4
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Here all functions are from $[0, 1] \to \mathbb R$.

Let $f_i$ be a sequence of continuous functions such that there exists some $M > 0$ such that $|f_i| < M$ for all $i$. Does there always exist some measurable $g$ such that

$\limsup_i \int |f_i - g| = \inf_{h \in L^1} \limsup_i \int |f_i - h|$?

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  • $\begingroup$ I believe $g(x) = (1/2) (\limsup f_i(x) + \liminf f_i(x))$ does the job? $\endgroup$ – Mateusz Kwaśnicki Sep 26 '19 at 12:05
  • $\begingroup$ Hmm surprisingly not, consider the “moving bump example” where you enumerate the dyadic intervals and let f_i = indicator of the ith dyadic interval. Then that formula gives you the constant function 1/2, but the minimum is in fact achieved by the constant 0 function. $\endgroup$ – James Baxter Sep 26 '19 at 15:53
  • $\begingroup$ In this case the f_i aren’t continuous, but you can rectify this by making them linearly increasing then decreasing for a little bit each time. It shouldn’t change that the minimising g is the 0 function. $\endgroup$ – James Baxter Sep 26 '19 at 15:54
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    $\begingroup$ Yes, I noticed that, and deleted my comment before you posted yours. $\endgroup$ – Mateusz Kwaśnicki Sep 26 '19 at 16:41
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    $\begingroup$ Another observation: It would be sufficient to find a closed subset $F$ of $\{h \in L^1 : \|h\|_\infty \le M\}$ such that the function $$ \phi(h) = \sup_{f \in F} \|f - h\|_1 $$ does not attain a global minimum. Indeed, consider a dense subset $\{g_1, g_2, g_2, \ldots\}$ of $F$ (with respect to the $L^1$ norm), and for each $g_n$ pick a sequence of continuous functions $f_{n,k}$ such that $\|f_{n,k}-g_n\|_1 < \tfrac{1}{n+k}$. By choosing $f_n$ to be the enumeration of $f_{n,k}$, we obtain $$\limsup_{n \to \infty} \|f_n - h\|_1 = \sup_{f \in F} \|f - h\|_1 = \phi(h).$$ $\endgroup$ – Mateusz Kwaśnicki Sep 27 '19 at 7:01

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