Timeline for Density of linear subspaces in $C(K)$
Current License: CC BY-SA 4.0
16 events
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| Nov 1, 2023 at 13:31 | comment | added | Pietro Majer | The advantage with the continuous measure is that the $x$-part of $f$ has $|\psi(g_{x,2m})|\le\frac1{2m}$, so in any case $\psi(f)$ can be put to zero adding $\lambda$ times the $y$-part, which is larger. | |
| Nov 1, 2023 at 13:14 | comment | added | Pietro Majer | Actually I see a (minor?) problem with the version with the discrete measure. The functional is now $$\Psi:=\sum_{k=1}^\infty\frac{(-1)^k}{k^2}\delta_{q_k}$$ for an enumeration $\{q_k\}_k$ of $\mathbb Q\cap [0,1]$. Then for all $k$ we have $\Psi(g_{q_k,n})=\frac{(-1)^k}{k^2}+o(1)$ as $n\to+\infty$. So the same construction with $f=g_{x,m}+\lambda g_{y,n}$ works, but for the case where $x=q_k$, for odd $k$: I do not see how to make $\Psi(f)\ge0$ and $f\ge0$ then. | |
| Nov 1, 2023 at 12:23 | vote | accept | Julian Hölz | ||
| Nov 1, 2023 at 12:12 | comment | added | Giorgio Metafune | And what about the counterexample with the discrete measure you added? I had a look and seemed to work? | |
| Nov 1, 2023 at 10:33 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Nov 1, 2023 at 10:21 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Nov 1, 2023 at 5:00 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Oct 31, 2023 at 23:36 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Oct 31, 2023 at 23:26 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Oct 31, 2023 at 23:12 | comment | added | Pietro Majer | b) Then one fixes $m$ so large that $(1-m|x-t|)_+$ has support in $U$, disjoint from the supports of $g_+$ and $g_-$, and such that $\psi(g_-)<\psi((1-m|x-t|)_+)<\psi(g_+)$ (note that $\psi((1-m|x-t|)_+)=o(1)$ as $m\to\infty$). Finally one chooses $g_-$ or $g_+$, and fices $0\le\epsilon<1$ | |
| Oct 31, 2023 at 23:02 | comment | added | Pietro Majer | a) The function $g$ can be taken as well of the form $g(t):=(1-n|y-t|)_+$, with $y\in U\setminus\{x\}$ a Lebesgue point of either $B$ of $K\setminus B$. THis make the sign of $\psi(g)$ positive resp. negative, if $n$ is large enough; one also takes $n$ so large that $g$ has support in $U\setminus\{x\}$. Say we do both: a $g_-$ and a $g_+$, with $\psi(g_-)<0<\psi(g_+)$ | |
| Oct 31, 2023 at 22:44 | comment | added | Giorgio Metafune | Can you explain a) how to choose $g$? b) how to fix $m, \epsilon$? | |
| Oct 31, 2023 at 22:09 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Oct 31, 2023 at 21:52 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Oct 31, 2023 at 21:36 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Oct 31, 2023 at 21:30 | history | answered | Pietro Majer | CC BY-SA 4.0 |