If $f:\mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^p$ is a bilinear function, then how to show that

Question

If $f:\mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^p$ is a bilinear function, then how to show that $$\lim\limits_{(h,k) \to (0,0)} \dfrac{|f(h,k)|}{|(h,k)|} = 0$$.

Answer 1

The general form of a bilinear function $f:R^n\times R^m\to R^p$ is $$ f_j(h,k)=\sum_{s,l}a_{jsl}h_sk_l $$ (where $f_j$, $h_s$, and $k_l$ are the coordinates of $f$, $h$, and $k$ in the standard bases in the respective spaces and $a_{jsl}$ are some constants). From this form, the claim is obvious, because $$ |h_sk_l|\le C|h||l|\le C_1|(h,l)||(h,l)|=C_1|(h,l)|^2 $$ with some constants $C, C_1$ depending on the specific choice of the norms.

Answer 2

For simplicity, I'll be using the euclidean norm for $\Bbb R^n$, $\Bbb R^m$ and $\Bbb R^n\times\Bbb R^m$, and the $\sup$ norm for the target space $\Bbb R^p$. The choice of norms is irrelevant to the result since all norms are equivalent in finite dimension.

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There are coefficients $a_{ij;\,l}\in\Bbb R$, where $(i,j)$ ranges over $\lbrace 1,\dots,n\rbrace\times\lbrace 1,\dots,m\rbrace$ and $k\in\lbrace 1,\dots,p\rbrace$ such that for all $h=(h_1,\dots,h_n)\in\Bbb R^n,k=(k_1,\dots,k_m)\in\Bbb R^m$, the $l$-th coordinate of $f(h,k)$ is equal to $$\left[f(h,k)\right]_l=\sum_{i,j}a_{ij;\,l}h_ik_j$$ If you let $A$ be the maximum of the absolute values $|a_{ij;\,l}|$, then for all $l$, $$|\left[f(h,k)\right]_l|\leq A\sum_{i,j}|h_ik_j|\leq \frac12A\sum_{i,j}h_i^2+k_j^2=\frac12A(m|h|^2+n|k|^2)\leq\frac12A(m+n)|(h,k)|^2$$ Thus, for all $(h,k)\neq(0,0)$ $$\frac{|f(h,k)|}{|(h,k)|}\leq C|(h,k)|$$ for $C=\frac12A(m+n)$. The right hand side tends to zero as $(h,k)$ tends to zero.