Why is $L^p$ isomorphic to $(L^p)^2$

Question

Is it possible to say why the spaces in the title are isomorphic as Banach spaces? Is their a Theorem that says this or is it even possible to find an explicit representation of this isomorphism?

Answer 1

With the understanding that we are talking about $E=L^p([0,1],\lambda)$, we can equip the Cartesian product $F = E\times E$ with the norm $$\Vert (f,g)\Vert_F=\root{p}\of{\frac{\Vert f\Vert_p^p+\Vert g\Vert_p^p}{2}}$$ With this notation the operator $$ \Phi:E\to F,\Phi(f)(x)=\left(f\left(\frac{x}{2}\right),f\left(\frac{x+1}{2}\right)\right) $$ Is an isometry.

Answer 2

For a general measure space $(\Omega,\mu)$, $L^p=L^p(\Omega,\mu)$ is not isomorphic to $(L^p)^2$. For example, consider $\Omega=\{ 0,1\}$ with the counting measure. Then $L^p$ has dimension $2$, but $(L^p)^2$ has dimension 4, so they cannot be isomorphic.

If $L^p$ is $L^p(I)$ for some nontrivial interval $I\subset \mathbb R$ (and Lebesgue measure), then there are very simple isometric isomorphisms between $L^p$ and the "$\ell^p$ direct sum" of $L^p$ with itself, i.e. $L^p\times L^p$ with norm $$\Vert (u,v)\Vert=\left( \Vert u\Vert_p^p+\Vert v\Vert_p^p\right)^{1/p}. $$

Write $I=I_1\cup I_2$ where $I_1$ and $I_2$ are nontrivial disjoint intervals. Then $L^p(I_1)$ and $L^p(I_2)$ are both isometric with $L^p=L^p(I)$, and the map $J=L^p(I)\to L^p(I_1)\times L^p(I_2)$ defined by $$J(f)=(f_{|I_1}, f_{|I_2}) $$ is an isometric isomorphism from $L^p$ onto the "$\ell^p$ direct sum" $L^{p}(I_1)\times L^p(I_2)$, which is isometric to the "$\ell^p$ direct sum" $L^p\times L^p$.

If $L^p$ is $\ell^p(\mathbb N)$, this (i.e. $L^p$ is isomorphic to $L^p\times L^p$) is also true. In fact, the result is true for $L^p(\Omega,\mu)$ as soon as one can write $\Omega=\Omega_1\cup \Omega_2)$, where $\Omega_1,\Omega_2$ are disjoint measurable sets such that $L^p(\Omega_i, \mu_{|\Omega_i})$ is isomorphic to $L^p(\Omega,\mu)$ for $i=1,2$.