. The exercise is about the behavior at the open end of the interval. If $a<0$ then for your example there is nothing to show.

the first term converges to $0$ and the second term oscilates between $-$ and $1$, so no limit.

Write down $f^\prime$. You get $2x\sin(x) - \sin(1/x)$ The limit of this function at $x=0$ does not exist.

$f$ is assumed to be differentiable on $(a,b]$. The exercise says that, if the limit of $f^\prime$ at the open side of the interval exists, then $f^\prime$ is continuous there. Once again, your example does not fulfil this assumption.

Calculate $f^\prime$. You'll get a $-\sin(\frac{1}{x})$ term, which is not continuos in $x=0$. At both ends of the interval the assumptions state that this limit exists. Differentiability alone is not sufficient.

The confusion comes from the fact that you created a "counterexample" on the uncritical side of the internval, which escaped me. $f$ is assumed to be differentiable in $b$, but not in $a$. There is nothing to show in $b$, and your example does not work cause the function is not differentiable in $b$ (and my first statement is still true, even though I've looked at the wrong end of the interval. If $f$ is differentiable in $b$ the corresponding limit exists).

@JoséCarlosSantos thanks, but no.

For this you shuold consult the first sentence.($C^1((a,b])$ with an interval closed on the rhs.

Not in the first sentence. After 'Show that'.

It does. Read it again.

The example you provided does not satisfy the hypothesis that $\lim_{x\rightarrow 0} f^\prime (x)$ exists. Apart from that the statement itself is true and is asked regularly here, so if you need to know the answer please search for it.

A ray is half of a straight line. Yes, with what you've done you can conclude it's a local diffeomorphism.

This is hard to read here in the comments section. The set $f(\mathcal{O})$ with your choices is the cone minus a ray on the cone which is perpendicular to the circles which make up the cone. The map $f$ is not a diffeomorphism since it covers the cone several times.

@MaryStar You have written down the parametriziation so you should decide, there is no general rule. I'd take $[0,2\pi)\times (0,\infty)$ (which is not nice since it has a boundary) or $(0,2\pi)\times (0,\infty)$ allowing the lack of one ray. If you take $(0,2\pi)\times (0,\infty)$ and $(-\pi,\pi)\times (0,\infty), $ say, you'd get your atlas.

@MaryStar Two subsets will do, one won't suffice in this case. You can also use more than two (as many as you want, actually), you only have to cover the manifold with the images of the maps, which should be defined on something diffeormorphic to Euclidean space (this is what is usually used in the definitions).

@MaryStar Because (each of) these line segments is/are diffeomorphic to an interval, while $S^1$ is not.

@MaryStar You have to look for coordinate patches which are homeomorphic to Euclidean space (e.g. Euclidean space itself, a unit disc, a product of intervals, ...). In the present case I already provided a possible choice, and the maps are also already given.

@MaryStar The circle $S^1$ is not diffeomorphic to $\mathbb{R}$, but it contains closed loops which cannot be deformed smoothly into a single point. Since it enters a factor of $\mathbb{R}\times S^1$ the product cannot be diffeomorphic to $\mathbb{R}^2$ either. Strictly speaking you don't need to know that for your task, but you have to write down the domain of definition for your charts. Even if you don't know that $\mathbb{R}\times S^1$ is not diffeomorphic to Euclidean space it is not clear that it is suitable as domain of definition for a chart. So you should try to find something else.

(have to leave now)

You're welcome

both

then boht $A_i-A_j$ and $C_i-C_j$ are small

now let $J_B := \max \{J_A, J_C\}$

You find $J_A, J_C$ such that the desired inequality holds for $A_i-A_j$ if $i,j >J_A$, and for $C_i -C_j $ if $i, j >J_C$.

this is a bit imprecise.

yes

you want me to continue?

as soon as $j, i> J_0$, say

correct

fine