Suppose L_p is the usual Lebesgue space over (0,1) if you wish. Suppose T_j:L_1-->L_2 defines a sequence of continuous linear operators. Suppose l_1(L_1) is the Banach space of sequences from L_1 with norm (f_j_j-->||f_1||+||f_2||+... . Suppose L_2(l_inf) is the Banach space of sequences (f_j)_j from L_2 with the norm (f_j)_j-->||sup_j|f_j||. Finally, suppose T:l_1(L_1)-->L^2(l_inf) is a linear map defined by
(f_j)_j-->(T_j(f_j))_j.
It seems to me that the fact that T is well-defined, i.e. all outputs are in L_2(l_inf), AND each T_j is continuous implies T is continuous by the closed graph theorem. This is because the candidate limit (f_j)_j when arguing T has a closed graph has to satisfy f_j=T_j(x_j) where (x_j^n)_j converges to (x_j)_j in L_1(l_1).
My uncertainty stems from the following example. Fix T_1 and let T_j=log(j+9)T_1 for j>2. Since this sequence (T_j)_j is not uniformly L_1-->L_2 bounded, the corresponding operator T cannot be bounded(continuous). However, the slow growth of the operator norms is slow enough so that for (f_j)_j in L_1(l_1),
||sup_{j\le N}|T_j(f_j)|||
Dear Geoff Diestel ,
Why is (\sum_j (\log(j+9))^2||f_j||^2)^{1/2} finite when \sum_j ||f_j|| is?
It seems to me that this is not correct.
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