If $1\leq p < \infty$ then show that $L^p([0,1])$ and $\ell_p$ are not topologically isomorphic

If $1\leq p < \infty$ then show that $L^p([0,1])$ and $\ell_p$ are not topologically isomorphic unless $p=2$.

Maybe I would have to use the Rademacher's functions.

Let me get rid of the cases $p = 1$ and $p=2$ first:

1. The space $\ell^1$ has the Schur property (every weakly convergent sequence is norm-convergent) while $L^1$ doesn't.
   [Alternatively, $\ell^1$ has the Radon–Nikodým property while $L^1$ doesn't, see also this thread]

2. The spaces $\ell^2$ and $L^2$ are isomorphic because they both are separable Hilbert spaces.

Now I think there's no way around discussing the cases $1 \lt p \lt 2$ and $2 \lt p \lt \infty$ separately.

I'll refer to some results in Albiac-Kalton, Topics in Banach space theory, Springer GTM 233, 2006.

It follows from Pitt's theorem⁽¹⁾ (Theorem 2.1.4, page 32) that every operator $\ell^2 \to \ell^p$ for $1 \lt p \lt 2$ is compact while Proposition 6.4.13 (page 155) shows that $\ell^2$ embeds isometrically in every $L^p$, $1\leq p \lt \infty$ (this isn't hard, it suffices to choose a sequence of independent normalized Gaussians on $[0,1]$).

This shows that $\ell^p$ and $L^p$ aren't isomorphic as Banach spaces if $1 \lt p \lt 2$ because $L^p$ admits a non-compact map from $\ell^2$ while $\ell^p$ doesn't.

The case $2 \lt p \lt \infty$ follows from this by duality: if $\ell^p$ and $L^p$ were isomorphic then their dual spaces would be isomophic and we've just excluded that.

⁽¹⁾ see also this note by Sylvain Delpech as well as this thread.

Later: Pitt's theorem also implies that $\ell^{p}$ and $\ell^{q}$ aren't isomorphic whenever $p \neq q$ and the above argument is easily adapted to show that $L^{p}$ and $\ell^{q}$ are only isomorphic if $p = q = 2$ (exercise). Moreover, $L^{p}$ and $L^{q}$ are non-isomorphic if $p \neq q$, see the MO thread Jonas linked to in a comment. To sum up:

The family $\ell^p, L^q$, $1 \leq p,q \lt \infty$ consists of pairwise non-isomorphic spaces, except for the obvious isomorphism between $\ell^2$ and $L^2$.