Proof that ℓp when p ≠ 2 is NOT a Hilbert Space
Proof that ℓp when p ≠ 2 is NOT a Hilbert Space
| Proposition 1: $\ell^p$ with $p \neq 2$ is not a Hilbert space. |
- Proof: On the Proof that ℓp when p ≠ 2 is an Inner Product Space page we saw that if $p \neq 2$ then $\ell^p$ is not an inner product space and hence it cannot be a Hilbert space either. $\blacksquare$
Recall from The Best Approximation Theorem for Hilbert Spaces page that if $H$ is a Hilbert space and $K$ is a nonempty, closed, convex subset of $H$ then if $h_0 \in H \setminus K$ there exists a unique $k_* \in K$ such that:
(1)\begin{align} \quad \| h_0 - k_* \| = \inf_{k \in K} \| h_0 - k_* \| \end{align}
We can use this theorem to prove that $\ell^{\infty}$ is NOT Hilbert spaces.
| Proposition 2: $\ell^{\infty}$ is not a Hilbert space. |
- Proof: Let $K = \{ x \in \ell^{\infty} : \| x \|_{\infty} \leq 1 \}$ be the closed unit ball in $\ell^{\infty}$. This set is nonempty, closed, and convex. Let $(h_n) = (2, 0, 0, ...)$. Then:
\begin{align} \quad \| (h_n) \|_{\infty} = \sup_{n \geq 1} |h_n| = 2 \end{align}
- Thus $(h_n) \in \ell^{\infty} \setminus K$.
- Note that:
\begin{align} \quad \inf_{(k_n) \in K} \| (h_n) - (k_n) \|_{\infty} = 1 \end{align}
- Let $(k_n) = (1, 1, 0, 0, ...)$ and let $(k_n') = (1, 1, 1, 0, 0, ...)$. Then $\| (k_n) \|_{\infty} = 1$ and $\| (k_n') \|_{\infty} = 1$, so $(k_n), (k_n') \in K$. Note that:
\begin{align} \quad \| (h_n) - (k_n) \|_{\infty} = 1 \end{align}
(5)
\begin{align} \quad \| (h_n) - (k_n') \|_{\infty} = 1 \end{align}
- So there is NOT a unique point in $K$ for which the distance between $(h_n)$ and that point is equal to the distance between $(h_n)$ and $K$. Thus $\ell^{\infty}$ is not a Hilbert space. $\blacksquare$
