# Show that if $T:E\times F\rightarrow G$ is a non-identically zero bilinear map, then there's two sequences with this property [closed]

Let $$E,F,G$$ be normed spaces, and let $$T:E\times F\rightarrow G$$ be a non-identically zero bilinear map, then there are two sequences $$(u_n)_{n\in\mathbb{N}}$$ and $$(v_n)_{n\in\mathbb{N}}$$ in $$E\times F$$, with $$\lim_{n\to \infty}\|u_n-v_n\|_{E\times F} = 0$$ and $$\lim_{n\to \infty}\|T(u_n)-T(v_n)\|_{G} > 0.$$

We are using the product norm: $$\|⋅\|_{ExF} = \max\{\|⋅\|_{E},\|⋅\|_{F}\}.$$

• What is $T(u_n)$ if $T$ is bilinear? Where is the other argument? Jun 18, 2021 at 1:58
• $u_n$ is in $E \times F$ so $u_n = (e_n,f_n)$ where $e_n\in E$ and $f_n\in F$ Jun 18, 2021 at 2:16

If $$T(e,f)\ne0$$ then define $$u_n = n (e,f)$$, $$v_n=(n+\frac1n)(e,f)$$.
This basically reduces everything to $$E=F=G=\mathbb R$$: the claim follows since $$\|u_n-v_n\| = \frac1n\|(e,f)\|$$ and $$\|T(u_n)-T(v_n)\|=\|T(e,f)\|\cdot (2 + \frac1{n^2})$$.