Let $F$ be a field. Then the Euclidean inner product on $F^n$ is defined as $\langle \mathbf{x},\mathbf{y}\rangle=x_1\overline{y_1}+\cdots+x_n\overline{y_n}$ for all $\mathbf{x}=(x_1,~x_2,~\cdots,~x_n),~\mathbf{y}=(y_1,~y_2,~\cdots,~y_n)\in F^n$. Then its induced norm, $\lVert x\rVert:=\sqrt{\langle\mathbf{x}, \mathbf{x}\rangle}$, is a $\ell^2$-norm on $F^n$. From this, I come up with a doubt that can it be possible that we define an inner product on $F^n$ such that its induced norm is $\ell^1$-norm, or $\ell^{\infty}$-norm, or even $\ell^{p}$-norm?
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5No. A norm in a normed space is induced by an inner product iff the parallelogram law holds. It is not hard to show that the parallelogram law fails for the $\ell^p$ norm except when $p=2$. – Ittay Weiss Oct 19 '17 at 15:31
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1This question has, in one form or another, been asked several times on this website. – Aweygan Oct 19 '17 at 15:40
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Or An example of a norm which can't be generated by an inner product – Oct 20 '17 at 19:02