L^p space as functor

 

$L^p$ space as a functorFrom $\mathcal M_1$ To $\mathrm{Ban}_{\mathrm{1}}$ From $\mathcal M$ to $\mathrm{Ban}$$\mathcal M$ is cocomplete$L^p$ functor maps colimit to limit.A short exact sequence.The deep structure hidden under the short exaxt sequence

$L^p$$L^p$ space as a functor

From ${\mathcal{M}}_1$$\mathcal M_1$ To ${\mathrm{Ban}}_1$$\mathrm{Ban}_{\mathrm{1}}$

Definition.

Let ${\mathcal{M}}_1$$\mathcal M_1$ be the sub category of measure space. The object is all the measure space $(M,\mathrm{\Sigma },\mu )$$(M,\Sigma,\mu)$.

The morphism is all the measurable function $f:(M,\mathrm{\Sigma },{\mu }_M)\to (N,\mathrm{\Omega },{\mu }_N)$$f:(M,\Sigma,\mu_M)\to(N,\Omega,\mu_N)$ such that

$$\mathrm{\forall }X\in \mathrm{\Omega },f_{\ast }{\mu }_M(X)={\mu }_M(f^{-1}(X))\le {\mu }_N(X)$$

Definition.

Let $(M,\mathrm{\Sigma },\mu )$$(M,\Sigma,\mu)$ be a measure space and $1\le p\le \mathrm{\infty }$$1\le p\le\infty$. When $p\ne \mathrm{\infty }$$p\ne\infty$, we define $L^p(M,\mathrm{\Sigma },\mu )$$L^p(M,\Sigma,\mu)$ to be the set of all the measurable functions from $M$$M$ to $\mathbb{K}$$\mathbb K$ (here $\mathbb{K}$$\mathbb K$ is real or complex number) such that:

$$L^p(M,\mathrm{\Sigma },{\mu }_M):=\{f:M\to \mathbb{K}|f\text{ is measurable },\parallel f{\parallel }_p<\mathrm{\infty }\}/K_M$$

Where $K_M$$K_M$ is the space of measurable function equal to $0$$0$ almost everywhere.

Proposition. $L^p(-):{\mathcal{M}}_1^{\mathrm{op}}\to {\mathrm{Ban}}_1$$L^p(-):\mathcal M_1^{\mathrm{op}}\to\mathrm{Ban}_{\mathrm{1}}$ is a functor.

Proof. Let $f:(M,\mathrm{\Sigma },{\mu }_M)\to (N,\mathrm{\Omega },{\mu }_N)$$f:(M,\Sigma,\mu_M)\to(N,\Omega,\mu_N)$ be a morphism in ${\mathcal{M}}_1$$\mathcal M_1$.

$$L^p(f)=f^{\ast }:L^p(N)\to L^p(M),h⟼h\circ f$$

First we prove $f^{\ast }$$f^*$ is well defined, then we check it is a contraction linear map.

To see $f^{\ast }$$f^*$ is well defined, we need to prove that $f^{\ast }(K_N)\subseteq K_M$$f^*(K_N)\subseteq K_M$.

Notice that $h\in K_N⟺X_h=\{x\in N:h(x)\ne 0\},{\mu }_N(X_h)=0$$h\in K_N\iff X_h=\{x\in N:h(x)\ne 0\},\mu_N(X_h)=0$, and $X_{h\circ f}=\{y\in M:h\circ f(y)\ne 0\}=f^{-1}(X_h)$$X_{h\circ f}=\{y\in M:h\circ f(y)\ne 0\}=f^{-1}(X_h)$

Then $f^{\ast }(h)\in K_M$$f^*(h)\in K_M$ follows from ${\mu }_M(X_{f^{\ast }h})={\mu }_M(f^{-1}(X_h))\le {\mu }_N(X_h)=0$$\mu_M(X_{f^*h})=\mu_M(f^{-1}(X_h))\le \mu_N(X_h)=0$.

Easy to see it is linear, and for contraction, if $p<\mathrm{\infty }$$p<\infty$, it is follows from the change of varible.

$${\int }_M|h(f(x)){|}^{p}d{\mu }_M={\int }_N|h(y){|}^{p}d(f_{\ast }{\mu }_M)\le {\int }_N|h(y){|}^{p}d{\mu }_N$$

 

For $p=\mathrm{\infty }$$p=\infty$, let's recall the definitions:

• For $(N,\mathrm{\Omega },{\mu }_N)$$(N,\Omega,\mu_N)$, $\parallel h{\parallel }_{L^{\mathrm{\infty }}(N)}=inf\{C\ge 0:|h(y)|\le C\text{ for }{\mu }_N\text{-a.e. }y\in N\}.$$\|h\|_{L^\infty(N)} = \inf\bigl\{C\ge0 : |h(y)|\le C \text{ for }\mu_N\text{-a.e.\ }y\in N\bigr\}.$

• For $(M,\mathrm{\Sigma },{\mu }_M)$$(M,\Sigma,\mu_M)$, $\parallel h\circ f{\parallel }_{L^{\mathrm{\infty }}(M)}=inf\{C\ge 0:|h(f(x))|\le C\text{ for }{\mu }_M\text{-a.e. }x\in M\}.$$\|h\circ f\|_{L^\infty(M)} = \inf\bigl\{C\ge0 : |h(f(x))|\le C \text{ for }\mu_M\text{-a.e.\ }x\in M\bigr\}.$

Fix a constant $C$$C$ and let $E_h=\{x\in N:h(x)>C\}$$E_h=\{x\in N:h(x)>C\}$, then $E_{h\circ f}=\{x\in M:h\circ f(x)>C\}=f^{-1}(E_h)$$E_{h\circ f}=\{x\in M:h\circ f(x)> C\}=f^{-1}(E_h)$.

If ${\mu }_N(E_h)=0$$\mu_N(E_h)=0$, then ${\mu }_M(E_{h\circ f})={\mu }_M(f^{-1}(E_h))\le {\mu }_N(E_h)=0$$\mu_M(E_{h\circ f})=\mu_M(f^{-1}(E_h))\le\mu_N(E_h)=0$. Hence if $|h(y)|\le C$$|h(y)|\le C$ almost everywhere, then $|h\circ f(x)|\le C$$|h\circ f(x)|\le C$ almost everywhere. Hence $\parallel h{\parallel }_N\ge \parallel h\circ f{\parallel }_M$$\|h\|_N\ge\|h\circ f\|_M$.

Hence $\parallel f^{\ast }\parallel \le 1$$\|f^*\|\le 1$ is a contraction map. Easy to see that $L^p(\mathrm{id})=\mathrm{id}$$L^p(\mathrm{id})=\mathrm{id}$ and $L^p(f\circ g)=L^p(g)\circ L^p(f)$$L^p(f\circ g)=L^{p}(g)\circ L^p(f)$. Hence $L^p$$L^p$ is a functor. $◻$$\square$

Remark. If ${\mu }_N={\mu }_M\circ f^{-1}$$\mu_N=\mu_M\circ f^{-1}$, then $f^{\ast }$$f^*$ is an isometry.

Well, ${\mathcal{M}}_1$$\mathcal M_1$ is not a good category, for example, we could not define coproduct. So Let us define a new category.

From $\mathcal{M}$$\mathcal M$ to $\mathrm{Ban}$$\mathrm{Ban}$

Definition.

Let $\mathcal{M}$$\mathcal M$ be the sub category of measure space. The object is all the measure space $(M,\mathrm{\Sigma },\mu )$$(M,\Sigma,\mu)$.

The morphism is all the measurable function $f:(M,\mathrm{\Sigma },{\mu }_M)\to (N,\mathrm{\Omega },{\mu }_N)$$f:(M,\Sigma,\mu_M)\to(N,\Omega,\mu_N)$ such that

$$\mathrm{\forall }X\in \mathrm{\Omega },f_{\ast }{\mu }_M(X)={\mu }_M(f^{-1}(X))\le K{\mu }_N(X)$$

Where $K\in (0,\mathrm{\infty })$$K\in(0,\infty)$ is a constant.

Proposition. $L^p(-)$$L^p(-)$ is a functor from $\mathcal{M}$$\mathcal M$ to $\mathrm{Ban}$$\mathrm{Ban}$.

First we prove $f^{\ast }$$f^*$ is well defined, then we check it is a bounded linear map.

To see $f^{\ast }$$f^*$ is well defined, we need to prove that $f^{\ast }(K_N)\subseteq K_M$$f^*(K_N)\subseteq K_M$.

Notice that $h\in K_N⟺X_h=\{x\in N:h(x)\ne 0\},{\mu }_N(X_h)=0$$h\in K_N\iff X_h=\{x\in N:h(x)\ne 0\},\mu_N(X_h)=0$, and $X_{h\circ f}=\{y\in M:h\circ f(y)\ne 0\}=f^{-1}(X_h)$$X_{h\circ f}=\{y\in M:h\circ f(y)\ne 0\}=f^{-1}(X_h)$

Then $f^{\ast }(h)\in K_M$$f^*(h)\in K_M$ follows from ${\mu }_M(X_{f^{\ast }h})={\mu }_M(f^{-1}(X_h))\le K{\mu }_N(X_h)=0$$\mu_M(X_{f^*h})=\mu_M(f^{-1}(X_h))\le K\mu_N(X_h)=0$.

Easy to see it is linear, and for bounded property , if $p<\mathrm{\infty }$$p<\infty$, it is follows from the change of varible.

$${\int }_M|h(f(x)){|}^{p}d{\mu }_M={\int }_N|h(y){|}^{p}d(f_{\ast }{\mu }_M)\le K{\int }_N|h(y){|}^{p}d{\mu }_N$$

 

For $p=\mathrm{\infty }$$p=\infty$, let's recall the definitions:

• For $(N,\mathrm{\Omega },{\mu }_N)$$(N,\Omega,\mu_N)$, $\parallel h{\parallel }_{L^{\mathrm{\infty }}(N)}=inf\{C\ge 0:|h(y)|\le C\text{ for }{\mu }_N\text{-a.e. }y\in N\}.$$\|h\|_{L^\infty(N)} = \inf\bigl\{C\ge0 : |h(y)|\le C \text{ for }\mu_N\text{-a.e.\ }y\in N\bigr\}.$

• For $(M,\mathrm{\Sigma },{\mu }_M)$$(M,\Sigma,\mu_M)$, $\parallel h\circ f{\parallel }_{L^{\mathrm{\infty }}(M)}=inf\{C\ge 0:|h(f(x))|\le C\text{ for }{\mu }_M\text{-a.e. }x\in M\}.$$\|h\circ f\|_{L^\infty(M)} = \inf\bigl\{C\ge0 : |h(f(x))|\le C \text{ for }\mu_M\text{-a.e.\ }x\in M\bigr\}.$

Fix a constant $C$$C$ and let $E_h=\{x\in N:h(x)>C\}$$E_h=\{x\in N:h(x)>C\}$, then $E_{h\circ f}=\{x\in M:h\circ f(x)>C\}=f^{-1}(E_h)$$E_{h\circ f}=\{x\in M:h\circ f(x)> C\}=f^{-1}(E_h)$.

If ${\mu }_N(E_h)=0$$\mu_N(E_h)=0$, then ${\mu }_M(E_{h\circ f})={\mu }_M(f^{-1}(E_h))\le K{\mu }_N(E_h)=0$$\mu_M(E_{h\circ f})=\mu_M(f^{-1}(E_h))\le K\mu_N(E_h)=0$. Hence if $|h(y)|\le C$$|h(y)|\le C$ almost everywhere, then $|h\circ f(x)|\le C$$|h\circ f(x)|\le C$ almost everywhere. Hence $\parallel h{\parallel }_N\ge \parallel h\circ f{\parallel }_M$$\|h\|_N\ge\|h\circ f\|_M$.

Hence indeed, $L^{\mathrm{\infty }}(f)$$L^\infty(f)$ is not only bounded but also contraction map! Therefore, $L^p:\mathcal{M}\to \mathrm{Ban}$$L^p:\mathcal M\to\mathrm{Ban}$ for $p<\mathrm{\infty }$$p<\infty$, and $L^{\mathrm{\infty }}:\mathcal{M}\to {\mathrm{Ban}}_1$$L^{\infty}:\mathcal M\to\mathrm{Ban}_1$. Easy to see that $L^p(\mathrm{id})=\mathrm{id}$$L^p(\mathrm{id})=\mathrm{id}$ and $L^p(f\circ g)=L^p(g)\circ L^p(f)$$L^p(f\circ g)=L^{p}(g)\circ L^p(f)$. Hence $L^p$$L^p$ is a functor. $◻$$\square$

$\mathcal{M}$$\mathcal M$ is cocomplete

To prove a category is cocomplete, we only need to prove the existence of coproduction and coequalizer.

Lemma. The coproduct exists in $\mathcal{M}$$\mathcal M$.

Proof. First let us prove that the empty coproduct exits, i.e. the initial object exists.

Let us consider $(\mathrm{\varnothing },\mathrm{\varnothing },0)$$(\empty,\empty,0)$. Since $\mathrm{\varnothing }$$\empty$ is the initial object in $\mathrm{Set}$$\mathrm{Set}$, for any measure space $(M,\mathrm{\Sigma },\mu )$$(M,\Sigma,\mu)$ we have a unique $f:\mathrm{\varnothing }\to M$$f:\empty\to M$.

Easy to see it is measurable and $f_{\ast }0=0(f^{-1}(X))=0\le {\mu }_M(X)$$f_*0=0(f^{-1}(X))=0\le \mu_M(X)$. Hence it is the initial object.

Now let us prove the binary product exits. Let $(M,\mathrm{\Sigma },{\mu }_M),(N,\mathrm{\Omega },{\mu }_N)$$(M,\Sigma,\mu_M),(N,\Omega,\mu_N)$ be two measure spaces.

The coproduct of $M$$M$ and $N$$N$ is $(M\coprod N,\mathrm{\Sigma }+\mathrm{\Omega },{\mu }_{M\coprod N})$$(M\coprod N,\Sigma+\Omega,\mu_{M\coprod N})$, here $\mathrm{\Sigma }+\mathrm{\Omega }$$\Sigma+\Omega$ is the greatest sigma algebra make $i_M,i_N$$i_M,i_N$ measurable and

$${\mu }_{M\coprod N}(X)={\mu }_M(X\cap M)+{\mu }_N(X\cap N)$$

We prove it is the coproduct.

Easy to see that $i_M,i_N$$i_M,i_N$ are morphisms in $\mathcal{M}$$\mathcal M$.

Let $h_M:M\to Z,h_N:N\to Z$$h_M:M\to Z,h_N:N\to Z$ be two morphisms in $\mathcal{M}$$\mathcal M$, i.e. two measurable functions such that ${\mu }_M(h_M^{-1}(V))\le K{\mu }_Z(V)$$\mu_M(h_M^{-1}(V))\le K\mu_Z(V)$ and ${\mu }_N(h_N^{-1}(V))\le K^{\prime }{\mu }_Z(V)$$\mu_N(h_N^{-1}(V))\le K'\mu_Z(V)$.

then we have a unique function $g=h_M\coprod h_N:M\coprod N\to Z$$g=h_M\coprod h_N:M\coprod N\to Z$.

It is measurable, since $g^{-1}(V)=h_M^{-1}(V)\cup h_N^{-1}(V)$$g^{-1}(V)=h_M^{-1}(V)\cup h_N^{-1}(V)$.

Also, $g_{\ast }{\mu }_{M\coprod N}(V)={\mu }_M(h_M^{-1}(V))+{\mu }_N(h_N^{-1}(V))\le K{\mu }_Z(V)+K^{\prime }{\mu }_Z(V)=(K+K^{\prime }){\mu }_Z(V)$$g_*\mu_{M\coprod N}(V)=\mu_M(h_M^{-1}(V))+\mu_N(h_N^{-1}(V))\le K\mu_Z(V)+K'\mu_Z(V)=(K+K')\mu_Z(V)$, hence we verify that the coproduction exists. $◻$$\square$

Lemma. The coequalizer exists in $\mathcal{M}$$\mathcal M$.

Proof. Let $f,g:(M,\mathrm{\Sigma },{\mu }_M)\to (N,\mathrm{\Omega },{\mu }_N)$$f,g:(M,\Sigma,\mu_M)\to (N,\Omega,\mu_N)$ be two measurebale functions such that $f_{\ast }{\mu }_M\le K{\mu }_N,g_{\ast }{\mu }_M\le K^{\prime }{\mu }_N$$f_*\mu_M\le K\mu_N,g_*\mu_M\le K'\mu_N$.

We take $\pi :N\to Q$$\pi:N\to Q$ to be the coequalizer in $\mathrm{Set}$$\mathrm{Set}$, and take the sigma algebra on $Q$$Q$ to be ${\mathrm{\Sigma }}_Q=\{A\subset Q\mid {\pi }^{-1}(A)\in {\mathrm{\Sigma }}_N\}.$$\Sigma_Q =\bigl\{\,A\subset Q \mid \pi^{-1}(A)\in\Sigma_N\bigr\}.$

Let ${\mu }_Q:={\pi }_{\ast }{\mu }_N$$\mu_Q:=\pi_*\mu_N$. Let us check the universal property of $(Q,\mathrm{\Sigma },{\mu }_Q)$$(Q,\Sigma,\mu_Q)$. Let $h:N\to Z$$h:N\to Z$ be a morphism such that $h\circ f=h\circ g$$h\circ f=h\circ g$, then it is uniquely passing through $Q$$Q$ by the universal property in $\mathrm{Set}$$\mathrm{Set}$, $h=k\circ \pi$$h=k\circ\pi$. By the definition of ${\mathrm{\Sigma }}_Q$$\Sigma_Q$, $k$$k$ is measurable.

Now we need to check that $k_{\ast }{\mu }_Q\le K{\mu }_Z$$k_*\mu_Q\le K\mu_Z$. That is follows from

$${\mu }_Q(k^{-1}(X))={\mu }_N({\pi }^{-1}{k}^{-1}(X))={\mu }_N(h^{-1}(X))=h_{\ast }{\mu }_N(X)\le K{\mu }_Z(X)◻$$

Hence $\mathcal{M}$$\mathcal M$ is cocomplete.

$L^p$$L^p$ functor maps colimit to limit.

We only need to check that $L^p(-)$$L^p(-)$ maps coproduct to product and coequalizer to equalizer.

Lemma. $L^p(X\coprod Y)\cong L^p(X)\times L^p(Y)$$L^p(X\coprod Y)\cong L^p(X)\times L^p(Y)$.

Proof. For $f:X\to Z,g:Y\to Z$$f:X\to Z,g:Y\to Z$ and $X\stackrel{i_X}{\to }X\coprod Y\stackrel{i_Y}{\leftarrow }Y$$X\xrightarrow{ i_X} X\coprod Y\xleftarrow{i_Y} Y$, we have $L^p(f):L^p(Z)\to L^p(X),L^p(g):L^p(Z)\to L^p(Y)$$L^p(f):L^p(Z)\to L^p(X),L^p(g):L^p(Z)\to L^p(Y)$

and

$$L^p(X)\stackrel{L^p(i_X)}{\leftarrow }{L}^p(X\coprod Y)\stackrel{L^p(i_Y)}{\to }{L}^p(Y)$$

By universal property of product, we have $\mathrm{\Phi }:L^p(X\coprod Y)\to L^p(X)\times L^p(Y)$$\Phi:L^p\left (X\coprod Y\right)\to L^p(X)\times L^p(Y)$.

Such that $\mathrm{\Phi }(u)=(i_X^{\ast }(u),i_Y^{\ast }(u))$$\Phi(u)=(i_X^*(u),i_Y^*(u))$, the inverse map is defined by $\begin{array}{c}\mathrm{\Psi }(a,b)=\{\begin{array}{l}a(t):t\in X\\ b(t):t\in Y\end{array}\end{array}$$\Psi(a,b)=\begin{cases}a(t):t\in X\\b(t):t\in Y\end{cases}$ $◻$$\square$

Lemma. $L^p$$L^p$ functor maps coequalizer to equalizer.

Proof. Let $h:N\to Z$$h:N\to Z$ coequalize $f$$f$ and $g$$g$, then $f^{\ast }\circ h^{\ast }=g^{\ast }\circ h^{\ast }$$f^*\circ h^*=g^*\circ h^*$, i.e. $h^{\ast }$$h^*$ equalize $f^{\ast }$$f^*$ and $g^{\ast }$$g^*$. Now we need to prove that

$(L^p(Q),{\pi }^{\ast })$$(L^p(Q),\pi^*)$ is isomorphic to the equalizer.

By the universal property of equalizer, we have a unique map $u:L^p(Q)\to \mathrm{Eq}(f^{\ast },g^{\ast })$$u:L^p(Q)\to\mathrm{Eq}(f^*,g^*)$. If we take $\mathrm{Eq}(f^{\ast },g^{\ast })$$\mathrm{Eq}(f^*,g^*)$ to be the kernel of $f^{\ast }-g^{\ast }$$f^*-g^*$, then $u={\pi }^{\ast }$$u=\pi^*$.

Let $h\in L^p(N)$$h\in L^p(N)$ such that $f^{\ast }(h)=h\circ f=h\circ g=g^{\ast }(h)$$f^*(h)=h\circ f=h\circ g=g^*(h)$. Then $h=h^{\prime }\circ \pi ={\pi }^{\ast }(h^{\prime })$$h=h'\circ \pi=\pi^*(h')$ by the universal property of coequalizer.

Hence ${\pi }^{\ast }:L^p(Q)\to \ker (f^{\ast }-g^{\ast })$$\pi^*:L^p(Q)\to\mathrm{ker}(f^*-g^*)$ is surjective. $◻$$\square$

Therefore, $L^p$$L^p$ maps colimit to limit.

A short exact sequence.

Let $(M,\mathrm{\Sigma },{\mu }_M)$$(M,\Sigma,\mu_M)$ be a measure space, $X\in \mathrm{\Sigma }$$X\in\Sigma$ then we have the following split short exact sequence.

$$0\stackrel{}{\to }{L}^p(X)\stackrel{i}{\to }{L}^p(M)\cong L^p(X)\oplus L^p(X^c)\stackrel{\pi }{\to }{L}^p(X^c)⟶0$$

Since we have $M\cong X\coprod X^c$$M\cong X\coprod X^c$.

The deep structure hidden under the short exaxt sequence

Key observation. Notice that each sigma algebra is a Boolean ring $(\mathrm{\Sigma },\mathrm{\Delta },\cap )$$(\Sigma,\Delta,\cap)$, hence a ${\mathbb{F}}_2$$\mathbb F_2$ algebra, thus a ${\mathbb{F}}_2$$\mathbb F_2$ vector space.

We could define a functor $F:{\mathrm{Meas}}^{\mathrm{op}}\to {\mathbb{F}}_2$$F:\mathrm{Meas}^{\mathrm{op}}\to\mathbb F_2$-$\mathrm{Mod}$$\mathrm{Mod}$ as follows.

For a measurable space $(M,\mathrm{\Sigma })$$(M,\Sigma)$, $F(M,\mathrm{\Sigma })=V_{\mathrm{\Sigma }}=\mathrm{\Sigma }$$F(M,\Sigma)=V_\Sigma=\Sigma$. For a measurable function $f:(M,\mathrm{\Sigma })\to (N,\mathrm{\Omega })$$f:(M,\Sigma)\to (N,\Omega)$, we define $F(f)=f^{-1}$$F(f)=f^{-1}$.

Easy to see that $f^{-1}:V_{\mathrm{\Omega }}\to V_{\mathrm{\Sigma }}$$f^{-1}:V_\Omega\to V_\Sigma$ is a ${\mathbb{F}}_2$$\mathbb F_2$ linear map.

Let $X\in \mathrm{\Sigma }$$X\in\Sigma$, we could define a sub sigma algebra on $X$$X$, $V_X=\{A\in \mathrm{\Sigma }:A\subseteq X\}$$V_X=\{A\in\Sigma:A\subseteq X\}$.

The coproduct $X\bigsqcup X^c$$X\sqcup X^c$ correspond to $V_X\oplus V_Y$$V_X\oplus V_Y$.

We also have the short exact sequence

$$0⟶V_X\stackrel{i}{\to }{V}_M\cong V_X\oplus V_{X^c}\stackrel{\pi }{\to }{V}_{X^c}⟶0$$