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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.

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possible duplicate of Operator norm on product space –  t.b. Jul 18 '11 at 22:28
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2 Answers 2

At least from the usual Lipschitz condition we can derive that $\|T(x,y)\|\leq C\sqrt{\|x\|\|y\|}$. Indeed, since $T(x,y) = T(x,y) - T(x,0) = T(x,y) - T(0,y)$, assuming Lip. of $T$ we obtain two independent abounds: $\|T\|\leq C_1\|x\|$ and $\|T\|\leq C_2\|y\|$. May be helpful to you.

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This is clearly incorrect. For example: T(x,y) = xy is bilinear, and $|T(x,y)| \leq |x||y|$ However, it isn't Lipschiz 1, since for example |T(1,2)-T(2,2)|=2, but |(1,2)-(2,2)|=1

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Indeed, a more-than-one-linear function is never Lipschitz. –  scineram Sep 30 '11 at 8:57
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