Non-equivalence $p$-norm on $\ell_p$, $L_p(X, \mu)$ to any inner product norm Show that the norm on the spaces $\ell_p$, $L_p(X, \mu)$ (where $(X, \mu)$ is a measure space containing infinitely many disjoint measurable sets of positive measure) is not equivalent to a norm generated by an inner product (unless $p = 2$).
I have trouble with this exercise. I don't know what to start from. In previous exercises I showed that the norm on the spaces $(\Bbb C^n, ∥ · ∥_p)$, $\ell_p$, $(C [a, b], ∥ · ∥_p)$, $L_p (X, \mu)$ is not generated by an inner product (unless $n = 1$, $p = 2$). I also generalized the inequality of the parallelogram on $n$ vectors. I think it's necessary. It is possible to construct many different inner products, but how to show non-equivalence?
 A: Alright, I realised that the only relevant part of my "proof" was the last one where I dismissively say that $L_q \not \simeq L_p$. This statement is actually proved in the same way as the original problem. I will leave my original post below (the use of duality was neat, I think, and, perhaps, one day I will be able to complete it in a satisfactory way) but, without further ado,
Proof Number 1 (due to my friend)
Suppose $p < 2$, let $e_n$ be the standard basis in $l^p$ (the case of $L^p$ is similar), $(\cdot, \cdot)_H$ and $||\cdot||_H$ be the inner product and the corresponding norm on $l^p$. Since $||e_n||_p = 1$ and the norms are equivalent, there is $C > 0$ such that $||e_n||_H \le C$ for every $n \in \mathbb{N}$. By induction, choose $\epsilon_n \in \{-1, 1\}$ so that $$||\epsilon_1e_1 + \ldots + \epsilon_ne_n||_H \le C\sqrt{n}$$ (take $\epsilon_{n+1} = -\mathrm{sign}(\epsilon_1e_1 + \ldots + \epsilon_ne_n, e_{n+1})_H$). Then
$$\frac{||\epsilon_1e_1 + \ldots + \epsilon_ne_n||_H}{||\epsilon_1e_1 + \ldots + \epsilon_ne_n||_p} \le \frac{C\sqrt{n}}{n^{1/p}} \to 0$$
which is a contradiction. If $p>2$ do the same but take $\epsilon_{n+1} = \mathrm{sign}(\epsilon_1e_1 + \ldots + \epsilon_ne_n, e_{n+1})_H$ so that 
$$||\epsilon_1e_1 + \ldots + \epsilon_ne_n||_H \ge c\sqrt{n}.$$
Proof Number 2 (outline)
Let $P_m$ be the standard projection on the first $m$ coordinates in $l^p$ and let $v_n$ be any orthonormal basis in $l^p$ as a Hilbert space. Choose $\epsilon > 0$. Since $P_m(v) \to v$ there is a function $M_\epsilon : \mathbb{N} \to \mathbb{N}$ such that 
$$\frac{||P_{M_\epsilon(n)}v_n||_p}{||v_n||_p} > 1 - \epsilon.$$ If for some $\epsilon$ the function $M_\epsilon$ is bounded, we can find a subsequence of $\{v_{n_k}\}$ that is almost the same in the first $\max M_\epsilon(n)$ coordinates, so $||\sum v_{n_k}||_p \sim k$ as $k \to \infty$. But $||\sum v_{n_k}||_H \sim \sqrt{k}$. That means $M_\epsilon(n)$ is unbounded for every $\epsilon$ and for every $d \in \mathbb{N}$ you can find $d$ vectors $\{v_{n_k}\}$ such that their norms in $l^p$ (up to $c\epsilon$) is concentrated in different coordinates. Then $||\sum v_{n_k}||_p \sim k^{1/p}$ but $||\sum v_{n_k}||_H \sim \sqrt{k}$.

Suppose there is an equivalent norm $||\cdot||_H$ generated by in inner product on $L_p$. By definition,
$$C_1||\cdot||_H \le ||\cdot||_p \le C_2||\cdot||_H.$$
The topologies they generate are the same so their dual spaces are the same (as sets). Moreover, for $f \in (L_p)^*$
$$||f||_{(L_p, ||\cdot||_p)^*} = \sup\limits_{x \in L_p} \frac{|f(x)|}{||x||_p}$$
so
$$\frac{||f||_{(L_p, ||\cdot||_H)^*}}{C_2} \le ||f||_{(L_p, ||\cdot||_p)^*} \le  \frac{||f||_{(L_p, ||\cdot||_H)^*}}{C_1}.$$
Therefore the norms induced on $(L_p)^*$ are equivalent as well. Thus, 
$$(L_q, ||\cdot||_q) \simeq (L_p, ||\cdot||_p)^* \simeq (L_p, ||\cdot||_H)^* \simeq (L_p, ||\cdot||_H) \simeq (L_p, ||\cdot||_p)$$
where "$\simeq$" means isomorphic as Banach spaces up to equivalent norms.  It remains to prove there is no isomorphism of this kind between $L_p$ and $L_q$.
