# Existence of function minimising L^1 distance to a sequence of functions

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

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

• I believe $g(x) = (1/2) (\limsup f_i(x) + \liminf f_i(x))$ does the job? – Mateusz Kwaśnicki Sep 26 '19 at 12:05
• 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. – James Baxter Sep 26 '19 at 15:53
• 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. – James Baxter Sep 26 '19 at 15:54
• Yes, I noticed that, and deleted my comment before you posted yours. – Mateusz Kwaśnicki Sep 26 '19 at 16:41
• 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 $$\limsup_{n \to \infty} \|f_n - h\|_1 = \sup_{f \in F} \|f - h\|_1 = \phi(h).$$ – Mateusz Kwaśnicki Sep 27 '19 at 7:01