Proof that ℓp when p ≠ 2 is an Inner Product Space

# Proof that ℓp when p ≠ 2 is an Inner Product Space

Recall from The Parallelogram Identity for Inner Product Spaces that if $X$ is a normed linear space then $X$ if an inner product space if and only if the norm on $X$ satisfies the parallelogram identity. We will use this very important result to show that $\ell^p$ when $p \neq 2$ is not an inner product space.

 Proposition 1: If $p \neq 2$ then $\ell^p$ is not an inner product space.
• Proof: Suppose that $p \neq 2$ and that $\ell^p$ is an inner product space. The the norm on $\ell^p$ must satisfy the parallelogram identity, that is, for all $(h_n), (k_n) \in \ell^p$:
(1)
\begin{align} \| (h_n) + (k_n) \|_p^2 + \| (h_n) - (k_n) \|_p^2 = 2 \| (h_n) \|_p^2 + 2 \| (k_n) \|_p^2 \end{align}
• Take $(h_n) = (1, 0, 0, ...)$ and $(k_n) = (0, 1, 0, ...)$. Then:
(2)
\begin{align} \quad \| (h_n) + (k_n) \|_p^2 = \left ( |1| + |1| \right )^{2/p} = 2^{2/p} \\ \quad \| (h_n) - (k_n) \|_p^2 = \left ( |1| + |-1| \right )^{2/p} = 2^{2/p} \\ \quad 2 \| (h_n) \|_p^2 = 2 \left ( |1| \right )^{2/p} = 2 \\ \quad 2 \| (k_n) \|_p^2 = 2 \left ( |1| \right )^{2/p} = 2 \end{align}
• And the parallelogram identity implies that:
(3)
\begin{align} \quad 4 &= 2^{2/p} + 2^{2/p} \\ &= 2(2^{2/p}) \\ &= 2^{2/p + 1} \\ &= 4^{1/p + 1/2} \end{align}
• Hence $p = 2$, a contradiction. Therefore $\ell^p$ when $p \neq 2$ is not an inner product space. $\blacksquare$