Two Notions of Convergence
Pointwise convergence examines function values at individual points. Weak convergence instead tests functions through bounded linear functionals. Neither notion generally determines the other: a sequence may converge almost everywhere without converging weakly, and weak convergence may occur even when pointwise convergence fails and the norms do not approach the norm of the limit.
Unless stated otherwise, work on the finite complete Lebesgue measure space \(((0,1),\mathcal{L},m)\), where \(m((0,1))=1\). Functions are real-valued, and \(L^p(0,1)\) denotes equivalence classes modulo equality almost everywhere. For \(1\le p<\infty\), \(\|f\|_p=(\int_0^1|f|^p\,dm)^{1/p}\). When \(1<p<\infty\), the conjugate exponent \(p'\) satisfies \(1/p+1/p'=1\).
These duality identifications are standard results for \(L^p\) spaces. Thus, in the indicated ranges, weak convergence means convergence of the integrals against every function in the stated testing space. It does not mean convergence of the functions pointwise, or convergence in the \(L^p\) norm.
Shrinking Supports: The Endpoint Difference
Worked Example: Spikes in \(L^1\) Converge Almost Everywhere but Not Weakly
Set \(f_n(x)=n\chi_{(0,1/n)}(x)\) on \((0,1)\). For each fixed \(x>0\), if \(n\ge 1/x\), then \(x\notin(0,1/n)\), so \(f_n(x)=0\). Thus \(f_n\to0\) at every \(x\in(0,1)\). The functions are measurable and nonnegative, and
Test weak convergence to zero using the bounded linear functional \(\Lambda(u)=\int_0^1u\,dm\), which is permitted since its testing function \(1\) belongs to \(L^\infty(0,1)\). For every \(n\), \(\Lambda(f_n)=1\), whereas \(\Lambda(0)=0\). Hence \(f_n\) does not converge weakly to its almost-everywhere limit. Check: the \(L^1\) norms are all \(1\), and the same test functional returns \(1\) for every term, so neither computation allows weak convergence to zero.
This example disproves the general implication from almost-everywhere convergence to weak convergence, even for a sequence bounded in \(L^1\). It also shows why testing against the constant function matters: a spike can become invisible at each fixed point while retaining a fixed total integral.
Worked Example: The Same Spikes in \(L^p\), \(1<p<\infty\)
Fix \(p\) with \(1<p<\infty\), and let \(u_n(x)=n^{1/p}\chi_{(0,1/n)}(x)\). As above, \(u_n(x)\to0\) for every \(x\in(0,1)\). Direct calculation gives
To check weak convergence, take any \(g\in L^{p'}(0,1)\). Hölder's inequality gives
The functions \(|g|^{p'}\chi_{(0,1/n)}\) converge to zero almost everywhere and are bounded by the integrable function \(|g|^{p'}\). The Dominated Convergence Theorem therefore gives \(\int_0^{1/n}|g|^{p'}\,dm\to0\). Its hypotheses hold because \(g\in L^{p'}\), so \(|g|^{p'}\) is integrable. The displayed pairing tends to zero for every \(g\), and duality implies \(u_n\rightharpoonup0\) in \(L^p(0,1)\). Check: the weak limit is zero, but \(\|u_n-0\|_p=1\) for every \(n\); weak convergence here is not norm convergence.
The difference between these two spike examples is the available testing space. In \(L^1\), the constant function \(1\) is an allowed test and detects the fixed mass. In \(L^p\) for \(p>1\), tests lie in \(L^{p'}\), and their integrals over shrinking intervals tend to zero. The example does not say that almost-everywhere convergence always implies weak convergence in \(L^p\); it shows that this particular normalized spike sequence is weakly null in that range.
Oscillation: Weak Convergence Without Pointwise Convergence
We next construct functions that keep oscillating rather than concentrating on smaller and smaller sets. For each positive integer \(n\), define \(r_n:(0,1)\to\{-1,1\}\) by \(r_n(x)=(-1)^{\lfloor 2^n x\rfloor}\), where \(\lfloor y\rfloor\) is the greatest integer at most \(y\). Values at dyadic endpoints do not affect any integral or almost-everywhere statement.
Proof. First, \(|r_n|=1\) everywhere, so \(r_n\in L^p(0,1)\) and \(\|r_n\|_p=(\int_0^1 1\,dm)^{1/p}=1\). We verify weak convergence by testing against an arbitrary \(g\in L^{p'}(0,1)\).
A dyadic interval of level \(m\) is an interval \([k2^{-m},(k+1)2^{-m})\), with \(k=0,\ldots,2^m-1\), intersected with \((0,1)\). A dyadic step function of level \(m\) is constant on each such interval. If \(n>m\), each level-\(m\) interval consists of \(2^{n-m}\) level-\(n\) intervals. This number is even, and \(r_n\) takes alternating values \(1,-1\) on consecutive level-\(n\) intervals. Therefore its integral over each level-\(m\) interval is zero. For any dyadic step function \(h\) of level \(m\), summing its constant multiples of these zero integrals gives \(\int_0^1 r_nh\,dm=0\) whenever \(n>m\).
Dyadic step functions are dense in \(L^{p'}(0,1)\). To recall why, simple functions are dense in \(L^{p'}\); each measurable set in a simple function can, by regularity of Lebesgue measure on \((0,1)\), be approximated in measure by a finite union of intervals. Such a finite union can in turn be approximated in measure by a union of dyadic intervals, by choosing a sufficiently fine level so that the dyadic intervals meeting its finitely many endpoints have arbitrarily small total measure. For indicators, the \(L^{p'}\) error is the measure of the symmetric difference raised to the power \(1/p'\). Applying this to the finitely many sets in a simple function proves the asserted density.
Given \(\varepsilon>0\), choose a dyadic step function \(h\) with \(\|g-h\|_{p'}<\varepsilon\), and let \(m\) be a level on which \(h\) is constant. For every \(n>m\), Hölder's inequality and the cancellation just proved give
Since this holds for all sufficiently large \(n\), the pairing tends to zero. This is true for every \(g\in L^{p'}\), so the duality characterization proves \(r_n\rightharpoonup0\).
For pointwise behavior, away from the dyadic rationals, the binary expansion of \(x\) has infinitely many digits equal to \(0\) and infinitely many equal to \(1\). If the digits were eventually all \(0\), or eventually all \(1\), then \(x\) would be dyadic. The parity of \(\lfloor2^n x\rfloor\) is the \(n\)-th binary digit, so \(r_n(x)\) takes both values \(1\) and \(-1\) infinitely often. It therefore does not converge at any such \(x\). The dyadic rationals form a countable, hence null, set; pointwise convergence fails almost everywhere. Finally, \(\|r_n-0\|_p=1\) for every \(n\), so convergence in norm fails. \(\square\)
Worked Example: Direct Cancellation Against a Dyadic Test
Let \(h=\chi_{(0,1/2)}-\chi_{[1/2,1)}\), a dyadic step function of level \(1\). Compute its pairing with \(r_3\). The four level-\(3\) intervals have length \(1/8\), and the values of \(r_3\) on them are, in order, \(1,-1,1,-1\). On the first half, \(h=1\), so the two contributions are \(1\cdot(1/8)\) and \((-1)\cdot(1/8)\). On the second half, \(h=-1\), so the contributions are \(1\cdot(-1/8)\) and \((-1)\cdot(-1/8)\). Thus
Check: the integral of \(r_3\) on each half is \(1/8-1/8=0\), so multiplying each half by its constant value of \(h\) still gives zero. This calculation is the finite-level cancellation used in the weak convergence proof.
Common Pitfall: Applying Dominated Convergence Without a Majorant
A tempting but incorrect argument says that \(f_n=n\chi_{(0,1/n)}\to0\) almost everywhere, so \(\int f_n\,dm\to0\) by the Dominated Convergence Theorem. The theorem requires a single integrable function \(G\) such that \(|f_n|\le G\) almost everywhere for every \(n\). No such \(G\in L^1(0,1)\) exists. Indeed, for \(0<x<1/4\), choose \(n=\lfloor1/(2x)\rfloor\). Then \(n\ge1/(4x)\) and \(nx\le1/2<1\), so \(x\in(0,1/n)\) and \(f_n(x)=n\ge1/(4x)\). Any common majorant would satisfy \(G(x)\ge1/(4x)\) throughout this interval, which is not integrable near zero.
The actual integral confirms the failure: \(\int_0^1f_n\,dm=n\cdot m((0,1/n))=1\) for every \(n\). Almost-everywhere convergence alone does not justify passing a limit through an integral, and it does not imply weak convergence. Conversely, weak convergence is defined by all bounded linear functionals, not by pointwise values or norms; the oscillating sequence demonstrates both distinctions at once.
Check Your Understanding
Use the definitions and hypotheses above to answer each question.
- For \(f_n=n\chi_{(0,1/n)}\), compute \(\|f_n\|_1\) and identify a single bounded linear functional that rules out weak convergence to zero.
- For \(u_n=n^{1/p}\chi_{(0,1/n)}\), \(1<p<\infty\), which convergence theorem shows that the testing integrals tend to zero, and what is the integrable dominating function?
- Why does cancellation of \(r_n\) against a dyadic step function \(h\) of level \(m\) hold whenever \(n>m\)?
- What are \(\|r_n\|_p\) and \(\|r_n-0\|_p\), and what do these values say about norm convergence?
- For which points can the sequence \(r_n(x)\) converge, and why does that exceptional set not affect the almost-everywhere conclusion?