Question about norm of bilinear form

Question

Let $\beta:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}$. In showing that $\frac{|\beta(x,y)|}{||(x,y)||}\to 0$ as $(x,y)\to (0,0)$, you use that $|\beta(x,y)|\leq ||\beta|| \cdot||x|| \cdot||y||\leq \beta(||x||+||y||)^2=||\beta||\cdot||(x,y)||^2$.

My question is, why can't we think of $\beta$ as a linear map from $\mathbb{R}^{n^2}$ to $\mathbb{R}$, setting $X=(x,y)$, so $\frac{|\beta(X)|}{||X||}\leq\frac{||\beta||\cdot||X||}{||X||}=\beta$, but this doesn't get you anywhere?

Answer 1

Well, $\beta$ is bilinear, not linear! Namely, we have in general

$$ \beta((x,y) + (z,w)) = \beta(x + z, y + w) = \beta(x,y) + \beta(x,w) + \beta(z,y) + \beta(z,w) \neq \beta(x,y) + \beta(z,w).$$

Answer 2

Let $\beta (x,y)$ be the dot product of and y. Then $\beta$ is bilinear but not linear: $\beta (2(x,y)) =\beta (2x,2y)=4 \beta (x,y)$, not $2 \beta(x,y)$.