Bounded in \(L^1\) Does Not Mean Weakly Compact
In \(L^1\), a bounded sequence need not have a weakly convergent subsequence. A common obstruction is concentration: a fixed amount of integral is packed into sets whose measures tend to zero. Another is disjointness: each function carries a fixed amount of integral on a different set. Both behaviors can be detected by uniform integrability, the condition that large values contribute uniformly little to the integral.
Throughout, let \(([0,1],\mathcal{L},m)\) be the complete finite Lebesgue measure space, and take functions to be real-valued and measurable. Elements of \(L^1(m)\) are identified when equal \(m\)-a.e. The weak topology on \(L^1(m)\) is defined by its continuous linear functionals. We use the standard dual identification \((L^1(m))^*=L^\infty(m)\): \(f_n\) converges weakly to \(f\) precisely when \(\int_0^1 f_n g\,dm\to\int_0^1 fg\,dm\) for every \(g\in L^\infty(m)\).
We will use the Dunford–Pettis criterion for a finite measure space: a bounded subset of \(L^1(m)\) is relatively weakly compact if and only if it is uniformly integrable. Here, relatively weakly compact means that its closure in the weak topology is compact. This is the standard uniform-integrability characterization of weak compactness in \(L^1\); we apply it below alongside direct tests that rule out weakly convergent subsequences.
Concentration: Normalized Spikes
Worked Example: Unit-Mass Spikes on Shrinking Intervals
For each integer \(n\ge1\), define \(f_n=n\chi_{(0,1/n)}\) on \([0,1]\). The interval has measure \(1/n\), so $$ \|f_n\|_1=\int_0^1|f_n|\,dm=n\,m((0,1/n))=n\cdot\frac1n=1. $$ At \(x=0\), every \(f_n(x)=0\). At each \(x>0\), choose \(N\) with \(1/N\le x\); then for every \(n\ge N\), \(x\notin(0,1/n)\), so \(f_n(x)=0\). Thus \(f_n\to0\) everywhere, but the \(L^1\) norms remain equal to \(1\).
For \(K\ge0\), if \(n>K\), then \(f_n=n>K\) throughout \((0,1/n)\), and it is zero outside that interval. Therefore $$ \int_{\{|f_n|>K\}}|f_n|\,dm=n\cdot\frac1n=1\qquad(n>K). $$ If \(n\le K\), the strict tail set is empty and the tail integral is zero. Since for every finite \(K\) there is an integer \(n>K\), the supremum over \(n\) of these tail integrals is \(1\). Hence \((f_n)\) is not uniformly integrable. Check: the full integral is \(1\), and when \(n>K\) the strict tail includes the whole support, so the tail integral is also \(1\).
In fact, no subsequence of these spikes converges weakly in \(L^1(m)\). The following argument proves this directly, without relying on the compactness criterion.
Proof. Suppose a subsequence \(f_{n_j}\) converges weakly to some \(f\in L^1(m)\). Fix an integer \(k\ge1\). For all sufficiently large \(j\), \(1/n_j<1/k\), so the support \((0,1/n_j)\) lies inside \((0,1/k)\) and $$ \int_{(0,1/k)} f_{n_j}\,dm=1. $$ The indicator \(\chi_{(0,1/k)}\) belongs to \(L^\infty(m)\). Weak convergence, tested against this fixed function, gives \(\int_{(0,1/k)} f\,dm=1\) for every \(k\).
But for every \(f\in L^1(m)\), these integrals must tend to zero as \(k\to\infty\). To verify this, let \(\varepsilon>0\). Integrability gives a finite \(M>0\) such that \(\int_{\{|f|>M\}}|f|\,dm<\varepsilon/2\). Then $$ \int_{(0,1/k)}|f|\,dm \le M\,m((0,1/k))+\int_{\{|f|>M\}}|f|\,dm <\frac{M}{k}+\frac{\varepsilon}{2}. $$ Choose \(k\) large enough that \(M/k<\varepsilon/2\). The integral of \(|f|\) over \((0,1/k)\) is then less than \(\varepsilon\), so \(\int_{(0,1/k)}f\,dm\to0\), contradicting its value \(1\) for every \(k\). Thus no such weakly convergent subsequence exists. \(\square\)
The proof identifies the obstruction precisely: weak convergence would force a limit \(f\in L^1\) to retain unit mass in every neighborhood \((0,1/k)\) of zero. An integrable function cannot retain that much mass on sets shrinking to a null set.
Disjointness: Another Failure of Weak Compactness
Worked Example: Disjoint Functions of Unit \(L^1\) Norm
For \(n\ge1\), set \(I_n=(2^{-n},2^{-(n-1)}]\) and \(u_n=2^n\chi_{I_n}\). These intervals are pairwise disjoint, and $$ m(I_n)=2^{-(n-1)}-2^{-n}=2^{-n},\qquad \|u_n\|_1=2^n\cdot2^{-n}=1. $$ For any finite \(K\ge0\), choose \(n\) with \(2^n>K\). Then the strict tail \(\{|u_n|>K\}\) is exactly \(I_n\), and $$ \int_{\{|u_n|>K\}}|u_n|\,dm=2^n m(I_n)=2^n\cdot2^{-n}=1. $$ Thus the family \((u_n)\) is bounded in \(L^1(m)\) but is not uniformly integrable.
There is also a direct weak-convergence test. Take any subsequence \((u_{n_j})_{j\ge1}\) and define \(g\) to equal \((-1)^j\) on \(I_{n_j}\), and \(0\) elsewhere. The intervals are disjoint, so this defines a measurable function with \(\|g\|_\infty\le1\). For every \(j\), $$ \int_0^1 g u_{n_j}\,dm=(-1)^j\,2^{n_j}m(I_{n_j})=(-1)^j. $$ This scalar sequence does not converge. Since weak convergence requires convergence of the pairing with every fixed \(g\in L^\infty(m)\), the chosen subsequence cannot converge weakly. The argument applies to every subsequence, so \((u_n)\) has no weakly convergent subsequence. Check: each pairing has magnitude \(2^{n_j}2^{-n_j}=1\), with alternating sign.
The test function is chosen after a subsequence is specified, which is legitimate: to show that a given subsequence is not weakly convergent, it suffices to find one fixed bounded measurable function for which its pairings fail to converge. The disjoint supports make that construction possible.
A Bounded Family That Does Behave Well
Worked Example: Shrinking Indicators Converge in \(L^1\)
Let \(h_n=\chi_{(0,1/n)}\) on \([0,1]\). Its norm is $$ \|h_n\|_1=m((0,1/n))=\frac1n\longrightarrow0. $$ For \(K\ge1\), the strict tail set \(\{|h_n|>K\}\) is empty for every \(n\), so all tail integrals are zero. For \(0\le K<1\), the strict tail is \((0,1/n)\), and its integral is \(1/n\); the supremum over \(n\ge1\) is \(1\), attained at \(n=1\). These values show that the tail supremum is zero for all \(K\ge1\), hence tends to zero as \(K\to\infty\). The family is uniformly integrable.
Moreover, \(h_n\to0\) in \(L^1(m)\), and therefore weakly: for every \(g\in L^\infty(m)\), $$ \left|\int_0^1 h_n g\,dm\right| \le \|g\|_\infty\|h_n\|_1 =\frac{\|g\|_\infty}{n}\longrightarrow0. $$ This contrasts with normalized spikes: the supports are the same shrinking intervals, but their heights are \(1\), not \(n\), so their total integrals tend to zero rather than staying fixed. Check: the bound on the pairing follows because \(|g|\le\|g\|_\infty\) \(m\)-a.e. and \(h_n\ge0\).
What the Endpoint Criterion Tells Us
The examples distinguish three properties that should not be conflated:
- Boundedness in \(L^1\) means \(\sup_n\|f_n\|_1<\infty\). Both the spikes and the disjoint unit-mass functions have norm exactly \(1\).
- Uniform integrability prevents a family from placing a fixed amount of integral in increasingly high-value tails. Both counterexample families fail this test, with tail supremum \(1\) at every finite threshold.
- Weak convergence requires convergence of the integrals against every fixed function in \(L^\infty(m)\). Norm boundedness alone does not ensure that any subsequence has this property.
By the Dunford–Pettis criterion, the bounded sets formed by either counterexample sequence are not relatively weakly compact because they are not uniformly integrable. The direct arguments establish the stronger sequence-level statement: neither sequence has a weakly convergent subsequence. This is the relevant \(L^1\) endpoint warning. Boundedness is necessary for weak compactness, but at this endpoint it is not sufficient; concentration or disjoint escape can prevent subsequential weak convergence.
Common mistake. A tempting but incorrect calculation is: “\(f_n\to0\) a.e. and \(\sup_n\|f_n\|_1=1\), so \(f_n\) converges weakly to \(0\).” For the normalized spikes, testing against the constant function \(g=1\) gives \(\int f_n g\,dm=1\) for every \(n\), whereas \(\int 0\cdot g\,dm=0\). Thus weak convergence to zero already fails. More strongly, the proposition proves that no other \(L^1\) function can be a weak subsequential limit.
Check Your Understanding
Use the tail definition and the weak pairing with \(L^\infty(m)\) to justify your answers.
- For \(v_n=3n\chi_{(0,1/(3n))}\), compute \(\|v_n\|_1\) and determine whether \((v_n)\) is uniformly integrable.
- For the disjoint functions \(u_n\), explain why testing only against the constant function \(1\) does not rule out weak convergence, and construct a test that does rule out convergence along a specified subsequence.
- Why does \(f_n\to0\) everywhere in the normalized-spike example fail to imply convergence in \(L^1(m)\)?
- For \(h_n=\chi_{(0,1/n)}\), compute the strict tail integral when \(K=1\), and state why this is consistent with uniform integrability.
- State the Dunford–Pettis criterion used here and identify which of the two unit-norm counterexample families violates its uniform-integrability condition.