My Math Forum Adherence of subset - Kernel

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 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)}
 November 25th, 2018, 12:07 PM #2 Senior Member   Joined: Oct 2009 Posts: 752 Thanks: 261 Use Hahn-Banach
 November 25th, 2018, 12:22 PM #3 Newbie   Joined: Nov 2018 From: Canada Posts: 4 Thanks: 0 Can you give more details please ?
November 25th, 2018, 03:09 PM   #4
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Quote:
 Originally Posted by whatsup123 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 $Y \in \text{ker}(x_0^*)$.

November 25th, 2018, 03:19 PM   #5
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Quote:
 Originally Posted by SDK 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....

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

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