Let $f_i \in L^1 ([0, 1])$ be a sequence of functions equibounded in $L^1$ norm - that is, there exists some $M > 0$ such that $\|f_i\|_{L^1} < M$.

Define the functional $F: L^1([0, 1]) \to \mathbb R$ by

$$F(h) = \limsup_{i \to \infty} \|f_i - h\|_{L^1}.$$

Question: Does this functional admit a minimiser? Is the minimiser unique whenever it exists?


What I have tried so far is to attempt to apply the direct method of the calculus of variations.

Since the $f_i$ are equibounded in $L^1$, it can be shown that $F$ is coercive, thus any minimising sequence is bounded in $L^1$ norm. In particular we have a weakly-* converging subsequence, say $h_n \overset{*}{\to} h$.

The result would follow if we had weak-* sequential lower semi continuity of $F$ at the minimiser - that is, that

$$\liminf_{n \to \infty} F(h_n) \geq F(h).$$

I could neither disprove this with a counterexample, nor prove it in generality.

Edit: As pointed out in the comments, weak-$*$ convergence to an $L^1$ function isn’t guaranteed, only convergence to a measure.

  • 3
    $\begingroup$ A minimizer is not unique in general. E.g., let $f_i=(-1)^i$. Then any $h$ with $|h|\le1$ and $\int h=0$ is a minimizer. $\endgroup$ Jul 22, 2021 at 15:03
  • $\begingroup$ $L^1(0,1)$ is not a dual space. How do you define weak-* convergence? $\endgroup$
    – gerw
    Jul 23, 2021 at 18:15
  • $\begingroup$ Oh you’re right, it needs to be defined in the sense of measures.. but then there is trouble in making sense of $F(\mu)$ for a measure $\mu$. Hmm... $\endgroup$
    – Nate River
    Jul 24, 2021 at 4:58

1 Answer 1


As it has been already noted in the comments, the minimizer doesn't need to be unique. However, it always exists. It is not terribly hard to show but it is not trivial either, so I wonder why the question attracted so few votes.

The proof consists of two independent parts. The first one is that the limit of every minimizing sequence that converges almost everywhere (or just in measure) is a minimizer and the second one is that there exists a minimizing sequence converging almost everywhere. I will use the fact that we deal with a finite measure space though it should, probably, be irrelevant. WLOG, we may assume that $\|f_j\|_1\le 1$ for all $j$.

Part 1

Assume that $h_n$ is a minimizing sequence and $h$ is its pointwise limit (or just limit in measure). Since we can assume WLOG that $\|h_n\|_1\le 3$ (to be a competitor, you need to perform not much worse than $0$, at the very least), we have $\|h\|_1\le 3$ as well (by Fatou). Then, by the definition of the convergence in measure, we can write $h_n=u_n+v_n$ where $u_n$ converge to $h$ uniformly and $v_n$ are supported on $E_n$ with $m(E_n)\to 0$ as $n\to\infty$. Also $\|v_n\|_1\le 7$, say.

Now, since $v_n\in L^1$, we can find $\delta_n>0$ such that for every set $E$ with $m(E)<\delta_n$, we have $\int_E|v_n|<\frac 1n$, say. By induction, we can now choose a subsequence $n_k$ such that $$ \sum_{q=k+1}^\infty m(E_{n_q})<\delta_{n_k}\,. $$ Then $\|v_{n_k}\chi_{\cup_{q>k}E_{n_q}}\|_1\to 0$ and, adding these parts to $u_{n_k}$, we see that we can (passing to a subsequence) assume that our minimizing sequence can be represented as $h_n=U_n+V_n$ where $U_n\to h$ in $L^1$ and $V_n$ have disjoint supports $G_n$.

Since correcting a minimizing sequence by a sequence tending to $0$ in $L^1$ results in a minimizing sequence, we can just as well assume that $h_n=h+V_n$.

If $\|V_n\|_1\to 0$ (even along a subsequence), we are done. Assume now that $\|V_n\|_1\ge \tau>0$. Then for every function $g$ of $L^1$-norm less than $4$, we have $$ \|g\|_1\le \max(\|g-V_n\|_1,\dots,\|g-V_{n+N-1}\|_1) $$ for all $n$ as soon as $N>8/\tau$. Indeed, since the supports $G_n$ of $V_n$ are disjoint, there is $q\in\{0,\dots,N-1)$ such that $\int_{G_{n+q}}|g|\le \frac 1N\|g\|_1< \frac\tau 2$, in which case subtracting $V_{n+q}$ can only drive the norm up.

Applying this to the functions $g_j=f_j-h$, we conclude that $$ \limsup_{j\to\infty}\|f_j-h\|_1\le \limsup_{j\to\infty}\max_{0\le q\le N-1}\|f_j-h_{n+q}\|_1= \max_{0\le q\le N-1}\limsup_{j\to\infty}\|f_j-h_{n+q}\|_1 $$ for every $n$, i.e. $h$ is a minimizer in this case as well.

Part 2

Let $I$ be the infimum of our functional. For every $\varepsilon>0$ consider the set $X_\varepsilon$ of all $L^1$-functions $h$ for which the value of the functional is at most $I+\varepsilon$. Clearly, it is a convex, non-empty, closed (in $L^1$) set and $X_{\varepsilon'}\supset X_{\varepsilon''}$ when $\varepsilon'\ge\varepsilon''$.

Fix some strictly convex non-negative function $\Phi(t)\le |t|$. To simplify the technicalities, I'll choose it by the conditions $\Phi(0)=\Phi'(0)=1$, $\Phi''(t)=\frac 2{\pi(1+|t|^2)}$ but pretty much any other choice will work as well.

Let $$ J_\varepsilon=\inf_{h\in X_\varepsilon}\int\Phi(h)\,. $$ Clearly, $J_\varepsilon$ is a bounded non-increasing function on $(0,1)$, say. Let $J$ be its limit at $0+$. Passing to an appropriate decreasing sequence $\varepsilon_n\to 0+$ and re-enumerating $X_n=X_{\varepsilon_n}, J_n=J_{\varepsilon_n}$, we can assume that $J_n\ge J-2^{-5n}$ We will choose a representative $h_n$ of $X_n$ for which $\int\Phi(h_n)$ is not more than $2^{-5n}$ above $J$ and, thereby, not more than $2\cdot 2^{-5n}$ above its infimum over $X_n$. Then for $m>n$, we have $h_{n,m}=\frac{h_n+h_m}2\in X_n$ (convexity of $X_n$ plus inclusion $X_n\supset X_m$) and $$ \int\Phi(h_{n,m})\le \int\frac 12(\Phi(h_n)+\Phi(h_m))-\int\frac{(h_n-h_m)^2}{4\pi (1+|h_n|^{2}+|h_m|^{2})} $$ (second order Taylor with the remainder in the Lagrange form), whence $$ \int\frac{(h_n-h_m)^2}{4\pi(1+|h_n|^2+|h_m|^2)}\le 2\cdot 2^{-5n} $$ regardless of $m>n$ (otherwise we would go below $J+2^{-5n}-2\cdot 2^{-5n}=J-2^{-5n}$, which is below the infimum over $X_n$). It remains to note that if $|h_n|\le 2^n$ and $|h_n-h_m|>2^{-n}$, then the integrand is at least $\rm{const}\, 2^{-4n}$, so we get $$ m(\{|h_n|\le 2^n, |h_n-h_m|>2^{-n}\})\le \rm{Const}\,2^{-n} $$ for all $m>n$ from where it follows at once that $h_n(x)$ is a Cauchy sequence for almost all $x$ (recall that $\|h_n\|_1\le 3$, so the first condition excludes only a set of measure $3\cdot 2^{-n}$, then use Borel-Cantelli).

That's it. Feel free to ask questions if anything is unclear :-)

  • $\begingroup$ Wow, this looks impressive. This part was a bit confusing to me - “ Then $\|v_{n_k}\chi_{\cup_{q>k}E_{n_q}}\|_1\to 0$ and, adding these parts to $u_{n_k}$, we see that we can (passing to a subsequence) assume that our minimizing sequence can be represented as $h_n=U_n+V_n$ where $U_n\to h$ in $L^1$ and $V_n$ have disjoint supports $G_n$.” Do you mean to add $v_{n_k}\chi_{\cup_{q>k}E_{n_q}}$ to $u_{n_k}$? I am not sure how to get the desired representation from this. $\endgroup$
    – Nate River
    Aug 2, 2021 at 4:04
  • 1
    $\begingroup$ @NateRiver Exactly as you said: $U_k=u_{n_k}+v_{n_k}\chi_{\cup_{q>k}E_{n_q}}$, $V_k=v_{n_k}\chi_{E_{n_k}\setminus \cup_{q>k}E_{n_q}}=v_{n_k}\chi_{G_k}$. $\endgroup$
    – fedja
    Aug 2, 2021 at 7:45
  • 2
    $\begingroup$ @NateRiver I just tried to solve the problem in $L^2$ first. There weak convergence is guaranteed, but doesn't seem to drop the value of the functional immediately, so I needed the norm convergence of a minimizing sequence. Fortunately, if one has a nested sequence of bounded closed convex sets in a Hilbert space, the smallest norm elements of the sets converge in norm (the proof is the same as above just using the parallelogram identity). I tried to mimic that idea in $L^1$ (which required strict convexity of something) and got the a.e. convergence this way, which turned out to be sufficient. $\endgroup$
    – fedja
    Aug 2, 2021 at 13:33
  • 1
    $\begingroup$ An alternative proof for Part 2 (existence of a minimizing sequence converging a.e.) Let $h_n$ be any minimizing sequence for $F$. As observed, it is bounded in $L_1$, so by the Komlós Theorem, up to extracting a subsequence, it is a.e. converging in Cesaro sense to some $h\in L^1$. Since $F$ is a convex functional, the sequence of the Cesaro means is still a minimizing sequence. $\endgroup$ Aug 2, 2021 at 13:38
  • 1
    $\begingroup$ Of course, and I wouldn't be surprised if you answer (which I'm still reading) contains as a byproduct an alternative proof of that Komlós theorem (which is a 10 page paper: link.springer.com/article/10.1007%2FBF02020976 ). $\endgroup$ Aug 2, 2021 at 13:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.