My Math Forum  

Go Back   My Math Forum > College Math Forum > Real Analysis

Real Analysis Real Analysis Math Forum


Reply
 
LinkBack Thread Tools Display Modes
November 25th, 2018, 11:57 AM   #1
Newbie
 
Joined: Nov 2018
From: Canada

Posts: 4
Thanks: 0

Adherence of subset - Kernel

Having Y, a subspace of X. How can we show that the adherence of Y can be expressed as :
Adherence of Y = intersection of { Ker(f) | f element of X* , Y contained in Ker(f)}
whatsup123 is offline  
 
November 25th, 2018, 12:07 PM   #2
Senior Member
 
Joined: Oct 2009

Posts: 752
Thanks: 261

Use Hahn-Banach
Micrm@ss is offline  
November 25th, 2018, 12:22 PM   #3
Newbie
 
Joined: Nov 2018
From: Canada

Posts: 4
Thanks: 0

Can you give more details please ?
whatsup123 is offline  
November 25th, 2018, 03:09 PM   #4
SDK
Senior Member
 
Joined: Sep 2016
From: USA

Posts: 578
Thanks: 345

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
Originally Posted by whatsup123 View Post
Can you give more details please ?
I'm assuming you are given that $X,Y$ are Banach spaces yes? Then fix $x_0 \in X \setminus \text{cl}(Y)$ and define a subspace, $X_0 = Y \bigcup \text{span}(x_0)$ and a linear functional, $x_0^* \in X_0^*$ by
\[x_0^*(x) = \frac{d(x,Y)}{d(x_0,Y)} \]
where $d: X \to \mathbb{R}$ is the set distance, $d(x,S) = \sup\{\left| \left|x-y \right| \right|: y \in S\}$.

Now, show that you can extend $x_0^*$ to a linear functional on the entire space and that this extension is a competitor in your intersection since $Y \in \text{ker}(x_0^*)$.
SDK is offline  
November 25th, 2018, 03:19 PM   #5
Newbie
 
Joined: Nov 2018
From: Canada

Posts: 4
Thanks: 0

Quote:
Originally Posted by SDK View Post
I'm assuming you are given that $X,Y$ are Banach spaces yes? Then fix $x_0 \in X \setminus \text{cl}(Y)$ and define a subspace, $X_0 = Y \bigcup \text{span}(x_0)$ and a linear functional, $x_0^* \in X_0^*$ by
\[x_0^*(x) = \frac{d(x,Y)}{d(x_0,Y)} \]
where $d: X \to \mathbb{R}$ is the set distance, $d(x,S) = \sup\{\left| \left|x-y \right| \right|: y \in S\}$.

Now, show that you can extend $x_0^*$ to a linear functional on the entire space and that this extension is a competitor in your intersection since $Y \in \text{ker}(x_0^*)$.


Actually, it doesn't say anywhere in the exercise that its Banach spaces....
whatsup123 is offline  
Reply

  My Math Forum > College Math Forum > Real Analysis

Tags
adherence, aherence, kernel, normed, space, subset, vector



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
kernel function mhhojati Linear Algebra 1 January 3rd, 2016 10:35 AM
Denumerability of a subset of a denumerable subset bschiavo Real Analysis 9 October 6th, 2015 11:17 AM
Subset of a Function g[a] subset g[b] redgirl43 Applied Math 1 April 21st, 2013 06:20 AM
Kernel of a function c.P.u1 Linear Algebra 1 January 6th, 2011 06:43 AM
The Kernel... DanielThrice Abstract Algebra 3 December 20th, 2010 03:03 PM





Copyright © 2019 My Math Forum. All rights reserved.