Suppose there is a ball such that every subball of is such that . How would this contradicts the definition of ?

Following the definition, one can see that is a nonempty open set, and thus there exists a ball which is disjoint from . It seems that the quoted observation above is stronger.

]]>By “disjoint”, do you mean the intersection is empty or the distance is positive?

]]>Example: X=L^1 and \phi_n \to \delta_0 (as n\to \infty) is a compactly supported, positive, smooth approximation of unity (as in http://en.wikipedia.org/wiki/Mollifier). Furthermore let Y=C(R,L^1), the continuous real valued functions on \R with bounded L^1 norm. This space is normed by L^1, and as it’s completion is L^1 it is not Banach. Now consider the operators: T_n: f\to f * \phi_n. By Young’s inequality and since \int \phi_n = 1, \phi_n>=0, then ||T_n(f)||_{L^1}=||f*\phi_n||_L_1 \leq ||f||_{L_1}||\phi_n||_L_1=||f||_L_1. In particular ||T_n|| \leq 1, and we have uniform boundedness. Moreover for any continuous function $f$ (which are dense in X), we have by standard results for mollifiers that T_n(f)=f*\phi_n \to f in L^1, and since the entire sequence is in C(R), then the convergence is strong in Y. Thus we have dense convergence. But now let $f=\xhi(0,1)$, the characteristic function of the unit interval. Clearly f \nin Y, but f \in X. Also, T_n(f)\to f in L^1, and hence the limiting function is “not in” Y, and there is no limit for the operators T_n:X\to Y.

Fixes: Obviously if you replace Y with a banach space L^1, then T_n\to Id (Identity). 3\to2 would be the only part requiring this. Another fix: Adjust the assumptions and assume in addition to convergence that there is some a priori limiting operator. Say if we assume that T_n \to T on a dense set and $T: X\to Y$ is some bounded linear operator, then the problem also disappears. In the above example e.g. take T_n f = f*\phi_n – f*\phi_{n+1}, then $T_n(y)\to 0$ on a dense set and in fact $T_n(x)\to 0$ everywhere.

]]>Example: X=L^1 and \phi_n \to \delta_0 (as n\to \infty) is a compactly supported, positive, smooth approximation of unity (as in http://en.wikipedia.org/wiki/Mollifier). Furthermore let Y=C(R,L^1), the continuous real valued functions on \R with bounded L^1 norm. This space is normed by L^1, and as it’s completion is L^1 it is not Banach. Now consider the operators: T_n: f\to f * \phi_n. By Young’s inequality and since \int \phi_n = 1, \phi_n>=0, then ||T_n(f)||_{L^1}=||f*\phi_n||_L_1<= ||f||_{L_1}||\phi_n||_L_1=||f||_L_1. In particular ||T_n||2 would be the only part requiring this. Another fix: Adjust the assumptions and assume in addition to convergence that there is some a priori limiting operator. Say if we assume that T_n \to T on a dense set and $T: X\to Y$ is some bounded linear operator, then the problem also disappears. In the above example e.g. take T_n f = f*\phi_n – f*\phi_{n+1}, then $T_n(y)\to 0$ on a dense set and in fact $T_n(x)\to 0$ everywhere.

]]>Example: X=L^1 and \phi_n \to \delta_0 (as n\to \infty) is a compactly supported, positive, smooth approximation of unity (as in http://en.wikipedia.org/wiki/Mollifier). Furthermore let Y=C(R,L^1), the continuous real valued functions on \R with bounded L^1 norm. This space is normed by L^1, and as it’s completion is L^1 it is not Banach. Now consider the operators: T_n: f\to f * \phi_n. By Young’s inequality and since \int \phi_n = 1, \phi_n>=0, then ||T_n(f)||_{L^1}=||f*\phi_n||_L_1<= ||f||_{L_1}||\phi_n||_L_1=||f||_L_1. In particular ||T_n||2 would be the only part requiring this. Another fix: Adjust the assumptions and assume in addition to convergence that there is some a priori limiting operator. Say if we assume that T_n \to T on a dense set and $T: X\to Y$ is some bounded linear operator, then the problem also disappears. In the above example e.g. take T_n f = f*\phi_n – f*\phi_{n+1}, then $T_n(y)\to 0$ on a dense set and in fact T_n(x)\to 0 everywhere.

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