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I am trying to show that $L^1(\mathbb{R})$ is not isometrically isomorphic to the dual of a Banach space. I know that one can show this by showing that the unit ball in $L^1(\mathbb{R})$ has no extremal points. Then the result follows by the Krein-Milnam and Banach-Alaoglu theorems.

As suggested in this question, if the measure space was finite you could find a constant such that $f\chi_{[0,c]}$ has exactly half the $L^1$ norm of $f$. Does this proof generalise to the infinite measure space with $\mathbb{R}$ as domain?

My idea was that you consider $F(t) := \int_{-t}^t|f(x)|\ dx$. This function is continuous and even though the intermediate value theorem does not necessarily apply here, the function converges to $||f||_{L^1}$ and is monotone, so there must be a $c$, such that $F(c) = \frac12 ||f||_{L^1}$. Then again I would just take $g_1 = 2*f\chi_{[-c,c]}$ and $g_2 = 2* f\chi_{[-c,c]^C}$ and I would have found a strict convex combination of $f$. Is this argument correct, or can you not simply copy the proof from finite intervals?

Any help would be appreciated

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    $\begingroup$ What's your objection to your argument? $\endgroup$
    – Elle Najt
    Jun 6, 2018 at 11:34
  • $\begingroup$ I would say it is correct. $\endgroup$ Jun 6, 2018 at 12:09
  • $\begingroup$ You are right, the method used in the answer of that other problem (about $L^1[0,1]$) works in cases other than $[0,1]$. If you want, try to do it even more generally. If $\mu$ is a measure, what are the conditions under which $L^1(\mu)$ is isometric to a dual space? $\endgroup$
    – GEdgar
    Jun 6, 2018 at 12:36

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