Radii and centers in Banach spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T19:49:42Z http://mathoverflow.net/feeds/question/20613 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20613/radii-and-centers-in-banach-spaces Radii and centers in Banach spaces David R. MacIver 2010-04-07T09:32:44Z 2010-04-08T18:52:39Z <p>Suppose I have a Banach space $V$ and a set $A \subseteq V$ such that for all $\epsilon > 0$ there exists $v$ such that $A \subseteq \overline{B}(v, r + \epsilon)$. Does there exist $c$ such that $A \subseteq \overline{B}(c, r)$?</p> <p>The answer is clearly yes for finite dimensional normed spaces: Define $T_\epsilon = \bigcap_{a \in A} \overline{B}(a, r + \epsilon)$. The $T_\epsilon$ form a chain of closed sets and for $\epsilon > 0$ are non-empty, so have the finite intersection property. Thus when $V$ is finite dimensional they have non-empty intersection, and any element of the intersection works as $c$.</p> <p>For more general Banach spaces I feel like you should be able to choose a cauchy sequence $x_n$ such that $x_n \in T_{\epsilon_n}$ with $\epsilon_n \to 0$, but I can't seem to make it work.</p> <p>Note that an arbitrary choice of $x_n \in T_{\epsilon_n}$ can't be guaranteed to be Cauchy: If $V$ is $l^\infty$ and $A = { x : x_0 = 0, ||x|| \leq 1 }$ then diam$(T_\epsilon) \geq 2$ because you can choose $c_0$ arbitrarily in $[-1, 1]$</p> <p>Note also that the assumption of $V$ a Banach space is essential: If $V$ is not Banach and $c$ is an element of the completion which is not in $V$ then $A = \overline{B}(c, 1) \cap V$ has no center. </p> http://mathoverflow.net/questions/20613/radii-and-centers-in-banach-spaces/20616#20616 Answer by Robin Chapman for Radii and centers in Banach spaces Robin Chapman 2010-04-07T10:14:28Z 2010-04-07T10:14:28Z <p>The answer may be affirmative when $V$ is a reflexive Banach space. Each $T_\epsilon$ is a closed bounded convex set. If the intersection of the $T_\epsilon$ is nonempty then each element of this intersection is an admissible centre. I found a paper online</p> <p>Sharma, B. K., Dewangan, C. L. Fixed point theorem in convex metric space, <em>Zb. Rad. Prirod.-Mat. Fak. Ser. Mat.</em> 25 (1995), no. 1, 9-18 <a href="http://www.emis.de/journals/NSJOM/framepaper.htm" rel="nofollow">http://www.emis.de/journals/NSJOM/framepaper.htm</a></p> <p>which asserts that every weakly compact convex subset of a Banach space has the property that a chain of nonempty closed convex subsets has nonempty intersection. Alas, they don't give a proof or reference for this.</p> <p>By a theorem of W. F. Eberlein,</p> <p>Weak Compactness in Banach Spaces, <em>Proc. Natl. Acad. Sci. USA</em> 33 (1947), 51–53</p> <p>the closed unit ball in a Banach space is weakly compact iff the Banach space is reflexive.</p> http://mathoverflow.net/questions/20613/radii-and-centers-in-banach-spaces/20672#20672 Answer by J Bytheway for Radii and centers in Banach spaces J Bytheway 2010-04-07T23:12:08Z 2010-04-08T06:41:25Z <p>I believe that the property does not hold for all Banach spaces, but my counterexample is a little involved. If you've the patience then follow me through...</p> <p>Let $V=\bigoplus_{n=1}^\infty \ell^n_2$ where $\ell^p_2$ is $\mathbb{R}^2$ with norm $\lVert\cdot\rVert_p$ (Note: $n$ is taking the role of $p$). For $i\geq1$ and $j\in{0,1}$ we have $e_{i,j}$, the $j^{th}$ standard basis vector of $\ell^i_2$ in $V$.</p> <p>Give $V$ the norm $\lVert v\rVert=\sup_n\lVert v_n\rVert_n$.</p> <p>Let <code>$W=\{v\in V:\lVert v_n\rVert_n\to 0\}$</code>. I assert that $W$ is a Banach space. Certainly every $e_{i,j}\in W$.</p> <p>Let <code>$A=\{e_{k,0}+e_{k,1}, e_{k,0}-e_{k,1}:k\geq 1\}$</code>.</p> <p>Fact: Let $r(A)$ be the infimum of radii of balls containing $A$. Then $r(A)\leq1$</p> <p>Proof:</p> <p>Let $c_N=\sum_{i=1}^n e_{i,0}$. We wish to compute the distance of each point of $A$ from $c_N$.</p> <p>For $k\leq N$ we have $\lVert c_N-e_{k,0}-e_{k,1}\rVert$ $=\lVert\sum_{i=1\ (i\not=k)}^Ne_{i,0}-e_{k,1}\rVert$ <code>$=\sup\{\lVert e_{i,0}\rVert_i:i\leq N,i\not=k\}\cup\{\lVert-e_{k,1}\rVert_k\}$</code> $=1$ and similarly for $\lVert c_N-e_{k,0}+e_{k,1}\rVert$.</p> <p>For $k>N$ we have $\lVert c_N-e_{k,0}-e_{k,1}\rVert$ $=\max(\lVert c_N\rVert,\lVert e_{k,0}+e_{k,1}\rVert_k)$ $= \max(1,(1+1)^\frac{1}{k})$ $= 2^\frac{1}{k}$ $\leq 2^\frac{1}{N}$ and similarly for $\lVert c_N-e_{k,0}+e_{k,1}\rVert$.</p> <p>Thus $A\subseteq \overline{B}(c_N,2^\frac{1}{N})$ and so $r(A)\leq2^\frac{1}{N}$. Letting $N\to\infty$ we have $r(A)\leq 1$.</p> <p>QED</p> <p>Fact: $A$ is not contained in a ball of radius $1$.</p> <p>Proof:</p> <p>Suppose $A\subseteq \overline{B}(c,1)$. Then in particular for every $n$ we have $\lVert c-e_{n,0}-e_{n,1}\rVert\leq 1$ and thus $\lVert c_n-e_{n,0}-e_{n,1}\rVert_n\leq 1$. Similarly $\lVert c_n-e_{n,0}+e_{n,1}\rVert_n\leq 1$.</p> <p>Simple consideration of $\ell^n_2$ shows that this implies $c_n=e_{n,0}$. Thus $\lVert c_n\rVert=1\not\to0$ and $c\not\in W$, contradicting the assumption.</p> <p>QED</p> http://mathoverflow.net/questions/20613/radii-and-centers-in-banach-spaces/20770#20770 Answer by Ady for Radii and centers in Banach spaces Ady 2010-04-08T18:52:39Z 2010-04-08T18:52:39Z <p>I think that <a href="http://www.ams.org/journals/tran/1982-271-02/S0002-9947-1982-0654848-2/S0002-9947-1982-0654848-2.pdf" rel="nofollow">http://www.ams.org/journals/tran/1982-271-02/S0002-9947-1982-0654848-2/S0002-9947-1982-0654848-2.pdf</a> [together with its references] provides us with several counterexamples [as well as with some remarkable examples], in the infinite-dimensional framework.</p>