## Norm attaining bilinear forms on $$L_1(\mu)$$.(English)Zbl 0964.46008

Given a real or complex Banach space $$X$$ and a natural number $$N$$, let $${\mathcal L}^N(X)$$ denote the space of all continuous $$N$$-linear forms on $$X$$ and we say that $$\phi\in{\mathcal L}^N(X)$$ attain its norm if there are $$x_1,x_2,\dots, x_N\in B_X$$ (the closed unit ball of $$X$$) such that $|\phi(x_1,x_2,\dots, x_N)|= \|\phi\|:= \sup\{|\phi(y_1,\dots, y_N)|: y_1,\dots y_N\in B_X\}.$ Let $${\mathcal A}{\mathcal L}^N(X)$$ be the set of norm attaining continuous $$N$$-linear forms on $$X$$. The question is whether $${\mathcal A}{\mathcal L}^N(X)$$ is dense in $${\mathcal L}^N(X)$$ or not. It is shown that given a finite measure $$\mu$$, the set of norm attaining bilinear forms is dense in the space of all continuous bilinear forms on $$L_1(\mu)$$ if and only if $$\mu$$ is purely atomic.
Main result: Given a finite measure $$\mu$$, the following statements are equivalent:
(1) $$\mu$$ is purely atomic.
(2) $${\mathcal A}{\mathcal L}^N(L_1(\mu))$$ is dense in $${\mathcal L}^N(L_1(\mu))$$ for any natural number $$N$$.
(3) $${\mathcal A}{\mathcal L}^N(L_1(\mu))$$ is dense in $${\mathcal L}^N(L_1(\mu))$$ for some $$N\geq 2$$.
(4) $${\mathcal A}{\mathcal L}^2(L_1(\mu))$$ is dense in $${\mathcal L}^2(L_1(\mu))$$.

### MSC:

 46B28 Spaces of operators; tensor products; approximation properties 46B20 Geometry and structure of normed linear spaces 46B22 Radon-Nikodým, Kreĭn-Milman and related properties
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