> [!definition] > > Let $E$ be a [[Banach Space|Banach space]]. Define the space $\symbi(E)$ as the set of all symmetric [[Bounded Multilinear Map|bounded bilinear]] maps from $E \to \real|\complex$. > [!definition] > > Let $E$ be a Banach space. $E$ is **self-dual** if there exists a bounded symmetric bilinear form > $ > \angles{\cdot, \cdot}: E^2 \to \real|\complex \quad (x, y) \mapsto \angles{x, y} > $ > such that the mapping $y \mapsto \angles{\cdot, y}$ is a [[Space of Toplinear Isomorphisms|toplinear isomorphism]]. > [!theorem] > > Let $E$ be a self-dual Banach space. An [[Bounded Linear Operator|bounded linear operator]] $T: E \to E$ is **symmetric** if > $ > \angles{Tx, y} = \angles{x, Ty} \quad \forall x, y \in E > $ > The symmetric linear maps form a [[Closed Set|closed]], and thus complete subspace of $L(E, E)$. > > *Proof*. Let $\seq{T_n}$ be a [[Cauchy Sequence|Cauchy sequence]] of symmetric bilinear maps such that $T_n \to T$, then > $ > \angles{Tx, y} =\limv{n}\angles{T_nx, y} = \limv{n}\angles{x, T_ny} = \angles{x, Ty} > $ > [!theorem] > > Let $E$ be a self-dual Banach space. If $\lambda \in \symbi(E)$, then there exists a bounded, symmetric linear [[Bounded Linear Map|operator]] $A: E \to E$ such that for any $x, y \in E$, > $ > \lambda(x, y) = \angles{Ax, y} = \angles{x, Ay} > $ > called the **associated operator** of $\lambda$. > > Thus there is a bijection between $\symbi(E)$ and the space of symmetric linear operators on $E$, which makes $\symbi(E)$ a Banach space with the [[Space of Bounded Linear Maps|operator norm]] as its norm. > > *Proof*. The mapping $y \to \lambda\paren{\cdot, y}$ is a bounded linear map. Composing this with the inverse of the dual representation yields the desired map.