$\ell_{p}$ space is not Hilbert for any norm if $p\neq 2$

Question

My question is motivated by this one: $\ell_p$ is Hilbert space if and only if $p=2$

Maybe it is a simple thing or im just confused but, suppose we are given any norm in $\ell_{p}$ for $p\neq 2$. How to show that this norm does not come from an inner product?

Thanks

Sorry if I do not post the problem with clarity.

Edit: $\ell_{p}=\{(x_{1},x_{2},...\}:(\sum_{i=1}^{\infty}|x_{i}|^{p})^{\frac{1}{p}}<\infty\}$

So that's my space and it is a vector space. Suppose I define on this space a norm (any norm). How can I show that this norm does not come from a inner product if $p\neq 2$?

Answer 1

I think you can turn any separable Banach space $(X,\Vert\cdot\Vert)$ into a Hilbert space. It is known$^1$ that every separable Banach space has a linear basis of cardinality $\mathfrak{c}$. Hence there exists a bijective linear operator $T:X \to\ell_2$. Given this operator, we define a new norm on $X$ by equality $$ \Vert x\Vert_\bullet=\Vert T(x)\Vert_{\ell_2} $$ It is an easy exercise to check that $(X,\Vert\cdot\Vert_\bullet)$ is a Hilbert space.

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$^1$Lacey, H. (1973). The Hamel dimension of any infinite-dimensional separable Banach space is c, Amer. Math. Montly, 80, 298