Timeline for Why isn't the theorem of approximation applicable in Banach spaces?
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| Aug 24, 2011 at 18:56 | comment | added | Margaret Friedland | @Zen: strict convexity (which is the crucial property here) is indeed defined as a property of norms; one of equivalent statements is the following: A Banach space $(X, \|\cdot\|)$ is strictly convex if and only if $x \neq y$ and $\| x \| = \| y \| = 1$ together imply that $\| x + y \| < 2$. In other words, balls are round in a strictly convex space. | |
| Aug 16, 2010 at 2:30 | history | edited | Zen Harper | CC BY-SA 2.5 |
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| Aug 13, 2010 at 10:33 | comment | added | Andreas Thom | Ian Morris has already pointed out a counterexample in his comment to your question. Take $V= {\mathbb R}^2$ with the $\infty$-norm $\|(x,y)\| = \max \lbrace |x|,|y|\rbrace$. Now set $B = \lbrace \xi \in V \mid \|\xi\| \leq 1 \rbrace= \lbrace(x,y) \mid -1 \leq x \leq 1, -1 \leq y \leq 1 \rbrace$ and observe that there is no unique point in $B$ which is closest to the vector $(2,0)$. Indeed, the points $(1,\alpha)$ have distance $1$ to the vector $(2,0)$ for all $\alpha \in [-1,1]$. | |
| Aug 13, 2010 at 9:39 | comment | added | Linda Raabe | I think I asked the wrong question. Of course the proof for Hilbert spaces makes essential use of the scalar product, but that doesn't explain why the proof doesn't work for Banach spaces. I think a counter example could be useful. | |
| Aug 13, 2010 at 9:11 | history | answered | Zen Harper | CC BY-SA 2.5 |