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The answer isIndeed, Lebesgue measure on $E\subset {\mathbb{R}}^n$ is a sigma-finite measure, and duality of Lp spaces gives $(L^1(E){)}^{\ast }=L^{\infty }(E)$. The restricted Lebesgue measure is a non-atomic measure and $E$ has positive measure, so the Dual of L infinity is larger than L1: the canonical integral representation does not exhaust $(L^{\infty }(E){)}^{\ast }$.

For $1<p<\infty$, the Holder conjugate exponent also lies strictly between one and infinity. Duality of Lp spaces then holds in both directions, in agreement with Reflexivity of Lp spaces. For $p=\infty$, the first required identification already fails.

The Lp norm of the forcing at the critical exponent is constant:To prove weak convergence in $L^{6/5}(B)$, the duality of Lp spaces says that we must test against an arbitrary $g\in L^6(B)$. Holder inequality, restricted to the shrinking measurable set $E_m$, givesThe last step is absolute continuity of the Lebesgue integral, since $|E_m|=r_m^3|E|\to 0$ and $|g{|}^6$ is integrable. ThusTesting only bounded or smooth functions would not by itself establish this weak convergence; the restricted Holder inequality handles every member of the full dual space.

We first construct a terminal integrable function without using the martingale convergence theorem. Put $G={\sup }_n|g_n|$ and $d\eta =Gdx$ on $(0,1{]}^d$. This is a finite positive measure. If $G=0$ almost everywhere the conclusion is immediate. Otherwise, for a bounded dyadic step function $h$ measurable with respect to ${\mathcal{F}}_n$ in the dyadic filtration, defineThe martingale property makes the definition independent of the choice of a sufficiently large $n$. Moreover $|\Lambda (h)|\le \int |h|d\eta$. Dyadic step functions are dense in $L^1(\eta )$: continuous functions on the closed unit cube are dense for this finite Borel measure (extend the measure by zero onto the omitted faces), and uniform continuity approximates each continuous function by step functions on fine dyadic partitions. Thus $\Lambda$ extends to a bounded linear functional on $L^1(\eta )$ of norm at most one. The duality of Lp spaces representation of this functional gives $H\in L^{\infty }(\eta )$ with $|H|\le 1$ and $\Lambda (h)=\int hHd\eta$. Set $g=HG$, taking $g=0$ where $G=0$. Then $g\in L^1(dx)$ and testing indicators of all dyadic atoms provesThis argument also works for complex-valued martingales, with a complex linear functional.

To prove convergence, choose a dyadic step function $h$ with $\Vert g-h{\Vert }_1$ arbitrarily small. For all $n$ beyond its dyadic level, $\mathbb{E}(h\mid {\mathcal{F}}_n)=h$. The L1 contraction of conditional expectation yieldsThis proves convergence in $L^1$. For convergence almost everywhere, Doob maximal inequality applied to the conditional expectations of $g-h$ givesLet the approximation error tend to zero, and then take positive rational $a$. Hence $g_n\to g$ both in $L^1$ and almost everywhere.

For the second assertion, write $M={\sup }_n\mathbb{E}|f_n|<\infty$ and take the given first-exit stopping time $\tau$. Each $g_n=f_{\tau \wedge n}$ is integrable, being a finite sum of restrictions of integrable variables. Its increments satisfyThe indicator is ${\mathcal{F}}_n$-measurable, so taking its conditional expectation proves that $(g_n)$ is a martingale.

The overshoot at $\tau$ need not be bounded by $N$. Instead, for a fixed $m$, on $\{\tau =j\}$ with $j\le m$ we have $|f_j|\le \mathbb{E}(|f_m|\mid {\mathcal{F}}_j)$. Integrating over this ${\mathcal{F}}_j$-measurable event, summing the disjoint events, and including $\{\tau >m\}$ givesThe variables $Z_m=|f_{\tau }|1_{\{\tau \le m\}}$ increase to $Z$, and $Z_m\le |g_m|$. The monotone convergence theorem gives $\mathbb{E}Z\le M$. Before stopping, the process has magnitude less than $N$; after stopping it remains at $f_{\tau }$. ConsequentlyThis is the integrable overshoot of a stopped L1-bounded martingale. The first part now proves that every such stopped process converges almost everywhere.

Finally, Doob maximal inequality gives $|\{{\sup }_n|f_n|\ge N\}|\le M/N$, by applying the finite-horizon estimate and then increasing the horizon. Thus ${\sup }_n|f_n|$ is finite almost everywhere. Outside the union of the null sets for all positive integer stopping thresholds, choose an integer $N$ larger than this supremum. No stopping occurs at that point, so $f_n=g_n$ for all $n$, and $f_n$ converges to a finite limit almost everywhere. We have not assumed, or concluded, $L^1$ convergence for this general $L^1$-bounded martingale.

The relevant Banach-Alaoglu theorem says that for $1<p<\infty$ the closed bounded balls of $L^p({\mathbb{R}}^n)$ are compact in the weak topology. In particular, every bounded sequence has a weakly convergent subsequence whose limit has norm no larger than the common bound. Applying the preceding duality of Lp spaces also identifies this as weak-star compactness of $L^p=(L^q{)}^{\ast }$; its weak and weak-star test spaces are both $L^q$.

Here is a direct proof of both compactness and the subsequence assertion. Let $\Vert f_j{\Vert }_p\le M$ and choose a countable dense set $\{g_m\}$ in the separable Lp space $L^q$. Each scalar sequence $\int f_j{g}_m$ is bounded. Successive subsequence selection followed by the diagonal subsequence argument gives $f_{j_k}$ for which these pairings converge for every $m$. For arbitrary $g\in L^q$,First approximate $g$ by $g_m$ and then take $k,l$ large. This proves convergence of every pairing, and the limit $\Lambda (g)={\lim }_k\int f_{j_k}g$ is linear with $|\Lambda (g)|\le M\Vert g{\Vert }_q$. Apply the already proved duality of Lp spaces with $q$ in place of $p$: $\Lambda (g)=\int fg$ for an $f\in L^p$ with $\Vert f{\Vert }_p\le M$. ThusTo obtain compactness rather than only a sequential statement, on the radius-$M$ ball defineDensity and Hölder's inequality show that this is a metric space distance and that its topology is exactly the weak topology on this ball: every remaining test pairing is uniformly approximated there by a dense-set pairing. The subsequence argument proves sequential compactness for this metric. A sequentially compact metric space is compact: otherwise either an infinite separated sequence contradicts total boundedness or a Cauchy sequence without a limit contradicts sequential compactness; completeness and total boundedness give compactness by successive finite coverings. This completes the proof. The same argument is the Sequential Banach-Alaoglu theorem for a separable predual.

We first construct a genuinely compact Besicovitch set; a sequence of unrelated small-area sets would not suffice. The compact Kakeya construction from small projections below gives the limiting compact set together with its segments.

Let $\mathcal{X}$ consist of nonempty compact sets $K\subset [0,1]\times \mathbb{R}$ whose horizontal projection is exactly $[0,1]$. With the Hausdorff metric this is a complete metric space. Indeed a Hausdorff-Cauchy sequence is eventually contained in one bounded closed rectangle and converges there to a nonempty compact set; horizontal projection is Lipschitz for Hausdorff distance, so its value $[0,1]$ is preserved in the limit.

For $t\in [0,1]$ put $L_t(x,y)=y+tx$ and $m_t(K)=|L_t(K)|$. The function $(K,t)\mapsto m_t(K)$ is upper semicontinuous. To verify this, cover the compact set $L_t(K)$ by finitely many open intervals with total length less than $m_t(K)+\epsilon$. Joint continuity of the compact images in the Hausdorff metric puts every sufficiently nearby image inside the same cover. This proves the required upper bound, uniformly over a compact parameter interval when a strict bound holds throughout it.

Fix $q\in [0,1]$ and $\epsilon >0$. The setis open. It is also dense, by the following explicit finite approximation. Given $K$ and tolerance $h>0$, partition $[0,1]$ into $N$ equal subintervals and choose $(j/N,y_j)\in K$ at each left endpoint. Above that subinterval put the segmentEach segment lies within $\sqrt{1+q^2}/N$ of its chosen point in $K$. Add a finite $h/2$-net of points of $K$ to their union, ensuring the reverse Hausdorff inclusion. For large $N$ the resulting compact set $K^{\prime }$ belongs to $\mathcal{X}$ and has Hausdorff distance less than $h$ from $K$. The image of each segment under $L_t$ has length $|t-q|/N$, and the added points contribute zero length. Henceso $K^{\prime }\in U(q,\epsilon )$.

Take a finite grid of $q$'s whose $\epsilon$-neighborhoods cover $[0,1]$. The finite intersection of the open dense sets $U(q,\epsilon )$ is dense and implies ${\sup }_{0\le t\le 1}{m}_t(K)<2\epsilon$. Thusis open and dense. The Baire category theorem gives a compact $K\in {\bigcap }_{r\ge 1}{V}_{1/r}$, and therefore $m_t(K)=0$ for every $t\in [0,1]$.

DefineThis is the continuous image of the compact set $K\times [0,1]$, hence compact. Its section at horizontal coordinate $t$ is $L_t(K)$, of measure zero. The Fubini theorem gives $|B|={\int }_0^1{m}_t(K)dt=0$. For every slope $x\in [0,1]$, horizontal surjectivity supplies $(x,y)\in K$, and $B$ contains the segment $(t,y+tx)$, $0\le t\le 1$, of length $\sqrt{1+x^2}\ge 1$. Four rotated copies, through angles $0,\pi /4,\pi /2,3\pi /4$, cover all unoriented directions. Their finite union $\mathcal{B}$ is compact, has area zero, and contains a unit subsegment in every direction:The point-line dual construction is the mechanism in Körner's construction; the density calculation above supplies the needed proof, rather than assuming that small-area sets automatically admit a suitable nested limit.

For the multiplier discussion, define $T_D$ by $\widehat{T_{D}g}(\xi )=1_{\{|\xi |\le 1\}}\hat{g}(\xi )$. The Plancherel theorem gives boundedness on $L^2$. We explain why a putative $L^p$ bound, $1<p<2$, would contradict Kakeya packing; duality of Lp spaces then treats $p>2$.

First, such a bound would force the vector-valued inequalityfor arbitrary directions $v_j$, where $P_v$ is the half-plane Fourier projection onto the half-plane $\xi \cdot v<0$. To see the important curvature step, let $T_R$ project onto the radius-$R$ disk and modulate by $e^{iR{v}_j\cdot x}$. Its conjugated symbol is $1_{\{|\xi +R{v}_j|\le R\}}$, tending almost everywhere to $1_{\{\xi \cdot v_j<0\}}$. Dilations give the same putative norm bound for every $R$. Apply this common operator to ${\sum }_j{ϵ}_j{e}^{iR{v}_j\cdot x}{g}_j$, average over independent random signs, and use the Khintchine inequality. Letting $R\to \infty$ gives the displayed estimate by $L^2$ convergence and Fatou lemma, first for finitely many smooth function functions and then by approximation. The disk has a boundary normal in every direction; a fixed polygon does not.

The geometric input for this overview is the standard finite Perron tree packing used in Fefferman's original proof: for arbitrarily small $\eta >0$ one obtains long thin rectangles $R_j$ with $A={\sum }_j|R_j|$ and $|{\bigcup }_j{R}_j|\le \eta A$, while fixed longitudinal translates $R_j^{\prime }$ are pairwise disjoint. This is a quantitative finite arrangement, stronger than merely having a compact null set. The splitting/sliding procedure subdivides an angular triangle into narrow triangles and overlaps their bases; the extensions along their axes separate in reverse order. Its area estimate has the form ${\alpha }^{2m}+2(1-\alpha )$ relative to the original total triangle area: the merged central part shrinks by ${\alpha }^2$ at each generation, and the accumulated caps contribute at most $2(1-\alpha )$. Taking $\alpha$ close to one and then $m$ large makes the ratio arbitrarily small. Inscribed rectangular portions retain a fixed proportion of the total area. The reflected extensions explain why the translates can be disjoint even though the original rectangles overlap strongly.

Here is the analytic contradiction, with the relevant powers explicit. Take $g_j=1_{R_j}$. In coordinates parallel and perpendicular to the long axis $v_j$, the directional Hilbert transform integrates only along that axis. If the longitudinal interval is $[0,\ell ]$, then outside itOn a fixed translate such as $s\in [5\ell ,6\ell ]$, this has magnitude at least ${\pi }^{-1}\log (6/5)$. Since $P_v=(I-i{H}_v)/2$ up to the harmless choice of half-plane sign, $|P_{v_j}{g}_j|\ge c_0>0$ on $R_j^{\prime }$. Their disjointness gives a lower bound $c_0{A}^{1/p}$ for the vector-valued output norm.

Writing $h={\sum }_j{1}_{R_j}$, Hölder's inequality for $p<2$ givesThe putative vector-valued inequality would implywhich is impossible as $\eta \downarrow 0$. The disk multiplier is self-adjoint, so a bound for $p>2$ would give one for its conjugate exponent below two. Thus the disk multiplier is bounded only at $p=2$ among $1<p<\infty$. This explains both the random-sign step and the precise geometric amplification; it is not an assertion that frequency rectangles by themselves produce a logarithmic norm blow-up.

It is enough to prove the stronger planar Kakeya maximal function boundWe first work with a $\delta$-separated net of unoriented directions $e_1,\dots ,e_M$, with $M\asymp {\delta }^{-1}$. Write ${\alpha }_{ij}$ for angular distance modulo $\pi$. Choose arbitrary length-one, width-$\delta$ rectangles $T_j$ in these directions. The permitted rectangle intersection fact isThe first alternative includes parallel rectangles. Their locations are arbitrary; only separation of their directions matters.

Define $Af(j)={\delta }^{-1}{\int }_{T_j}f$, and put $\Vert a{\Vert }_{{\ell }_{\delta }^2}^2=\delta {\sum }_j|a_j{|}^2$. With these weights the adjoint operator is $A^{\ast }a={\sum }_j{a}_j{1}_{T_j}$. Since a separated angular net has only a bounded number of directions at each distance scale $k\delta$ from $e_i$, its overlap matrix satisfiesThe diagonal term is of size $\delta$. Using $|a_i{a}_j|\le (|a_i{|}^2+|a_j{|}^2)/2$ and symmetry gives the Schur test estimateBy duality of Lp spaces, $\Vert Af{\Vert }_{{\ell }_{\delta }^2}\lesssim \sqrt{\log (2/\delta )}\Vert f{\Vert }_2$. This is uniform over every choice of the translated rectangles. For each direction choose a rectangle approaching the supremum for $|f|$, then take the limit. That gives the same estimate for the discretized Kakeya maximal function.

To recover all directions, partition the direction circle into arcs of length comparable to $\delta$, each with a net direction. A tube in an arc is contained in a rectangle in its net direction with width $C\delta$ and length at most two. A bounded subdivision in the length direction reduces this to the same averaging operators; the wider tubes obey the same overlap estimate with fixed-factor changes. ThereforeFinally $\sqrt{\log (2/\delta )}\le C_{\epsilon }{\delta }^{-\epsilon }$ for every $\epsilon >0$. The planar Kakeya maximal estimate therefore has the required arbitrary small power loss.

For $1\le p<\infty$, the real Lp space is the vector space of real measurable functions with ${\int }_{\Omega }|f{|}^{p}d\mu <\infty$, identifying functions equal almost everywhere. Its Lp norm is $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. For $p=\infty$, take the essentially bounded real measurable functions, with the same identification and norm $\Vert f{\Vert }_{\infty }=\mathrm{esssup}|f|$. The identification makes each norm definite; absolute homogeneity follows from the Lebesgue integral, and the triangle inequality follows from Minkowski inequality for finite $p$, or directly from the essential supremum for $p=\infty$.

Here is a completeness proof valid on any measure space. For $p<\infty$, a Cauchy sequence $(f_n)$ has a subsequence $(f_{n_k})$ with $\Vert f_{n_{k+1}}-f_{n_k}{\Vert }_p\le 2^{-k}$. Choose measurable function representatives and put $h_k=f_{n_{k+1}}-f_{n_k}$. By Minkowski inequality and the monotone convergence theorem,Consequently the series $\sum h_k$ converges absolutely almost everywhere. Define $f=f_{n_1}+\sum h_k$ there, and define it to be zero on the measurable exceptional null set. Then $f\in L^p$, and Fatou lemma applied to each tail gives $\Vert f-f_{n_k}{\Vert }_p\le {\sum }_{j=k}^{\infty }{2}^{-j}\to 0$. The original Cauchy sequence also converges in Lp norm, by the triangle inequality. For $p=\infty$, choose the same subsequence using the essential supremum norm. Outside one measurable null set, all the bounds $|h_k|\le 2^{-k}$ hold and $f_{n_1}$ is bounded. The series then converges uniformly there, with an essentially bounded measurable function limit and the same tail estimate in essential supremum norm. Thus all these spaces are Banach spaces.

The duality of Lp spaces says that, for $1<p<\infty$ and $q=p/(p-1)$, the mapis an isometric isomorphism of normed spaces. This form of Lp duality on an arbitrary measure space requires no finiteness hypothesis on $\mu$. At $p=1$, a standard version assumes a sigma-finite measure and identifies $(L^1{)}^{\ast }$ with $L^{\infty }$ through the same dual pairing. That endpoint assertion must not be made without a suitable measure-space hypothesis.

We first prove the required duality of Lp spaces for a finite measure. Let $F\in (L^p{)}^{\ast }$ and set $\nu (E)=F(1_E)$. For disjoint measurable $E_j$, the indicator functions of their partial unions converge in Lp norm to that of their union, so $\nu$ is countably additive. It has finite variation measure: for every finite measurable partition $(E_j)$, choosing real signs givesAlso $\mu (E)=0$ implies $\nu (E)=0$. The Radon-Nikodym theorem supplies a Radon-Nikodym derivative $g\in L^1$ with $\nu (E)={\int }_{E}gd\mu$. Linearity gives $F(f)=\int fgd\mu$ for simple functions. Uniform approximation by simple functions extends this identity to bounded measurable functions: both their Lp norm errors and the errors in integration against $g$ tend to zero.

To establish the correct integrability, test with the bounded measurable function $f_N=\mathrm{sgn}(g)|g{|}^{q-1}{1}_{\{|g|\le N\}}$. Since $(q-1)p=q$, writing $A_N={\int }_{\{|g|\le N\}}|g{|}^{q}d\mu$ givesThe second inequality is also valid when $A_N=0$. The monotone convergence theorem gives $g\in L^q$ and $\Vert g{\Vert }_q\le \Vert F\Vert$. Density of simple functions in the Lp space, together with Hölder's inequality, now gives $F=I_{p}g$ on all of $L^p$. Conversely Hölder's inequality gives $\Vert I_{p}g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, testing againstgives $\Vert f{\Vert }_p=1$ and $(I_{p}g)(f)=\Vert g{\Vert }_q$, so $\boxed{\Vert I_{p}g\Vert =\Vert g{\Vert }_q}$. This also proves uniqueness of the representing Radon-Nikodym derivative.

For completeness, the passage to an arbitrary measure space can be made without losing a hypothesis in the reflexive Banach space argument below. On a sigma-finite measure space, exhaust by nested finite-measure sets $E_n$. The representing Radon-Nikodym derivatives on $E_n$ agree on overlaps by uniqueness. Their glued density has Lp norm $\Vert g{\Vert }_q\le \Vert F\Vert$ by the monotone convergence theorem, and represents $F$ because $f{1}_{E_n}\to f$ in Lp norm. Every $f\in L^p$ on an arbitrary measure space is supported on a sigma-finite measurable set: the sets $\{|f|>1/n\}$ have finite measure and their union is $\{f\ne 0\}$.

Use support localization of an Lp functional as follows. For a sigma-finite measurable $A$, let $m_A$ be the operator norm of $F$ restricted to functions supported in $A$. The support observation gives ${\sup }_A{m}_A=\Vert F\Vert$. Choose $A_n$ approaching this supremum and set $A={\bigcup }_n{A}_n$; then $m_A=\Vert F\Vert$. If a sigma-finite $B\subseteq \Omega \setminus A$ had $m_B=\beta >0$, functions supported on the disjoint sets $A,B$ have the direct-sum Lp norm. Optimizing their two scalar coefficients by Hölder's inequality, and using functions approaching the two restriction operator norms, would givea contradiction. Thus $F$ vanishes on functions supported outside $A$. The density on $A$, extended by zero, represents $F$ globally. The case $F=0$ simply uses $g=0$. This proves the stated Lp duality on an arbitrary measure space.

Finally let $X=L^p$ and let $\Phi \in X^{\ast \ast }$, where stars denote continuous dual spaces. Compose with $I_p$ to obtain the bounded linear functional $g\mapsto \Phi (I_{p}g)$ on $L^q$. Applying duality of Lp spaces with the exponents reversed gives $f\in L^p$ with $\Phi (I_{p}g)=\int fgd\mu$. For the canonical embedding into the bidual $J_{X}f$, its value on $I_{p}g$ is also $\int fgd\mu$. Since $I_p$ is onto, $\Phi =J_{X}f$. Its norm is $\Vert f{\Vert }_p$ by the same dual pairing norm identity. Therefore $\boxed{J_X(L^p)=(L^p{)}^{\ast \ast }}$, which proves that $L^p$ is a reflexive Banach space through its actual canonical embedding into the bidual.

The Radon-Nikodym theorem for positive measures says: if $\mu$ and $\nu$ are sigma-finite measures on the same measurable space, and $\nu$ is absolutely continuous with respect to $\mu$, then there is a nonnegative measurable function $g$, unique $\mu$-almost everywhere, such thatHere sigma-finiteness means that the space is a countable union of measurable sets of finite measure; absolute continuity of measures, written $\nu \ll \mu$, means that $\mu (E)=0$ implies $\nu (E)=0$. The function $g=d\nu /d\mu$ is the Radon-Nikodym derivative. For a finite signed or complex measure $\nu$ of finite total variation norm of a measure, absolutely continuous with respect to a sigma-finite $\mu$, the corresponding density belongs to $L^1(\mu )$. This follows by applying the positive theorem to the positive and negative parts of the real and imaginary parts of $\nu$.

For every measure space and $1<p<\infty$, $(L^p{)}^{\ast }$ is isometrically $L^q$, where $1/p+1/q=1$. We use the complex-linear pairingWith the convention $\int h\overline{g}$, the same identification is conjugate-linear in $g$. By Hölder's inequality, ${\Phi }_g$ is a bounded linear functional with $\Vert {\Phi }_g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, setinterpreting the numerator as zero where $g=0$. The identity $(q-1)p=q$ gives $\Vert h{\Vert }_p=1$ and ${\Phi }_g(h)=\Vert g{\Vert }_q$. ThusIt remains to represent an arbitrary $\Phi \in (L^p{)}^{\ast }$, rather than merely produce functionals from $L^q$.

We prove Lp duality on an arbitrary measure space without imposing sigma-finiteness on $\mu$. Write $C=\Vert \Phi \Vert$. For every measurable set $E$ with $\mu (E)<\infty$, define a complex measure on $E$ byIt is countably additive: for disjoint $F_j\subseteq E$, the partial sums of their indicator functions tend in $L^p$ to $1_{\bigcup F_j}$, because the measure of the omitted tail tends to zero. It is absolutely continuous with respect to $\mu {|}_E$, since indicator functions of null sets represent zero in $L^p$.

Its total variation norm of a measure is finite. For any finite measurable partition $E={\bigcup }_j{E}_j$, choose scalars $c_j$ of modulus $1$ with $c_j{\nu }_E(E_j)=|{\nu }_E(E_j)|$. ThenTaking the supremum over partitions gives the variation bound. Since $\mu {|}_E$ is finite, the Radon-Nikodym theorem supplies $g_E\in L^1(E)$ with ${\nu }_E(F)={\int }_F{g}_{E}d\mu$. By uniform approximation with simple functions,for every bounded measurable $h$ supported in $E$.

To improve $g_E$ from $L^1$ to $L^q$, test with the bounded functionIf $I_N={\int }_{E\cap \{|g_E|\le N\}}|g_E{|}^{q}d\mu$, thenThus $I_N^{1/q}\le C$ when $I_N>0$, and the same bound is trivial when it is zero. The monotone convergence theorem givesIf $E,F$ both have finite measure, the densities $g_E,g_F$ agree almost everywhere on $E\cap F$: their integrals over every measurable subset of the intersection equal the same functional value. This is uniqueness in the Radon-Nikodym theorem.

We now perform support localization of an Lp functional. SetChoose finite-measure sets $E_n$ whose displayed integrals tend to $M$, and let $D_n=E_1\cup \cdots \cup E_n$, $S={\bigcup }_n{D}_n$. If $M=0$, take $E_n=\varnothing$. Compatibility allows us to define a measurable function $g$ on $S$ by taking $g_{D_n}$ on the disjoint measurable sets $D_n\setminus D_{n-1}$, and put $g=0$ off $S$. It agrees almost everywhere with $g_{D_n}$ on every $D_n$. Moreover,Indeed each integral is at most $M$, and it is at least the integral over $E_n$, which tends to $M$.

For any finite-measure set $F\subseteq \Omega \setminus S$, compatibility on the disjoint union $D_n\cup F$ givesLetting $n\to \infty$ forces $g_F=0$ almost everywhere. For an arbitrary finite-measure set $E$, compatibility on $E\cap D_n$ and the preceding conclusion on $E\setminus S$ show that $g_E=g$ almost everywhere on $E$. Consequently $\Phi (h)=\int hgd\mu$ for every simple function supported on a finite-measure set.

Those simple functions are dense in $L^p$ even for this arbitrary measure space. To see the needed finite-support property, for $h\in L^p$ the sets $\{|h|>1/n\}$ have finite measure, bounded by $n^p\Vert h{\Vert }_p^p$. First truncate $h$ to such sets and to bounded values, then approximate by simple functions; the discarded $L^p$ integral tends to zero. Continuity of $\Phi$ and Hölder's inequality therefore extend the representation to every $h\in L^p$.

We have constructed $g\in L^q$ with $\Phi ={\Phi }_g$, and the previously proved norm identity gives $\Vert g{\Vert }_q=\Vert \Phi \Vert$. It also proves uniqueness: if ${\Phi }_g={\Phi }_{\tilde{g}}$, then $\Vert g-\tilde{g}{\Vert }_q=0$. This completes the isometric duality of Lp spaces.

The dominated sequence actually converges to zero in norm, and hence weakly. The assumptions give $|f_n{|}^p\le |f{|}^p\in L^1$ and $|f_n{|}^p\to 0$ almost everywhere. The dominated convergence theorem yieldsFor every $\Phi \in (L^p{)}^{\ast }$, $|\Phi (f_n)|\le \Vert \Phi \Vert \Vert f_n{\Vert }_p\to 0$. Therefore

Let $q$ be conjugate to $p$, so $1/q=1-1/p$, including $q=1$ for $p=\infty$. Hölder's inequality implies, for every measurable finite-measure set $E$,This proves integrability on $A$ and gives uniform integrability there. Choose a set $B\subseteq A$ with $\lambda (A\setminus B)\le \delta$ on which convergence is uniform, using Egorov theorem. ThenFirst take $n\to \infty$ and then $\delta \downarrow 0$. Thus even the local $L^1$ difference tends to zero, and in particularFor $1<p<\infty$, simple functions supported on finite-measure Borel sets are dense in $L^q(\mathbb{R})$. The displayed convergence proves pairing convergence for such functions. Approximate any $h\in L^q$ by a simple $h_0$ and useBy the duality of Lp spaces, these are all continuous linear functionals on $L^p$, so $f_n$ converges weakly to $f$. It need not converge strongly: $f_n=1_{[n,n+1]}$ converges pointwise to zero but has $L^p$ norm one. For $p=\infty$ the pairing argument gives weak-star convergence against $L^1$, rather than asserting weak convergence against every functional in the much larger full dual.

Let $q$ be the Holder conjugate exponent of $p$. On ${\mathbb{R}}^n$ with Lebesgue measure, the duality of Lp spaces isunder the pairing ${\Lambda }_g(f)=\int fg$. Thus $(L^1{)}^{\ast }=L^{\infty }$, while for $1<p<\infty$ the dual exponent satisfies $1/p+1/q=1$.

A Banach space is reflexive when its canonical map into its bidual is surjective. The spaces $L^p({\mathbb{R}}^n)$ are reflexive exactly forFor these exponents, applying $L^p$ duality twice identifies the bidual with $L^p$. The endpoint spaces $L^1$ and $L^{\infty }$ are not reflexive; for example, disjoint unit cubes give closed copies of ${\ell }^1$ and ${\ell }^{\infty }$, respectively.

Integration gives an isometric embedding $L^1\hookrightarrow (L^{\infty }{)}^{\ast }$, but it is not onto. On the subspacedefine $\Lambda (c+h)=c$, the limit at infinity. Since $|\Lambda (v)|\le \Vert v{\Vert }_{\infty }$, the Hahn-Banach theorem extends $\Lambda$ to all of $L^{\infty }$. If this extension were integration against some $g\in L^1$, its vanishing on every compactly supported bounded function would force $g=0$ almost everywhere, but $\Lambda (1)=1$. ThereforeThis is the Dual of L infinity is larger than L1.

No. In the Lebesgue space $L^1([0,2\pi ])$, letFor every $g\in L^{\infty }([0,2\pi ])$, finiteness of the interval gives $g\in L^1([0,2\pi ])$, and the Riemann-Lebesgue lemma yieldsThe duality of Lp spaces therefore shows that $f_n\rightharpoonup 0$ in $L^1$. On the other hand,for every $n$. Thus weak convergence does not imply norm convergence in this $L^1$ space; the Weakly null sine sequence in L1 is the required counterexample.

The continuous dual space $X^{\prime }$ consists of all bounded linear maps $\Lambda :X\to \mathbb{C}$, withIf $Y=L^p({\mathbb{R}}^n)$ with $1\le p<\infty$ and $q$ is the conjugate exponent, then duality of Lp spaces identifies $Y^{\prime }$ isometrically with $L^q({\mathbb{R}}^n)$, including $q=\infty$ when $p=1$. Every functional has the formand $\Vert {\Lambda }_h\Vert =\Vert h{\Vert }_q$.

Decompose $f={\sum }_{j=0}^n{f}^{(=j)}$ into its homogeneous Fourier levels. The Bonami lemma and the triangle inequality giveApply this estimate to the $m$-fold tensor power $f^{\otimes m}$. Tensor products multiply both relevant norms and commute with the noise operator on the Boolean hypercube, soTaking $m$th roots and the limit $m\to \infty$ proves the hypercontractive inequality on the Boolean hypercube