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Working with measurable functions modulo equality almost everywhere, the sequence is Cauchy in Lp space if for every $ϵ>0$ there exists $N$ such that $\Vert f_n-f_m{\Vert }_p<ϵ$ for all $m,n\ge N$. This is a statement about integral norm distance, not pointwise convergence. Without quotienting null functions the displayed quantity is a seminorm, so the conventional $L^p$ normed space uses these equivalence classes.

Completeness means that every norm-Cauchy sequence has a limit belonging to the same Lp space, with $\Vert f_n-f{\Vert }_p\to 0$. To prove it, choose a subsequence $f_{n_j}$ with $\Vert f_{n_{j+1}}-f_{n_j}{\Vert }_p\le 2^{-j}$. The functionshave $p$-norm bounded by $\Vert f_{n_1}{\Vert }_p+1$, by Minkowski inequality. They increase to $H$, and Fatou's lemma gives $\int H^p\le (\Vert f_{n_1}{\Vert }_p+1{)}^p$. Thus $H$ is finite almost everywhere, so the subsequence telescopes absolutely there to a measurable limit $f$ with $|f|\le H$ and $f\in L^p$.

The same argument on the tails gives $\Vert f-f_{n_j}{\Vert }_p\le {\sum }_{k\ge j}{2}^{-k}\to 0$. Given $ϵ$, choose such a subsequence term beyond the Cauchy threshold and use $\Vert f_n-f{\Vert }_p\le \Vert f_n-f_{n_j}{\Vert }_p+\Vert f_{n_j}-f{\Vert }_p$. This proves convergence of the full sequence and completeness for every $1\le p<\infty$, on an arbitrary measure space.

For $1\le p<\infty$, the Lp norm is $\Vert f{\Vert }_p=({\int }_E|f{|}^{p}d\mu {)}^{1/p}$; for $p=\infty$ it is the essential supremum of $|f|$. The Lp space consists of measurable functions with finite norm, identifying functions equal almost everywhere. This identification makes the quantity a norm rather than only a seminorm. Convergence in $L^p$ means $\Vert f_n-f{\Vert }_p\to 0$, including the essential-supremum norm when $p=\infty$.

First let $1\le p<\infty$, and take a Cauchy sequence $f_j$ in the Lp space, with functions identified almost everywhere. Choose a subsequence $f_{j_k}$ with $\Vert f_{j_{k+1}}-f_{j_k}{\Vert }_p\le 2^{-k}$ and put $a_k=f_{j_{k+1}}-f_{j_k}$. The triangle inequality givesBy the monotone convergence theorem, $A={\sum }_{k\ge 1}|a_k|$ belongs to $L^p$, and is finite almost everywhere. The series $f_{j_1}+{\sum }_{k\ge 1}{a}_k$ therefore converges almost everywhere to an $L^p$ function $f$. Applying the same argument to each absolute tail yieldsThe original sequence is Cauchy, so the triangle inequality extends convergence from this subsequence to the whole sequence.

For $p=\infty$, choose the same subsequence in essential supremum norm. Outside one null set, all its difference bounds hold simultaneously:The series then converges uniformly on that full-measure set to a bounded measurable $f$, and $\Vert f-f_{j_k}{\Vert }_{\infty }\le {\sum }_{\ell \ge k}{2}^{-\ell }$. Again the full sequence converges. Thus every $L^p$, $1\le p\le \infty$, is a Banach space, proving completeness of Lp spaces.

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.

The Lp space is the vector space of equivalence classes, modulo equality almost everywhere, of measurable functions $f$ with $\int |f{|}^{p}d\mu <\infty$, equipped with $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. To prove it is a Banach space, let $(f_n)$ be Cauchy in this norm. Choose a subsequence $(f_{n_j})$ such thatTake measurable representatives and put $g_j=|f_{n_{j+1}}-f_{n_j}|$, $G_M={\sum }_{j=1}^M{g}_j$. The Minkowski inequality gives $\Vert G_M{\Vert }_p\le {\sum }_j{2}^{-j}\le 1$. By the monotone convergence theorem, $G={\sum }_{j\ge 1}{g}_j$ satisfies $\int G^{p}d\mu \le 1$. Consequently $G$ is finite almost everywhere, and the telescoping seriesconverges absolutely almost everywhere. Define it arbitrarily on the null exceptional set. The bound $|f|\le |f_{n_1}|+G$ puts $f$ in $L^p$.

For the tails, monotone convergence and Minkowski again giveSince the original sequence is Cauchy, the triangle inequality then gives $f_n\to f$ in norm, not merely along the subsequence. HenceThis proof needs no finiteness or sigma-finiteness assumption on the measure space.

The finite-$p$ assertion needs the zero-divergence hypothesis from the preceding part, which is not repeated in this part's printed hypotheses. The norm is on ${\mathbb{R}}_x^3\times {\mathbb{R}}_v^3$, and finite conservation requires $f_0$ to belong to the corresponding Lp space in addition to its smoothness.

Writing ${\Phi }_t$ for the characteristic flow map, the general solution is $f_t=f_0\circ {\Phi }_t^{-1}$. A change of variables givesThus the Lp conservation for incompressible transport isWithout that condition the requested conclusion is false: the force in part (c), together with any nonzero smooth integrable initial datum given by a Gaussian function, gives $\Vert f_t{\Vert }_p=e^{3t/p}\Vert f_0{\Vert }_p$. The essential supremum is still conserved by a complete invertible flow, because composition does not change the range of values.

Choose $f_0(x,v)=e^{-(x^2+v^2)}$ and $h(t,x,v)=x^2+v^2$. The initial Gaussian function is smooth and belongs to every finite Lp space, and is also bounded. The source is smooth and nonzero. The accumulated source along the backward characteristic curve isIts quadratic-form eigenvalues are $(e^{2t}-1)/2$ and $(1-e^{-2t})/2$. Both are positive for $t>0$, soConsequently $\Vert f_t{\Vert }_p=\infty$ for every $t>0$ and every finite $p\ge 1$; it is unbounded, so its $L^{\infty }$ norm is infinite as well. This spatially nonintegrable forcing in transport example avoids any ambiguity about whether the last endpoint is included in “all $p$”.

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.

All vector spaces in this solution are complex, while convex combinations use real coefficients. An extreme point $x$ of a convex set $C$ is one for which $x=ty+(1-t)z$, $y,z\in C$ and $0<t<1$, forces $y=z=x$. Equivalently, $x$ is not the midpoint of two distinct points of $C$.

The Krein-Milman theorem says that every nonempty compact convex set $K$ in a Hausdorff locally convex space is the closed convex hull of its extreme points. Here is a proof. A face of a convex set $F\subseteq K$ is a convex set such that whenever an interior point of a segment in $K$ belongs to $F$, both endpoints belong to $F$. Consider nonempty compact faces, ordered by reverse inclusion. A chain has nonempty intersection by compactness and the finite intersection property; that intersection is again a compact face. The Zorn lemma therefore gives a minimal nonempty compact face $F$.

If $F$ contains distinct $x,y$, a continuous real linear functional $\ell$ on the underlying real locally convex space distinguishes them. Such a linear functional exists because the space is Hausdorff and locally convex, by the Hahn-Banach theorem. The maximizers of $\ell$ on $F$ form a nonempty proper compact face of $F$. A face of a face is a face of $K$, contradicting minimality. Therefore $F$ is a singleton, yielding an extreme point. The same argument inside any nonempty compact face of $K$ supplies an extreme point of $K$ lying in that face.

Let $D$ be the closed convex hull of the extreme points of $K$. It is a nonempty closed subset of compact $K$, hence compact. If $x_0\in K\setminus D$, the Hahn-Banach separation theorem provides a continuous real linear functional $\ell$ with $\ell (x_0)>{\sup }_D\ell$. Its maximizer set on $K$ is a nonempty compact face, which contains an extreme point $e$ of $K$. But $e\in D$ and $\ell (e)\ge \ell (x_0)>{\sup }_D\ell$, a contradiction. Thus $\boxed{K=\overline{\mathrm{co}}(\mathrm{ext}K)}$, proving Krein-Milman theorem.

For a nonempty compact Hausdorff space $K$, the extreme points of the dual unit ball of C(K) areThis is the permitted description without proof, with $C(K)$ denoting the complex space of continuous functions on a compact space equipped with the supremum norm. If $K$ is empty, $C(K)=\{0\}$ and the closed unit ball of its continuous dual space has the single extreme point $0$ instead.

The complex Banach–Stone theorem states that a surjective complex-linear isometric isomorphism of normed spaces $T:C(K)\to C(L)$ has the formwhere $u\in C(L)$ and $\phi :L\to K$ is a homeomorphism. Conversely, every such map is a surjective complex-linear isometric isomorphism of normed spaces for the supremum norm.

To prove this, suppose first that $K,L$ are nonempty. The Banach-space adjoint $T^{\ast }$ is a bijective isometric isomorphism of normed spaces on the continuous dual spaces, so it bijects their closed unit balls and preserves extreme points. The displayed extreme point description gives uniquelyUniqueness follows by evaluating at the constant function $1$, and then using that continuous functions separate points of a compact Hausdorff space. Evaluation gives the required formula for $Tf$, and $u=T1$ is continuous. Surjectivity of $T^{\ast }$ on extreme points proves surjectivity of $\phi$: the preimage of any ${\delta }_x$ is $\alpha {\delta }_y$ for some $y$. If $\phi (y_1)=\phi (y_2)$, every function in the range of $T$ has equal values at these two points after division by $u$; surjectivity of $T$ and separation of points force $y_1=y_2$. Thus $\phi$ is bijective.

For every $f\in C(K)$, $f\circ \phi =(Tf)/u$ is continuous. The evaluation map $K\to {\mathbb{C}}^{C(K)}$, $x\mapsto (f(x){)}_f$, is a continuous injection of a compact Hausdorff space into a Hausdorff product topology, hence a homeomorphism onto its image. Continuity of every coordinate $f\circ \phi$ proves continuity of $\phi$. A continuous bijection between compact Hausdorff spaces is a homeomorphism. Conversely the weighted-composition formula plainly preserves the supremum norm, and its inverse isIf one compact space is empty, a surjective isometric isomorphism of normed spaces forces the other to be empty, and the empty homeomorphism gives the corresponding trivial case. This completes Banach–Stone theorem.

Neither the space of sequences converging to zero $c_0$ nor $L^1[0,1]$ can be isometrically a continuous dual space of a Banach space. Indeed, every nonzero continuous dual space has a nonempty weak-star compact closed unit ball by Banach-Alaoglu theorem, and Krein-Milman theorem then guarantees an extreme point. A bijective linear isometry preserves extreme points of closed unit balls.

The closed unit ball of $c_0$ has no extreme points. Given $x\in c_0$ with $\Vert x{\Vert }_{\infty }\le 1$, choose $n$ with $|x_n|<1$ and $0<\epsilon <1-|x_n|$. The two distinct elements $x\pm \epsilon e_n$ remain in that closed unit ball and have midpoint $x$.

The closed unit ball of the complex Lp space $L^1[0,1]$ also has no extreme points. An element of norm less than one can be perturbed by a sufficiently small nonzero Lp space element in both directions. If $\Vert f{\Vert }_1=1$, the non-atomic measure $|f(t)|dt$ admits a measurable set $A$ of mass $1/2$; for example the continuous function $s\mapsto {\int }_0^s|f(t)|dt$ attains $1/2$. Put $h=f(1_A-1_{A^c})$. For $0<\epsilon <1$, the distinct functions $f\pm \epsilon h$ both have Lp normwith the two coefficients interchanged for the minus sign. Their midpoint is $f$. Thus neither proposed space is isometrically a Banach dual.

Finally, $[0,1]$ is a connected space, whereas $[0,1]\cup [2,3]$ is disconnected. They cannot be homeomorphic. By Banach–Stone theorem, $C[0,1]$ and $C([0,1]\cup [2,3])$ are not isometrically isomorphic as complex Banach spaces.

For the globally invertible Hamiltonian flow from (a), the general characteristic formula isEvery $\Phi (s,t)$ preserves Lebesgue measure, by (b). Thus composition with it is an isometry of each Lp space. The Minkowski integral inequality givesApplying the Minkowski inequality also in time yields the finite-time Lp bound for Hamiltonian transport:The smoothness assumptions allow the characteristic construction; the norm estimate itself only uses the Lp space data and volume preservation.

For a concrete failure on infinite time, choose $\omega =1$, $f_0=0$, andThis is smooth, time independent and in every finite Lp space; it is invariant under the isotropic harmonic oscillator flow. Therefore the solution grows linearly:This invariant-source secular growth in Hamiltonian transport supplies the counterexample even with zero initial data.

On a bounded Lipschitz domain $\Omega \subset {\mathbb{R}}^d$, the Rellich-Kondrachov compactness theorem states that $W^{1,p}(\Omega )\hookrightarrow L^q(\Omega )$ is compact for $1\le p<\infty$ andwith $1/q=0$ for $q=\infty$. Thus $q<p^{\ast }$ when $p<d$, every finite $q$ is allowed when $p=d$, and the embedding into continuous functions with their uniform norm is compact when $p>d$.

For the main proof, a Sobolev extension operator puts a bounded sequence into a common compactly supported region of ${\mathbb{R}}^d$. The Sobolev fundamental theorem of calculus on lines gives the uniform translation estimate $\Vert u(\cdot +h)-u{\Vert }_p\le |h|\Vert \nabla u{\Vert }_p$. Convolution with a mollifier therefore approximates the sequence uniformly in Lp space. At any fixed smoothing scale, its derivatives and supremum are uniformly bounded, so the Arzelà-Ascoli theorem gives a convergent subsequence. A diagonal choice and the uniform approximation give strong Lp space convergence. The finite measure of $\Omega$ handles $q<p$, while Lp interpolation inequality with the bounded Sobolev embeddings upgrades this to the larger subcritical finite Lp spaces; for $p>d$, Morrey's inequality gives uniform equicontinuity directly and the Arzelà-Ascoli theorem gives uniform convergence. Boundedness of the domain is essential.

A quasi-norm on a vector space is nonnegative, vanishes only at zero, is absolutely homogeneous, and satisfies a triangle inequality with one constant factor $C\ge 1$. For $0<p<1$, $(\int |f{|}^p{)}^{1/p}$ on an Lp space is a quasi-norm: $\Vert f+g{\Vert }_p^p\le \Vert f{\Vert }_p^p+\Vert g{\Vert }_p^p$ gives a quasi-triangle inequality. Measure-preserving composition preserves this quantity even though the Minkowski inequality is unavailable below exponent one.

On $[0,1]$, the continuous functions $f_n(x)=\sin (nx)$ do not form a Cauchy sequence in the Lp space with $p=1$. Put $d_n=f_n-f_{2n}$. The identity ${\int }_0^1{d}_n^{2}dx=1-\sin (2n)/(4n)-\sin (4n)/(8n)-\sin n/n+\sin (3n)/(3n)$ tends to $1$. Since $|d_n|\le 2$, one has ${\int }_0^1|d_n|dx\ge \frac{1}{2}{\int }_0^1{d}_n^{2}dx$, which stays bounded away from zero. Thus arbitrarily late pairs are separated in the integral norm; in particular the sequence cannot converge in that norm.