Duality identifies continuous linear functionals on \(L^p\) spaces. Weak convergence then becomes convergence against all dual test functions.
Continuous functionals from \(L^q\)
H?lder gives the basic dual pairing.
Proposition. If \(1\le p\lt \infty\), \(1/p+1/q=1\), and \(g\in L^q\), then \(T_g(f)=\int fg\,\mathrm{d}\mu\) defines a bounded linear functional on \(L^p\) with \(\|T_g\|\le\|g\|_q\).
Proof. Linearity is immediate. H?lder gives \(|T_g(f)|\le\|f\|_p\|g\|_q\), so \(\|T_g\|\le\|g\|_q\). For \(1<p<\infty\), truncate \[ f=\operatorname{sgn}(\overline g)|g|^{q-1} \] to finite-measure, bounded level sets and normalize it in \(L^p\); the resulting pairings increase to \(\|g\|_q\). For \(p=1,q=\infty\), test on sets where \(|g|>\|g\|_\infty-\varepsilon\), with the phase chosen to align \(fg\). Thus \(\|T_g\|=\|g\|_q\).
Duality for \(1\lt p\lt \infty\)
For sigma-finite measure spaces, every functional arises from an \(L^q\) function.
Theorem. If \(1\lt p\lt \infty\) and the measure space is sigma-finite, then \((L^p)^*=L^q\), isometrically under the pairing \(\int fg\,\mathrm{d}\mu\).
Proof. First suppose \(\mu(X)<\infty\). Define \(\nu(A)=T(\mathbf1_A)\). If \(A_n\downarrow\varnothing\), then \[ \|\mathbf1_{A_n}\|_p=\mu(A_n)^{1/p}\longrightarrow0, \] so \(\nu(A_n)\to0\); finite additivity and this continuity prove that \(\nu\) is a finite signed measure. It is absolutely continuous with respect to \(\mu\), hence \(\nu=g\mu\) by Radon-Nikodym. For simple \(s\), \[ T(s)=\int sg\,\mathrm{d}\mu, \] and density of simple functions extends the identity to all \(L^p\).
It remains to prove \(g\in L^q\). On \(E_n=\{|g|\le n\}\), take \[ f_n=\operatorname{sgn}(\overline g)|g|^{q-1}\mathbf1_{E_n}. \] Then \[ \int_{E_n}|g|^q =T(f_n) \le\|T\|\left(\int_{E_n}|g|^q\right)^{1/p}. \] Therefore \((\int_{E_n}|g|^q)^{1/q}\le\|T\|\), and monotone convergence gives \(\|g\|_q\le\|T\|\). The reverse inequality was proved above. For sigma-finite \(X\), apply the finite argument on an increasing exhaustion and use uniqueness to paste the local densities.
Theorem (endpoint \(p=1\)). On a sigma-finite measure space, \[ (L^1)^*=L^\infty \] isometrically. In contrast, the dual of \(L^\infty\) is generally strictly larger than \(L^1\); finitely additive functionals appear.
Weak convergence in \(L^p\)
Weak convergence is convergence under all continuous linear tests.
Definition. \(f_n\rightharpoonup f\) weakly in \(L^p\) if \(\int f_ng\,\mathrm{d}\mu\to\int fg\,\mathrm{d}\mu\) for every \(g\in L^q\), where \(q\) is conjugate to \(p\).
Proposition. Strong \(L^p\) convergence implies weak \(L^p\) convergence.
Proof. If \(\|f_n-f\|_p\to0\), then for each \(g\in L^q\), H?lder gives \(|\int(f_n-f)g|\le\|f_n-f\|_p\|g\|_q\to0\).
Proposition. Every weakly convergent sequence is norm bounded.
Proof. For each \(g\in L^q\), the scalar sequence \(\int f_ng\) converges and is therefore bounded. Regard \(f_n\) as bounded linear functionals on \(L^q\). The uniform boundedness principle yields \[ \sup_n\|f_n\|_p<\infty. \]
Example (weak without strong convergence). In \(L^2(0,2\pi)\), the orthonormal sequence \[ e_n(x)=(2\pi)^{-1/2}e^{inx} \] converges weakly to \(0\), because the Fourier coefficients \(\langle g,e_n\rangle\) of every \(g\in L^2\) tend to zero. Yet \(\|e_n\|_2=1\), so there is no strong convergence.
Reflexivity
Reflexivity gives weak compactness of bounded sets.
Theorem. For \(1\lt p\lt \infty\), \(L^p\) is reflexive.
Proof. The dual of \(L^p\) is \(L^q\), and the dual of \(L^q\) is \(L^p\). Under these identifications, the canonical embedding of \(L^p\) into its double dual maps \(f\) to the functional \(g\mapsto\int fg\), which is onto.
Weak lower semicontinuity
Norms behave well under weak limits.
Proposition. If \(f_n\rightharpoonup f\) in \(L^p\), then \(\|f\|_p\le\liminf_n\|f_n\|_p\).
Proof. For \(f\ne0\), choose \(g\in L^q\) with \(\|g\|_q=1\) and \(\int fg\) arbitrarily close to \(\|f\|_p\). Weak convergence gives \(\int f_ng\to\int fg\), while H?lder gives \(|\int f_ng|\le\|f_n\|_p\). Take the liminf and then optimize over \(g\).
Corollary (Radon-Riesz property). For \(1<p<\infty\), if \(f_n\rightharpoonup f\) in \(L^p\) and \(\|f_n\|_p\to\|f\|_p\), then \(f_n\to f\) strongly in \(L^p\).
Proof outline. The spaces \(L^p\), \(1<p<\infty\), are uniformly convex. If \(\|f_n-f\|_p\) stayed above \(\varepsilon\), uniform convexity would force the midpoint norms \[ \left\|\frac{f_n+f}{2}\right\|_p \] to stay strictly below the common limiting norm. But the midpoints converge weakly to \(f\), and weak lower semicontinuity gives the opposite inequality. Hence the norm difference must vanish.