Why Separation Controls Weak Closedness
The weak topology tests vectors using continuous linear functionals rather than the norm itself. It is therefore weaker than the norm topology: a weakly open set is norm-open, but a norm-open set need not be weakly open. In general, a norm-closed set need not be weakly closed. Convexity changes the picture: continuous linear functionals can separate any point outside a norm-closed convex set from that set, and the resulting separating half-spaces are weakly open.
Let \(X\) be a real Banach space and let \(X^*\) denote its continuous dual. The weak topology on \(X\), written \(\sigma(X,X^*)\), is the coarsest topology for which every \(x^*\in X^*\) is continuous. Equivalently, neighborhoods are generated by finitely many conditions on the values \(x^*(x)\). A set \(C\subseteq X\) is weakly closed if it is closed in this topology. In a complex Banach space the same definition uses complex-linear functionals; separation is expressed using their real parts.
The strict inequality at \(x_0\) is essential: it gives a weakly open half-space containing \(x_0\) and missing \(C\). The theorem applies to a point outside a norm-closed convex set; it does not assert separation from an arbitrary nonconvex set.
Closed Convex Sets Are Weakly Closed
Proof. First suppose \(C\) is norm-closed. If \(C=\varnothing\), it is weakly closed, so assume it is nonempty. For each \(x_0\in X\setminus C\), separation gives \(x^*\in X^*\) and \(\alpha\in\mathbb R\) with \(x^*(c)\le\alpha<x^*(x_0)\) for every \(c\in C\). The set \(H=\{x\in X:x^*(x)>\alpha\}\) is weakly open because \(x^*\) is weakly continuous. It contains \(x_0\) and does not meet \(C\). Thus \(X\setminus C\) is the union of such weakly open sets, one for each of its points; it is weakly open, so \(C\) is weakly closed.
Conversely, suppose \(C\) is weakly closed. The weak topology is weaker than the norm topology, so every weakly open set is norm-open. Hence every weakly closed set is norm-closed. This direction does not require convexity. If \(X\) is complex, apply the real separation theorem to the underlying real space; a continuous real-linear separator is the real part of a continuous complex-linear functional, giving the same weakly open half-space argument. \(\square\)
A useful equivalent formulation is that, for every convex \(A\subseteq X\), its weak closure equals its norm closure. Indeed, the norm closure \(\overline A^{\,\|\cdot\|}\) is convex, hence weakly closed by the theorem; it contains \(A\), so the weak closure of \(A\) is contained in it. Conversely, the weak closure contains the norm closure because the weak topology is weaker than the norm topology.
Worked Example: Separating a Point from a Closed Disk
In \(X=\mathbb R^2\) with the Euclidean norm, let \(C=\{(u,v):u^2+v^2\le1\}\) and \(x_0=(2,0)\). Choose the functional \(x^*(u,v)=u\). For each \((u,v)\in C\), \(u\le\sqrt{u^2+v^2}\le1\), so \(x^*(c)\le1\). At the point \(x_0\), \(x^*(x_0)=2\). Thus \(\alpha=1\) gives \(\sup_Cx^*\le1<2=x^*(x_0)\).
Check: the half-space \(\{(u,v):u>1\}\) contains \((2,0)\), and no point of the unit disk has first coordinate greater than \(1\). The functional is continuous since \(|x^*(u,v)|=|u|\le\sqrt{u^2+v^2}\).
Applications to \(L^p\) Spaces
For the applications below, take \(([0,1],\mathcal L,m)\) with Lebesgue measure, and let \(1\le p<\infty\). The real space \(L^p(m)\) consists of equivalence classes, modulo equality \(m\)-a.e., of real-valued measurable functions with finite \(p\)-norm; it is a Banach space. Write \(q=\infty\) if \(p=1\), and \(q=p/(p-1)\) if \(1<p<\infty\). Hölder's inequality makes \(h\mapsto\int_0^1 hg\,dm\) a continuous linear functional whenever \(g\in L^q(m)\). The standard duality theorem for these \(L^p\) spaces says that all continuous linear functionals have this form. We use that earlier duality result, not a new proof of it.
Worked Example: The Nonnegative Cone Is Weakly Closed
Let \(P=\{h\in L^p(m):h\ge0\ m\text{-a.e.}\}\). This set is convex: if \(h,k\ge0\) a.e. and \(0\le t\le1\), then \(th+(1-t)k\ge0\) a.e. We show directly how a functional separates any \(f\notin P\).
Since \(f\notin P\), the set \(\{f<0\}\) has positive measure. It is the countable union of \(E_n=\{f\le-1/n\}\), so at least one \(E_n\) has positive measure; fix such an \(n\) and write \(E=E_n\). Set \(g=-\chi_E\). Since \(m(E)\le1\), \(g\in L^q(m)\), including when \(q=\infty\). For every \(h\in P\), \(\int hg\,dm=-\int_Eh\,dm\le0\). But $$ \int fg\,dm=-\int_E f\,dm\ \ge\ \frac{m(E)}{n}\ >0. $$ Therefore the continuous functional \(h\mapsto\int hg\,dm\) separates \(f\) from \(P\), with threshold \(0\). The cone is norm-closed and convex, hence weakly closed by the theorem; the displayed separation also gives the weakly open neighborhood of each point outside it.
Check: take \(p=2\) and \(f=-2\chi_{[0,1/2]}+\chi_{(1/2,1]}\). Then \(E=\{f\le-1\}=[0,1/2]\), \(g=-\chi_E\), and \(\int fg\,dm=(-2)(-1)(1/2)=1\); for every \(h\in P\), \(\int hg\,dm=-\int_Eh\,dm\le0\). Both sides obey the stated separation inequality.
The cone’s norm-closedness can also be verified without separation. If \(h_j\in P\) and \(h_j\to h\) in \(L^p\), then \(h^-=\max(-h,0)\) satisfies \(0\le h^-\le |h-h_j|\) a.e. for every \(j\), because \(h_j\ge0\). Thus \(\|h^-\|_p\le\|h-h_j\|_p\to0\), so \(h^-=0\) a.e. and \(h\in P\). This check confirms the norm-closed hypothesis used in the theorem.
Worked Example: An Integral Constraint Is Weakly Closed
In \(L^p([0,1],m)\), let \(C=\{h:\int_0^1h\,dm=1\}\). The constant function \(1\) belongs to \(L^q\), since its \(L^q\)-norm is \(1\) for finite \(q\) and its essential supremum is \(1\) for \(q=\infty\). Hence \(T(h)=\int_0^1h\,dm\) is a continuous linear functional by Hölder's inequality. The set \(C=T^{-1}(\{1\})\) is weakly closed because \(T\) is weakly continuous and \(\{1\}\) is closed in \(\mathbb R\). It is also convex: if \(h,k\in C\) and \(0\le t\le1\), then \(\int(th+(1-t)k)\,dm=t+(1-t)=1\).
For a concrete point outside \(C\), take \(h_0=0\). The functional \(T\) gives \(T(h)=1\) for every \(h\in C\) and \(T(h_0)=0\), so \(-T(h)=-1<0=-T(h_0)\) separates the point from the constraint set (with the separating functional oriented toward the set). Check: \(T(1)=\int_0^1 1\,dm=1\), while \(T(0)=0\), and the affine condition is preserved by every convex combination.
Worked Example: Norm Balls Are Weakly Closed
For \(r\ge0\), the closed ball \(B_r=\{h\in L^p:\|h\|_p\le r\}\) is convex by the triangle inequality. It is norm-closed because the reverse triangle inequality gives \(|\|h\|_p-\|k\|_p|\le\|h-k\|_p\). Therefore it is weakly closed by the theorem. The case \(r=0\) is the singleton \(\{0\}\), also covered. For a concrete check at \(p=2\), \(r=1\), the function \(h=\chi_{[0,1/2]}\) has \(\|h\|_2=(\int_0^1\chi_{[0,1/2]}\,dm)^{1/2}=(1/2)^{1/2}\le1\), so it belongs to the ball; the norm calculation uses \(h^2=h\). The ball’s weak closedness follows from convexity and norm closedness, not from any claim that the norm itself is weakly continuous.
Why Convexity and Closedness Matter
Convexity is not a cosmetic assumption. On \(\ell^2\), the set \(S=\{x:\|x\|_2=1\}\) is norm-closed but is not convex. Let \(e_n\) be the sequence with \(1\) in coordinate \(n\) and \(0\) elsewhere. For every \(y\in\ell^2\), the standard duality of \(\ell^2\) gives the weak test \(x\mapsto\sum_jx_jy_j\); evaluating at \(e_n\) gives \(y_n\to0\), since every square-summable sequence tends to zero. Thus \(e_n\to0\) weakly, although every \(e_n\in S\) and \(0\notin S\). So \(S\) is not weakly closed.
Closedness also matters: separation requires a closed convex set to obtain the strict gap at an outside point. In applications, first verify convexity and norm-closedness, then invoke the theorem. In \(L^p\), this turns norm-closed convex constraints—such as positivity, balls, and linear or affine moment constraints—into constraints stable under weak limits. This is useful when a sequence of admissible functions has a weak limit: if every member lies in such a set, the limit remains in it.
A frequent mistake is to infer that every norm-closed set is weakly closed because the weak topology is weaker. The topology comparison gives the reverse implication: weakly closed sets are norm-closed. To conclude norm-closed implies weakly closed, one needs convexity and separation, as the unit sphere example demonstrates.
Check Your Understanding
For each question, identify the relevant topology and state any convexity or duality fact you use.
- Explain why the set \(\{x\in X:x^*(x)\le a\}\) is weakly closed for \(x^*\in X^*\) and \(a\in\mathbb R\).
- For \(1\le p<\infty\) on \([0,1]\), construct a functional separating a function \(f\) that is negative on a set of positive measure from the nonnegative cone.
- Show that \(\{h\in L^p:\int_0^1h\,dm=0\}\) is convex and weakly closed.
- Why does weak closedness imply norm closedness even when the set is not convex?
- Verify that \(e_n\) converges weakly to \(0\) in \(\ell^2\), and explain why this disproves weak closedness of the unit sphere.