Show that a (not identically zero) billinear map $T: E \times F \rightarrow G$ is not uniformly continuous

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.

• But $T(x,y) = x+y$ is not a bilinear map. $T(x_1 + x_2, y) = x_1 + x_2 + y \neq (x_1+y)+(x_2+y) = T(x_1,y) + T(x_2,y)$ Commented Jun 11, 2021 at 20:41
• Yes , you are right, Matheus, but still, the statement seems to be incorrect. Commented Jun 11, 2021 at 20:43
• @azif00 How did you get this inequality? Commented Jun 11, 2021 at 20:45
• @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 $\mathbb{R}^2$. Commented Jun 11, 2021 at 21:16
• @KeeperOfSecrets You're right, my mistake. Let me delete my comment. Commented Jun 11, 2021 at 21:31

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.