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4h
comment Show that the operator sequence $ A_n = 1/2(A_{n-1} + A^{-1}_{n-1})$ converges strongly, $A_0 = I+T$, where $T$ is compact and $||T|| \le 1/2$.
Right, that works. I'm wondering if there's a way to see it without appealing to the diagonalization, though. (Because if you're going to diagonalize, why not just do that in the first place and be more explicit?)
4h
answered Show that the operator sequence $ A_n = 1/2(A_{n-1} + A^{-1}_{n-1})$ converges strongly, $A_0 = I+T$, where $T$ is compact and $||T|| \le 1/2$.
4h
comment Show that the operator sequence $ A_n = 1/2(A_{n-1} + A^{-1}_{n-1})$ converges strongly, $A_0 = I+T$, where $T$ is compact and $||T|| \le 1/2$.
Honest question: Must a self-adjoint operator with $1$ as the only point in its spectrum be the identity?
4h
comment Show that the operator sequence $ A_n = 1/2(A_{n-1} + A^{-1}_{n-1})$ converges strongly, $A_0 = I+T$, where $T$ is compact and $||T|| \le 1/2$.
I see what you're trying to do, but is this enough? All you've shown is that the spectrum of the limit is just $1$, which I don't think uniquely characterizes the identity. Or does it?
7h
comment How to write down the pull back of a differential form by exponential map?
Why is the integral expression you give equivalent to the more standard one (found here, for example: en.wikipedia.org/wiki/Derivative_of_the_exponential_map)?
1d
accepted Function that decays faster than any polynomial, but not in the Schwartz space?
1d
revised Function that decays faster than any polynomial, but not in the Schwartz space?
added 72 characters in body
1d
asked Function that decays faster than any polynomial, but not in the Schwartz space?
1d
comment Is a meromorphic function satisfying $f(2z)=\frac{f(z)}{1+f(z)^2}$ constant?
@mwomath Doesn't that just lead to a longer version of the answers below? You see it has to be at most cubic then do the same computation as Hagen von Eitzen...
1d
accepted Analytic continuation of a certain Dirichlet series
1d
comment Analytic continuation of a certain Dirichlet series
@Winther Great, thanks! If you post that as an answer, I will accept it.
1d
accepted Is a meromorphic function satisfying $f(2z)=\frac{f(z)}{1+f(z)^2}$ constant?
1d
asked Is a meromorphic function satisfying $f(2z)=\frac{f(z)}{1+f(z)^2}$ constant?
1d
asked Analytic continuation of a certain Dirichlet series
1d
comment When does “every closed path is homotopic to a point” imply the space is path connected?
@MickG This is valid in any $\mathbb R^n$ and some more general spaces. In one direction, path connectedness implies connectedness. In the other, connectedness plus a property called local path connectedness implies path connectedness. There are spaces that are connected but not path-connected. (Google "Topologist's Sine Curve.")
1d
comment When does “every closed path is homotopic to a point” imply the space is path connected?
@MickG Your "pseudo-starshaped" is just the property of being path connected, it seems.
1d
comment When does “every closed path is homotopic to a point” imply the space is path connected?
@MickG Contractibility does not imply a space is star-shaped! You only produce a continuous path from every $y$ to $x$. In a star-shaped domain, the connecting paths must be straight lines.
1d
comment When does “every closed path is homotopic to a point” imply the space is path connected?
@MickG I'm sure he has the usual definition in mind. Generally when we speak of regions in $\mathbb C$ in complex analysis, we are implicitly speaking of open connected (which implies path connected) regions.
1d
comment When does “every closed path is homotopic to a point” imply the space is path connected?
You're correct that he needs to include path connectedness in the definition of "simply connected." In general, being path connected neither implies or is implied by the property that every closed path can be contracted to a point. This is true even for subsets of $\mathbb R^n$.
2d
comment How to evaluate the derivatives of matrix inverse?
Very helpful answer. Thank you.