The inner product is
\begin{equation*} \langle u, v \rangle = \sum\limits_{j \in J} u_{j} \overline{v_{j}} \end{equation*} where $u,v$ are vectors and $J$ is the countably infinite set $J = \mathbb{N}$.
New try for $J = \mathbb{N}$
For $k,t \in \mathbb{N}$, consider $e_{k}, e_{t} \in l_{p}$. By the parallegrom law, we have for $1 \leq p < \infty$ \begin{equation*} \| e_{k} + e_{t} \|_{p}^{2} + \| e_{k} - e_{t} \|_{p}^{2} = 2 \Big( \| e_{k} \|_{p}^{2} + \| e_{t} \|_{p}^{2} \Big), \end{equation*} so $2^{2/p} + 2^{2/p} = 2^{2}$, $2^{1+2/p} = 2^{2}$, i.e. $p = 2$. So $(l^{2}, \| \cdot \|)$ is a Hilbert space. $\square$
Source
- Exercises in Functional Analysis, Constantin Costara and Dumitru Popa
I think the use of parallegrom theorem is convenient. One possibility is with Cauchy-Schwarz directly. Probably more rigorous proofs still?
[depreciated; wrong because $J = \mathbb{C}^{n}$ i.e. countably finite set]
Conjugate symmetric. The inner product is conjugate symmetric: \begin{equation*} (\forall u,v \in J) \, \langle u, v \rangle = \overline{v, u}. \end{equation*}
Linearity in one of the arguments. The inner product satisfies the properties where it is linear in the first variable:
- Positive-definite: $(\forall u \in J) \, \langle u, u \rangle \geq 0$; $\langle u, u \rangle = 0$ iff $u = 0.$
- Distributive law: $(\forall u,v,h \in J) \langle u + h, v \rangle = \langle u, v \rangle + \langle h, v \rangle$.
- Homegenous under scaling for each variable: $(\forall c \in \mathbb{C}) \, (\forall u,v \in J), \, \langle c u, v \rangle = c \langle u,v \rangle = \langle u, c v \rangle$.
Assume there is a bounded bilinear form $b$: \begin{equation*} \exists C \geq \quad b(u,v) \leq C \|u\| \|v\| \, \forall u,v \in J. \end{equation*} For all $u \in J$, $b(u, \cdot)$ defines a bounded linear functional $J$, so that the Riesz theorem implies $b(u,\cdot) = (w(u), \cdot)_{J}$ for a unique $w(u) \in J$. $u : w(J \to J')$ defines a bounded linear operator on $J$ where $J'$ is a space of continuous linear functionals. Riesz theorem says what it means for an operator to be self-adjoint and positive-definite, when the operator is bounded, symmetric and positive-definite. The mapping is also surjective because $\forall u \in J, \exists w(u) \in J'$ such that $u = w(u)$. If the mapping is surjective, it must a Hilbert space.
Note $b$ is symmetric iff $w$ is self-adjoint. Recall that $b$ is positive-definite, so $b$ is a weighted inner product. $b$ is positive-definite and thus a weighted inner product iff $w$ is positive definite. So the bilinear form is sesquilinear: \begin{equation*} (\forall c \in \mathbb{C}) \, (\forall u,v \in J) \, \overline{ \langle u,v \rangle } = \langle v,u \rangle \Rightarrow \langle u, c v \rangle = \bar{c} \langle u,v \rangle. \end{equation*} where the last three points imply that the inner product is conjugate symmetric in the second variable: \begin{equation*} (\forall \alpha, \beta \in \mathbb{C})\, (\forall u,v,h \in J) \, \langle h, \alpha u + \beta v \rangle = \bar{\alpha}\langle h, u \rangle + \bar{\beta} \langle h, v \rangle. \end{equation*}
The $l_{2}(J)$ is Hilbert space with the inner product. $\square$
Sources
- Hilbert spaces definition: 555notes5.ps_pages.pdf
- Hilbert Space and Completeness: lec_11.pdf
- Is every symmetric bilinear form on a Hilbert space a weighted inner product?
How can you show rigorously that $l_{2}(J)$ is Hilbert space for the countably infinite set $J$ with the inner product?