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In a normed vector space I know that for a linear map $L:E\rightarrow F$ that there exists an $M\in \mathbb{R}$ such that $\forall x\in E$ $||L(x)||\leq M||x||$. The proof is this is quite straightforward but I am unsure how to generalize to a bilinear map.

Let $A:E\times F\rightarrow G$. I would like to show that $\exists N \in \mathbb{R}$ such that $\forall x\in E,y\in G$ that $||A(x)(y)||\leq N||x||||y||$. I am unsure if my generalization is permissible:

Attempt: $A(x):F\rightarrow G$, $A(x)$ linear, so $\exists B$ such that $||A(x)(y)||\leq B||y||$. Let $N=\frac{B}{||x||}$, it then follows that $||A(x,y)||\leq N||x||||y||$ but this isn't right because N shouldn't depend on $||x||$.

Edit: I guess i should say that I am considering only finite dimensional spaces.

Mael
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    Note: If $E$, $F$ are finite dimensional vector spaces, then also $E\times F$ is. – J.R. Jan 23 '14 at 17:52
  • My previous proof reads: $||L(x)||\leq ||\sum_{i=1}^{n}x_{i}L(e_{i})||$ $<\sum_{i=1}^{p} |x_{i}|M<nM||x||$ where $M=max(L(e_{1}),...)$. To extend this as you say I'd have something like: $<n_{x}M_{x}||x||+n_{y}M_{y}||y||$ – Mael Jan 23 '14 at 17:59

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