Reflexivity of Lp for 1 < p < Infinity

From Duality to Reflexivity

A Banach space is reflexive when its elements account for every bounded linear functional on its dual. For \(L^p\), with \(1<p<\infty\), this property follows by applying the \(L^p\)-duality theorem twice: first to identify the dual of \(L^p\), and then to identify the dual of that dual. The key point is to check that the second representation is exactly evaluation at an element of \(L^p\), not merely an abstract representation by another function.

Throughout, \((X,\mathcal{M},\mu)\) is a sigma-finite measure space; it need not have finite measure or be complete. All functions and Banach spaces in this tutorial are real-valued. The space \(L^r(\mu)\), for \(1\le r<\infty\), consists of equivalence classes up to equality \(\mu\)-a.e. of measurable functions with finite norm \(\|f\|_r=(\int_X|f|^r\,d\mu)^{1/r}\). The continuous dual \(E^*\) of a real Banach space \(E\) is the space of bounded linear maps from \(E\) to \(\mathbb{R}\), with its operator norm; its bidual is \(E^{**}=(E^*)^*\).

We use the \(L^p\)-duality theorem, established earlier by the Radon–Nikodym argument: if \((X,\mathcal{M},\mu)\) is sigma-finite and \(1<r<\infty\), with conjugate exponent \(r'\) defined by \(1/r+1/r'=1\), then every \(\Lambda\in(L^r(\mu))^*\) has a unique representing function \(g\in L^{r'}(\mu)\) such that \(\Lambda(f)=\int_X fg\,d\mu\) for every \(f\in L^r(\mu)\). Moreover, \(\|\Lambda\|=\|g\|_{r'}\). The integral pairing here is bilinear because the scalar field is real.

The Double-Dual Identification

Proof. Fix \(p\) with \(1<p<\infty\), and let \(q\) be its conjugate exponent, so \(1/p+1/q=1\). In particular \(1<q<\infty\). Apply the \(L^p\)-duality theorem to \(L^p(\mu)\). The map \(\mathcal{R}:L^q(\mu)\to(L^p(\mu))^*\) given by \(\mathcal{R}(g)(f)=\int_X fg\,d\mu\) is an onto linear isometry. It is well-defined and bounded by Hölder’s inequality, and the duality theorem gives both its surjectivity and the equality of norms.

Now take an arbitrary \(\Psi\in(L^p(\mu))^{**}\). Define \(F:L^q(\mu)\to\mathbb{R}\) by \(F(g)=\Psi(\mathcal{R}(g))\). This is linear. Since \(\mathcal{R}\) is an isometry, for each \(g\in L^q(\mu)\) we have \(|F(g)|\le\|\Psi\|\,\|\mathcal{R}(g)\|=\|\Psi\|\,\|g\|_q\); hence \(F\in(L^q(\mu))^*\). Apply the \(L^r\)-duality theorem again, this time with \(r=q\). Its hypotheses hold because the measure space is sigma-finite and \(1<q<\infty\); the conjugate exponent of \(q\) is \(p\). Thus there is an \(h\in L^p(\mu)\) such that \(F(g)=\int_X gh\,d\mu\) for every \(g\in L^q(\mu)\).

For any \(\Lambda\in(L^p(\mu))^*\), surjectivity of \(\mathcal{R}\) gives a \(g\in L^q(\mu)\) with \(\Lambda=\mathcal{R}(g)\). Therefore \(\Psi(\Lambda)=\Psi(\mathcal{R}(g))=F(g)=\int_X gh\,d\mu=\mathcal{R}(g)(h)=\Lambda(h)=J_{L^p}(h)(\Lambda)\). This holds for every \(\Lambda\in(L^p(\mu))^*\), so \(\Psi=J_{L^p}(h)\). Since \(\Psi\) was arbitrary, \(J_{L^p}\) is onto and \(L^p(\mu)\) is reflexive. \(\square\)

The order of the two duality applications matters. The first says that a functional on \(L^p\) is represented by a function in \(L^q\). The second says that a functional on \(L^q\) is represented by a function in \(L^p\). The displayed identity in the proof verifies that this final \(L^p\) function is the one at which the original bidual functional evaluates every element of \((L^p)^*\).

Checking the Canonical Map

The proof shows surjectivity. The canonical map’s isometry also helps explain what the representation means: it preserves the norm of each element. In fact, for real \(L^p(\mu)\) on any measure space and \(1<p<\infty\), the \(L^p\)-duality theorem gives the following direct verification.

Proposition. For \(f\in L^p(\mu)\), \(\|J_{L^p}(f)\|=\|f\|_p\).

Proof. If \(f=0\) as an \(L^p\) element, then \(\Lambda(f)=0\) for every \(\Lambda\in(L^p)^*\), so \(\|J_{L^p}(f)\|=0=\|f\|_p\). Otherwise define \(g=\operatorname{sgn}(f)|f|^{p-1}/\|f\|_p^{p-1}\), where \(\operatorname{sgn}(f)\) is \(1\) where \(f>0\), \(-1\) where \(f<0\), and \(0\) where \(f=0\). With \(q=p/(p-1)\), \(|g|^q=|f|^p/\|f\|_p^p\), so \(\|g\|_q=1\). The duality theorem gives \(\|\mathcal{R}(g)\|=1\), and \(J_{L^p}(f)(\mathcal{R}(g))=\int_X fg\,d\mu=\|f\|_p\). Thus \(\|J_{L^p}(f)\|\ge\|f\|_p\). Conversely, for every \(\Lambda\in(L^p)^*\) with \(\|\Lambda\|\le1\), \(|J_{L^p}(f)(\Lambda)|=|\Lambda(f)|\le\|f\|_p\). Taking the supremum over such \(\Lambda\) gives \(\|J_{L^p}(f)\|\le\|f\|_p\). The two inequalities prove equality. \(\square\)

A Common Misidentification

A frequent error is to replace the conjugate exponent \(q\) by \(p\) in the first duality step, as though every functional on \(L^p\) were represented by an \(L^p\) function. This is not what the duality theorem states: it represents functionals on \(L^p\) by elements of \(L^q\), where \(1/p+1/q=1\). For instance, on \([0,1]\) with Lebesgue measure and \(p=3\), the constant function \(1\) represents the bounded functional \(f\mapsto\int_0^1 f\,dm\) and belongs to \(L^{3/2}\). Its \(L^{3/2}\) norm is \((\int_0^1 1\,dm)^{2/3}=1\). The mistaken exponent assignment would instead claim that the representing space is \(L^3\); that particular function happens to belong to both spaces on this finite-measure domain, so it does not expose the error. The correction is to track the exponents in both applications: \(L^p\) has dual \(L^q\), and \(L^q\) has dual \(L^p\).

The hypotheses also matter. The proof uses \(1<p<\infty\), so that both exponents lie strictly between \(1\) and \(\infty\), and it uses sigma-finiteness for the stated \(L^r\)-duality theorem. It establishes reflexivity under those assumptions; it does not assert the same conclusion at either endpoint. Reflexivity is useful because it turns an abstract bidual functional into evaluation at a genuine \(L^p\) element, allowing questions about bidual elements to be handled through the original function space.