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Let $E,F,G$ be normed spaces, and let $T:E\times F \rightarrow G$ be a bilinear map (not indentically zero). Show that T is not uniformly continuous. As usual, we are using the product norm $\lvert\lvert \cdot \rvert\rvert_{E\times F} = \textrm{max}\{\lvert\lvert \cdot \rvert\rvert_E, \lvert\lvert \cdot \rvert\rvert_F\}$.

I know this follows from the Hahn-Banach theorem, but it's an introduction to an undergrad class, so we are not allowed to use such results.

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  • $\begingroup$ But $T(x,y) = x+y$ is not a bilinear map. $T(x_1 + x_2, y) = x_1 + x_2 + y \neq (x_1+y)+(x_2+y) = T(x_1,y) + T(x_2,y)$ $\endgroup$ Commented Jun 11, 2021 at 20:41
  • $\begingroup$ Yes , you are right, Matheus, but still, the statement seems to be incorrect. $\endgroup$ Commented Jun 11, 2021 at 20:43
  • $\begingroup$ @azif00 How did you get this inequality? $\endgroup$ Commented Jun 11, 2021 at 20:45
  • $\begingroup$ @azif00 But this does not make it uniformly continuous since $||(x,y)||_{E \times F}$ cannot be uniformly approximated by $||x||_E \cdot ||y||_F$. For example $T(x,y) = xy$ is not unifromly continuous on $\mathbb{R}^2$. $\endgroup$ Commented Jun 11, 2021 at 21:16
  • $\begingroup$ @KeeperOfSecrets You're right, my mistake. Let me delete my comment. $\endgroup$
    – azif00
    Commented Jun 11, 2021 at 21:31

1 Answer 1

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We will first reduce the problem to a nonzero bilinear map from $\mathbb{R}^2$ to $\mathbb{R}$.

Take any $x \in E$ and $y \in F$ such that $T(x,y) = z$ for some $0 \neq z \in G$ and consider the subspaces $A = \operatorname{span} \{x\} \times \operatorname{span} \{y\}$ of $E \times F$, and $B = \operatorname{span} z$ of $G$. Now just identify $A$ with $\mathbb{R}^2$ and $B$ with $\mathbb{R}$ and consider the restriction of $T$ on $A$ with range in $B$. If we show that this restriction is not uniformly continuous, then so cannot be the original map $T$.

So it remains to show that any nonzero bilinear map $T$ from $\mathbb{R}^2$ to $\mathbb{R}$ is not uniformly continuous. Note that any such map is of the form $T(x,y) = c \cdot xy$ for some nonzero constant $c$ (just take $c = T(1,1)$). It is now an easy excercise to show that this $T$ is not unifromly continuous.

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