Skip to main content
PPB's user avatar
PPB's user avatar
PPB's user avatar
PPB
  • Member for 2 years, 1 month
  • Last seen this week
comment
What makes the generalized projection different than metric on a Banach space?
Sure. I have difficulty in understanding why the generalized projection operator has been employed even though we always find a metric for every Banach space. Also, in one paper, the concept of best proximinality has been introduced by using the generalized projection (see Guan, Wei-Bo, and Wen Song. "W-approximative compactness and continuity of the generalized projection operator in Banach spaces." Journal of approximation theory). So far, I have understood that the concept is restrictive. So, what is the motivation behind using this operator anyway? I hope my question is clear.
Loading…
comment
Kadec-Klee property of an equivalent norm on a Hilbert space
Thank you for your comment, sir. I have made the question clear in my recent edit. Also, it is not a homework question. This example can be found in the paper Mosco convergence and the kadec property-DOI- 10.1090/S0002-9939-1989-0969313-4. This equivalent norm does not have the dual kadec property, as shown in the paper. However, I am confused regarding whether this norm has kadec property on the space itself or not.
revised
Loading…
Loading…
awarded
comment
Verifying the Cauchy behavior of a sequence
Indeed, I am talking about the convergence of $\{x_n\}$ here. You are right, sir. It may need some other criteria to be fulfilled for the convergence of $\{x_n\}$.
revised
Verifying the Cauchy behavior of a sequence
added 210 characters in body
Loading…
comment
Verifying the Cauchy behavior of a sequence
Also, I have made some changes to clarify my question. Thank you.
revised
Verifying the Cauchy behavior of a sequence
added 210 characters in body
Loading…
comment
Verifying the Cauchy behavior of a sequence
@Willie Wong, my question is: Under what assumptions on T or any other hypothesis does the sequence $\{x_n \}$ become Cauchy?
Loading…
revised
Loading…
comment
Strong sub-differentiability of an equivalent strictly convex norm
Ok sir. My confusion is to check whether the result that $\ell_1$ with the new norm is SSD or not. Is there a posibility that, if we take the original norm $\Vert \cdot \Vert_1$ to be SSD, does this imply the norm $\Vert \cdot \Vert$ is SSD for $f \in NA(c_0(\mathbb{N})$. Also, I have mentioned the author names in my recent edits.
comment
Strong sub-differentiability of an equivalent strictly convex norm
Sir, here the $\Vert \cdot \Vert$ defines a dual norm on $\ell_1(\Gamma)$, which is a strictly convex dual norm, which will imply the predual norm on $c_0(\Gamma)$ is gateaux differentiable. And for a strictly convex Banach space $(X_2, \Vert \cdot \Vert_2)$, we can find a continuous, linear and one-one operator $T: (X_1, \Vert \cdot \Vert_1) \to (X_2, \Vert \cdot \Vert_2)$ such that $\Vert x \Vert = \Vert x \Vert_1+\Vert x \Vert_2, x \in X_1$ defines a strictly convex norm on $X_1$.
comment
Strong sub-differentiability of an equivalent strictly convex norm
As per my knowledge, we can extend the result to $\ell_1(\Gamma)$ if we could prove it for $\ell_1(\mathbb{N})$. However, knowing the SSD of the norm for $\ell_1(\mathbb{N})$ solves my query.
revised
Loading…
comment
Strong sub-differentiability of an equivalent strictly convex norm
Thank you, sir. I have made edits to my question.
revised
Loading…
Loading…
1
2 3 4 5