Banach spaces not isomorphic to $\ell^p(S)$?

Question

We know that every Hilbert space is unitarily equivalent to $\ell^2(S)$, for a set $S$ of suitable cardinality.

Is there a Banach space which is NOT isomorphic to $L^p(X)$, for any $1\leq p \leq \infty$, and any measure space $X$?

Answer 1

Note the following three facts

• $c_0$ is not a dual space, because otherwise it must be complemented in its second dual, but it is not.

• $L_p(X)$ is dual for $1<p<\infty$ because $L_p(X)=L_\frac{p}{p-1}(X)^*$

• $L_\infty(X)$ is dual because $L_\infty(X)=L_1(X)^*$

So $c_0$ is not isomorphic to $L_p(X)$ for any measure space $X$ and $1< p\leq\infty$. It is remains to handle the case $p=1$.

Note another two facts

• $c_0^*=\ell_1$, and $\ell_1$ is infinite dimensional and separable

• $L_1(X)^*=L_\infty(X)$ and $L_\infty(X)$ is infinite dimensional and nonseparable.

Thus duals of $c_0$ and $L_1(X)$ can't be isomorphic, and a fortiori neither does $c_0$ and $L_1(X)$.

P.S.

Almost always there is no simple way to show that two Banach spaces are not isomorphic. Deal with it.