SoG

n=1, 2,...,m

(a_n) is a sequence of real numbers and f_n : [0, 1] \to R is a sequence of continuous functions. Does there exist any Borel measure s such that a_n=\int_{0^1}f_n ds?

Taken from the book TVS by Narici and Beckenstein.

This 1958 paper of Morisuke Hasumi described the said isomorphism.

A nls satisfying metric extension property is isomorphic to C(S) where S is Stonean.

@leslietownes Any hint?

@leslietownes

Could you give an example of a Banach space satisfying finite metric extension property that doesn't satisfy metric extension property?

If dim(M) is finite, we call finite metric extension property.

A Banach space Y has the metric extension property if for every Banach space X and every vector subspace M of X, Y has the metric extension property from M to X.

How to upload image here( chat)?

Is it homeomorphic to joint spectrum of (N_1, N_2)?

N_1, N_2 are commuting normal operators

What is the maximal ideal space of C*(N_1, N_2) ?

Give an example of a non degenerate representation of a C* algebra that is not faithful.

where $f_i\in H^2(B(W)) and $U_1U_2=U=U_2U_1$ , W is a wandering subspace of V=V_1V_2 , U is unitrary part of Wold decomposition of V=V_1V_2

If (V_1, V_2) is a pair of commuting isometries on a Hilbert space H, then $V_i= diag( M_{f_i}, U_i) $

What is the C* algebra generated by a pair of commuting isometries?

@leslietownes Could you recommend any books, journals, or papers on dilation theory in finite dimensions?

@leslietownes Nice:)

How to prove that any rank 1 operator on l^2 space is of the form S^m(I-SS*) S*^n ?

@Jakobian It's not continuous at $0$.

@Jakobian Sorry $[0, \infty) $

Q. Let $f\in C[0, 1]$ and $f\ge 0$ and suppose that $\int_{0}^{\infty} f $ is finite. Does this imply $\int_{0}^{\infty} f^2$ also finite?

@leslietownes Thanks:)

Here is the SVD.

How to find the explicit polar form?

Write the matrix (col 1( 1 , 0) col 2 (1, 1)) as U•P where U is an isometry and P is positive.

Got it. Please ignore :)

Is it always true that $T$ is unitarily equivalent to $T*$ for any bounded operator $T$ on a Hilbert space?

And I am still working on it :(

I thought isometry could add something weaker than commutativity to imply the normality of the product.

@leslietownes Got it, thanks :)

An isometry is necessarily normal, and if they commute, then the product is normal.

What can be said about the product of an isometry and a normal operator?

@ThomasFinley Yes. "Continuous image of a compact set is compact ".

$f(cl(A)) $ is compact (implies bounded).

@ThomasFinley $cl(A) $ is compact.

Then...How to proceed ?

$trace(P+Q)(P+Q) ^t<1$

$I-(P+Q) $ is invertible iff $\|P+Q\|<1$

rank(P) =trace(P) and rank(Q) =trace(Q)

I have found 2 different ways to prove it. But I am interested to prove it as follows:

$P, Q \in M_n(\Bbb{R}) $ and $P^2=Q^2=I $ $I-(P+Q) $ is invertible. Then show that rank(P) =rank(Q)

@TedShifrin You mean Goursat theorem?

@TedShifrin Yes. $f$ is continuous on an open set $U$ and integral of $f$ over every triangle is $0$ then $f$ is holomorphic on $U$.

Is it possible to have a non-convex open set and a holomorphic function such that the integral over every triangle is 0, but there exists a closed curve such that the integral is nonzero?

Is $K[x, y]/(x, y) $ free?

Consider $M=R=K[x, y]$ and $N=(x, y) $