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2 votes
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Representing an $L^2$-functional by a non-$L^2$-function on a dense subspace - Part II

Gro-Tsen's answer to your previous question provides a counterexample if you define $D$ to be all vectors in $\ell_2$ that are of the form $\sum_n a_n f_n$, where $f_n = e_n + e_{n+1}$, $(e_n)$ is the ...
6 votes

Representing an $L^2$-functional by a non-$L^2$-function on a dense subspace

I believe the following is a simple counterexample: Let $(X,\mu)$ be $\mathbb{N}$ with the counting measure (so I will be writing $\ell^2$ for $L^2(X,\mu)$. Let $g=0$ and $\tilde g(n) = (-1)^n$. Let ...
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6 votes
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Representing an $L^2$-functional by a non-$L^2$-function on a dense subspace

The answer is no and the following result provides a quite interesting counterexample. This is a known result, but I am not sure where to find it in the literature. Theorem. If $f\in L^1_{\rm loc}(\...
4 votes
Accepted

Left and right eigenvectors are not orthogonal

Yes, this is always true if $\lambda \not= 0$. The subsequent theorem shows a more general result. To formulate it, we need the following terminology: For an eigenvector $\lambda$ of a bounded linear ...
2 votes

Motivation for Heisenberg's modeling of observables

Sorry, for self answer, but I think this is what's happening. I don't know how this is related to Connes' explanation though. Any measurement can be interpreted as a combination of 'yes-no' ...
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0 votes

Function monotony between [0,T] and $L^2$

First, since you have $H^1(0,T)$ imbedds in $\mathscr{C}^0([0,T])$, $z$ can be seen as an element of $\mathscr{C}^0([0,T];L^2(\Omega))$ and you can speak without ambiguity of $z(t_1)$ and $z(t_2)$. ...
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