Mazur's Lemma and Convex Combinations

From Weak Convergence to Norm Approximation

Weak convergence tests a sequence using continuous linear functionals; norm convergence controls the distance between the sequence elements themselves. These notions need not coincide. Mazur's lemma gives a useful bridge: even when a weakly convergent sequence does not approach its limit in norm, finite convex combinations selected from any sufficiently late part of the sequence do approach that limit in norm. The combinations need not be consecutive averages or have equal weights; the theorem guarantees that suitable finite combinations exist.

We first work in a real normed vector space \(X\). Weak convergence \(x_n\to x\) means that \(F(x_n)\to F(x)\) for every continuous real-linear functional \(F:X\to\mathbb{R}\). A finite convex combination of points \(z_1,\ldots,z_r\) is a sum \(\sum_{j=1}^r\alpha_jz_j\), where \(\alpha_j\ge0\) and \(\sum_{j=1}^r\alpha_j=1\). For a set \(A\subseteq X\), its convex hull \(\operatorname{conv}(A)\) is the set of all its finite convex combinations.

The lemma itself does not require completeness, reflexivity, or boundedness as an additional hypothesis. Its proof uses the Hahn–Banach theorem through a separation argument. We give the needed argument explicitly so that the uniform gap used in the proof is visible.

Proof. Since \(C\) is closed and \(x\notin C\), there is an \(r>0\) such that the open ball \(B(x,r)=\{z:\|z-x\|<r\}\) misses \(C\). Set \(U=C+B(0,r/2)\), the set of all \(c+v\) with \(c\in C\) and \(\|v\|<r/2\). This set is nonempty, open, and convex. Also \(x\notin U\): otherwise \(x=c+v\) would give \(\|x-c\|<r/2<r\) for some \(c\in C\).

Here is the point-versus-open-convex-set consequence of Hahn–Banach that we need. If \(U\) is a nonempty open convex set and \(x\notin U\), there is a nonzero continuous real-linear \(F\) with \(F(u)\le F(x)\) for every \(u\in U\). To see this consequence, translate to \(V=U-x\), so \(V\) is open and convex and \(0\notin V\). Choose \(a\in V\) and let \(W=V-a\). Then \(W\) is open, convex, contains \(0\), and does not contain \(-a\). Its gauge \(p_W(z)=\inf\{t>0:z\in tW\}\) is finite and sublinear; because \(W\) is open and convex and contains \(0\), \(W=\{z:p_W(z)<1\}\). Thus \(p_W(-a)\ge1\). On the line spanned by \(-a\), define \(G(t(-a))=t\,p_W(-a)\). For \(t\ge0\), \(G(t(-a))=p_W(t(-a))\); for \(t<0\), \(G(t(-a))\le0\le p_W(t(-a))\). The Hahn–Banach theorem extends \(G\) to a real-linear \(F\) on \(X\) with \(F(z)\le p_W(z)\). This extension is continuous because \(W\) contains a ball about \(0\), so \(p_W(z)\le A\|z\|\) for some finite \(A\), and applying the domination to \(z\) and \(-z\) bounds \(|F(z)|\). For \(v\in V\), \(v-a\in W\), so \(F(v-a)<1\), while \(F(a)=-F(-a)=-p_W(-a)\le-1\). Therefore \(F(v)<0\), which gives \(F(u)<F(x)\) for \(u\in U\).

Apply this consequence to the set \(U=C+B(0,r/2)\). For every \(c\in C\) and every \(v\) with \(\|v\|<r/2\), it gives \(F(c+v)\le F(x)\). Taking the supremum over such \(v\) yields \(F(c)+(r/2)\|F\|\le F(x)\), since the supremum of \(F(v)\) over the open ball of radius \(r/2\) is \((r/2)\|F\|\). The functional is nonzero, so \(\delta=(r/2)\|F\|>0\). Hence \(F(c)\le F(x)-\delta\) for every \(c\in C\), proving the separation fact. \(\square\)

Proof of Mazur's Lemma

Proof. Fix \(k\ge1\), and set \(C_k=\overline{\operatorname{conv}\{x_n:n\ge k\}}\), where the closure is in the norm. This is a nonempty, norm-closed, convex set. Suppose, for contradiction, that \(x\notin C_k\). By the separation fact, there are a continuous real-linear \(F\) and \(\delta>0\) such that \(F(c)\le F(x)-\delta\) for every \(c\in C_k\). In particular, \(x_n\in C_k\) for each \(n\ge k\), so \(F(x_n)\le F(x)-\delta\) for every \(n\ge k\). But weak convergence gives \(F(x_n)\to F(x)\). Thus for some \(n\ge k\), \(|F(x_n)-F(x)|<\delta/2\), implying \(F(x_n)>F(x)-\delta/2\), a contradiction. Therefore \(x\in C_k\).

By the definition of norm closure, for each \(k\) there is a point \(y_k\in\operatorname{conv}\{x_n:n\ge k\}\) with \(\|y_k-x\|<1/k\). Each \(y_k\) is a finite convex combination of terms from that tail. Since \(1/k\to0\), the inequality gives \(\|y_k-x\|\to0\). This proves both assertions. \(\square\)

For a complex normed space, apply the real-space result after treating the space as a real vector space. Weak convergence with respect to continuous complex-linear functionals implies convergence under every continuous real-linear functional: such a functional can be written as the real part of a continuous complex-linear functional. The resulting combinations have real nonnegative coefficients summing to one, so they are convex combinations in the complex space as well.

Applying the Lemma in Lp

Let \((X,\mathcal{M},\mu)\) be any measure space, and let \(1\le p<\infty\). The space \(L^p(\mu)\) consists of equivalence classes, up to equality \(\mu\)-a.e., of measurable real-valued functions with finite norm \(\|f\|_p=(\int_X|f|^p\,d\mu)^{1/p}\). For complex-valued functions, use the same norm and the complex version of the lemma. Weak convergence in \(L^p(\mu)\) means convergence under every continuous linear functional on \(L^p(\mu)\). The theorem applies because this is a normed vector space; the measure space need not be finite or sigma-finite for Mazur's lemma itself.

A Common Convexity Mistake

Averaging vectors does not mean averaging their norms. For the first example, a mistaken calculation would be \(\|\frac1{16}\sum_{j=k}^{k+15}e_j\|_2=\frac1{16}\sum_{j=k}^{k+15}\|e_j\|_2=1\). This equality is invalid: a norm is not additive for arbitrary vectors. The correct computation uses the coordinates of the sum: there are \(16\) entries equal to \(1/16\), so the squared norm is \(16(1/16)^2=1/16\), and the norm is \(1/4\). Convexity gives an upper bound by the weighted sum of norms, not a general equality.

Mazur's lemma is useful whenever a weak limit is available but an argument needs norm approximation. It supplies finite convex combinations from each tail, not necessarily a norm-convergent subsequence of the original sequence. In applications, one must keep this distinction: the combinations are new elements of the space, built using nonnegative coefficients that sum to one.