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For a bilinear function $T$, it can be shown that $\lVert T(x,y)\rVert\leq C \lVert x\rVert \lVert y \rVert$

I saw some books say a bilinear function T is Lipschitz with Lipschitz constant $C$ given the above inequality holds.

Now I'm confused because a function T is Lipschitz if $\lVert T(\alpha)-T(\beta)\rVert \leq C \lVert \alpha - \beta \rVert$. So for a bilinear function $T:R^2 \rightarrow R$ to be Lipschitz, it means $\lVert T(x_1,y_1)- T(x_2,y_2)\rVert \leq C \lVert \langle x_1-x_2,y_1-y_2\rangle \rVert$. I can't figure out how this inequality is equivalent to the above one.

  • 1
    possible duplicate of [Operator norm on product space](http://math.stackexchange.com/questions/46355/operator-norm-on-product-space)2011-07-18

2 Answers 2