Testing Weak Convergence With Standard Sequences

Weak Convergence Is Detected by Every Functional

A sequence can keep the same norm while converging weakly: weak convergence tests each vector against every bounded linear functional, rather than measuring the distance between vectors. The outcome depends on both the sequence and the Banach space. Coordinate vectors in sequence spaces and Rademacher functions give explicit examples where dual pairings reveal the difference.

For a Banach space \(Y\), a sequence \(y_n\) converges weakly to \(y\), written \(y_n\rightharpoonup y\), if \(\Lambda(y_n)\to\Lambda(y)\) for every bounded linear functional \(\Lambda\in Y^*\). This is convergence in the weak topology; it is not convergence in norm, which would require \(\|y_n-y\|\to0\). In the examples below the candidate weak limit is zero, so it is enough to compute \(\Lambda(y_n)\) and show that it tends to zero for every allowed \(\Lambda\).

Coordinate Vectors in Sequence Spaces

Let \(\ell^p\), for \(1\le p<\infty\), be the space of real sequences \(x=(x_k)_{k\ge1}\) with \(\sum_{k=1}^\infty |x_k|^p<\infty\), with its usual norm. Regard it as the \(L^p\) space for counting measure on \(\mathbb{N}\). Let \(e_n\) have value \(1\) at coordinate \(n\) and \(0\) elsewhere. For \(1<p<\infty\), write \(q=p'=p/(p-1)\). The standard duality theorem for sequence spaces says that every bounded linear functional on \(\ell^p\) has the form \(\Lambda_a(x)=\sum_{k=1}^\infty a_kx_k\) for some \(a\in\ell^q\). For \(\ell^1\), the corresponding dual is \(\ell^\infty\).

Proof. First let \(1<p<\infty\), and let \(\Lambda_a\) be any bounded linear functional, with \(a\in\ell^q\). Then \(\Lambda_a(e_n)=a_n\). Since \(\sum_k|a_k|^q<\infty\), its terms satisfy \(a_n\to0\): otherwise some \(\varepsilon>0\) would have \(|a_n|\ge\varepsilon\) for infinitely many \(n\), forcing the series to diverge. Thus every bounded linear functional tends to zero on \(e_n\), which is precisely weak convergence to zero.

For \(\ell^1\), the functional \(\Lambda(x)=\sum_{k=1}^\infty x_k\) is bounded, since \(|\Lambda(x)|\le\sum_k|x_k|=\|x\|_1\), and \(\Lambda(e_n)=1\) for every \(n\). Hence \(e_n\) is not weakly convergent to zero in \(\ell^1\).

Finally, consider any bounded linear functional \(\Lambda\) on \(\ell^\infty\). Its restriction to the closed subspace \(c_0\) of sequences tending to zero is bounded. Put \(a_k=\Lambda(e_k)\). For any \(N\), choose signs \(s_k\in\{-1,1\}\) so that \(s_ka_k=|a_k|\) when \(a_k\ne0\), choosing either sign when \(a_k=0\). The finite-support vector \(x=\sum_{k=1}^N s_ke_k\) belongs to \(c_0\), has \(\|x\|_\infty=1\), and satisfies \(\sum_{k=1}^N|a_k|=\Lambda(x)\le|\Lambda(x)|\le\|\Lambda\|\). Thus the partial sums of \(\sum_k|a_k|\) are bounded, so \(a_k\to0\). In particular, \(\Lambda(e_n)=a_n\to0\) for every \(\Lambda\in(\ell^\infty)^*\). This proves the asserted weak convergence even without identifying the full dual of \(\ell^\infty\). In each space \(\|e_n\|=1\), since exactly one coordinate has absolute value \(1\); therefore \(\|e_n-0\|\not\to0\). \(\square\)

Rademacher Functions on the Unit Interval

Now use Lebesgue measure \(m\) on \([0,1]\), and identify functions equal \(m\)-a.e. The \(n\)th Rademacher function \(r_n\) takes value \(1\) on the first half and \(-1\) on the second half of each interval \([j2^{-(n-1)},(j+1)2^{-(n-1)})\), for \(j=0,\ldots,2^{n-1}-1\). Values at the finitely many interval endpoints may be assigned arbitrarily, since those endpoints form a null set. In particular \(|r_n|=1\) a.e.

Proof. First take \(1<p<\infty\), set \(q=p'\), and let \(\Lambda\in(L^p(m))^*\). By the \(L^p\) duality theorem, there is \(g\in L^q(m)\) such that \(\Lambda(f)=\int_0^1 fg\,dm\). The theorem applies because \(1<p<\infty\) and Lebesgue measure on \([0,1]\) is finite. For \(p=1\), the duality theorem for \(L^1\) gives the same pairing with \(g\in L^\infty(m)\).

Dyadic step functions are dense in \(L^q([0,1],m)\) when \(1\le q<\infty\), and are dense in \(L^1([0,1],m)\). Here a dyadic step function of level \(j\) is constant on each interval \([k2^{-j},(k+1)2^{-j})\), \(0\le k<2^j\). This density is the standard density theorem for \(L^q\) on the unit interval. If \(h\) is such a function and \(n>j\), then \(\int_0^1 h r_n\,dm=0\): on each level-\(j\) interval, \(h\) is constant and the \(+1\) and \(-1\) portions of \(r_n\) have equal measure, so their contributions cancel.

For \(1<p<\infty\), choose any \(\varepsilon>0\). By density, select a dyadic step function \(h\) with \(\|g-h\|_q<\varepsilon\), and let \(j\) be its level. For every \(n>j\), Hölder's inequality gives \(\left|\Lambda(r_n)\right|=\left|\int_0^1(g-h)r_n\,dm\right|\le\|g-h\|_q\|r_n\|_p<\varepsilon\), because \(\int h r_n\,dm=0\) and \(\|r_n\|_p=1\). The same argument for \(p=1\) uses \(\|g-h\|_1<\varepsilon\) and \(|r_n|\le1\), giving \(\left|\Lambda(r_n)\right|\le\|g-h\|_1<\varepsilon\). Thus, in either finite-\(p\) case, for every \(\varepsilon>0\) all sufficiently large \(n\) satisfy \(|\Lambda(r_n)|<\varepsilon\). This proves weak convergence to zero.

To show failure for \(L^\infty\), consider the finite linear span of the \(r_n\). For any \(N\) and real coefficients \(c_1,\ldots,c_N\), \(\left\|\sum_{n=1}^N c_nr_n\right\|_\infty=\sum_{n=1}^N|c_n|\). Indeed, the triangle inequality gives the upper bound. For the lower bound, choose signs \(\sigma_n\in\{-1,1\}\) with \(c_n\sigma_n=|c_n|\). The first \(N\) Rademacher functions attain the sign pattern \((\sigma_1,\ldots,\sigma_N)\) on a dyadic interval of positive measure, so the sum equals \(\sum_n|c_n|\) there.

Consequently the rule \(\Lambda_0(\sum_{n=1}^N c_nr_n)=\sum_{n=1}^N c_n\) is well-defined and bounded on their span: the norm identity implies linear independence and \(|\Lambda_0(f)|\le\sum_n|c_n|=\|f\|_\infty\). The Hahn–Banach theorem extends \(\Lambda_0\) to a bounded linear functional on all of \(L^\infty(m)\). This extension satisfies \(\Lambda_0(r_n)=1\) for every \(n\), so the sequence cannot converge weakly to zero in \(L^\infty(m)\). \(\square\)

What the Tests Do—and Do Not—Show

The pairings explain the central distinctions. In \(\ell^p\) for \(1<p<\infty\), every dual test reads off a coefficient of an \(\ell^{p'}\) sequence, and those coefficients vanish at infinity. In \(\ell^1\), the constant coefficient test prevents weak convergence to zero. For Rademacher functions, cancellation against approximating dyadic step functions proves weak convergence for finite \(p\); in \(L^\infty\), a bounded functional on their span extends to a test that keeps every value equal to \(1\).

A common mistake is to test only the standard integral pairing when deciding weak convergence in \(L^\infty\). For example, the dyadic cancellation argument shows \(\int r_ng\,dm\to0\) for every \(g\in L^1(m)\). That establishes convergence against the \(L^1\) pairings, but weak convergence in \(L^\infty\) requires convergence against every bounded linear functional on \(L^\infty\). The Hahn–Banach functional above is such a functional and satisfies \(\Lambda(r_n)=1\), so the stronger claim is false. Likewise, one successful test cannot prove weak convergence: all bounded linear functionals must be considered.