= Solution
If a <bilinear map> is continuous, <continuity> at $(0,0)$ gives $\delta>0$ such that $\|F(x,y)\|\leq1$ whenever $\|x\|,\|y\|<\delta$. For nonzero $x,y$, apply this to $\delta x/(2\|x\|)$ and $\delta y/(2\|y\|)$. Bilinearity gives
$$
\boxed{\|F(x,y)\|\leq\frac4{\delta^2}\|x\|\|y\|.}
$$
If either input is zero, the inequality follows from bilinearity directly.
Conversely, suppose the product bound holds with $C\geq0$. At a fixed point $(x,y)$,
$$
F(x+h,y+k)-F(x,y)=F(h,y)+F(x,k)+F(h,k).
$$
Its <norm> is at most $C(\|h\|\|y\|+\|x\|\|k\|+\|h\|\|k\|)$, which tends to zero as $(h,k)\to(0,0)$. Thus \b[the product bound is equivalent to joint <continuity>]. This is <boundedness criterion for a bilinear map>.
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