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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: $$ \|⋅\|_{ExF} = \max\{\|⋅\|_{E},\|⋅\|_{F}\}. $$

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

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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 $\|T(u_n)-T(v_n)\|=\|T(e,f)\|\cdot (2 + \frac1{n^2})$.

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