> [!theorem] > > Let $\seqf{E_j}$ and $F$ [[Banach Space|Banach spaces]]. A [[Multilinear Map|multilinear]] map $T: \prod_j E_j \to F$ is **bounded** if there exists $C \ge 0$ such that > $ > \norm{Tx} \le C\prod_j \norm{x_j} > $ > for all $x \in \prod_j E_j$. > [!theorem] > > Let $\seqf{E_j}$ and $F$ be Banach spaces, and $T: \prod_j E_j \to F$ be a bounded multilinear map. Define > $ > Lo: L^n\paren{\prod_j E_j, F} \to [0, \infty) \quad Lo(T) = \inf_{\norm{x_j} = 1 \forall j}\norm{Tx} > $ > then $Lo$ is continuous. > > *Proof*. Let $S \in L^n$, then > $ > \begin{align*} > \norm{(T + S)x} &\ge \norm{Tx} - \norm{Sx} \ge \norm{Tx} - \norm{S} \\ > &\le \norm{Tx} - \norm{Sx} \le \norm{Tx} + \norm{S} > \end{align*} > $ > so $Lo$ is continuous. > [!theorem] > > Let $\cx, \cy, \mathcal{Z}$ be Banach spaces and $B: \cx \times \cy \to \mathcal{Z}$ be a separately continuous bilinear map: $B(\cdot, y) \in L(\cx, \mathcal{Z})$ is [[Bounded Linear Map|continuous]] for all $y \in \cy$ and $B(x, \cdot) \in L(\cy, \mathcal{Z})$ for all $x \in \cx$. > > Then $B \in L(\cx \times \cy, \mathcal{Z})$ is jointly continuous, that is, > $ > B(x, y) \le C \cdot \norm{x} \cdot \norm{y} > $ > for some $C \in \real$. > > *Proof*. Use uniform boundedness on one coordinate with respect to the other. > > Let $T_y \in L(\cx, \mathcal{Z})$ with $x \mapsto B(x, y)$, and consider > $ > \alg = \bracs{T_y: y \in \cy, \norm{y} = 1} > $ > For any $x \in \cx$, > $ > \norm{T_y(x)} = \norm{B(y, x)} \le \norm{B(\cdot, x)} < \infty > $ > By the [[Uniform Boundedness Principle]], $M = \sup_{\norm{y} = 1}\norm{T_y} < \infty$. Therefore > $ > \begin{align*} > \norm{B(x, y)} &= \norm{B(\cdot, y)(x)} \\ > &\le \norm{B(\cdot, y)} \cdot \norm{x} \\ > &= \norm{T_y} \cdot \norm{x} \\ > &\le M \cdot \norm{y} \cdot \norm{x} > \end{align*} > $ > where $\norm{T_y} = \norm{T_{y/\norm{y}}} \cdot \norm{y}$ for $y \ne 0$. > >