For a bilinear map, boundedness is equivalent to joint continuity

Assume $B$ is bounded, with constant $C$ from [L1]. Bilinearity gives $B(x_0+h,y_0+k)-B(x_0,y_0)=B(h,k)+B(h,y_0)+B(x_0,k)$. If $\max \{\mathrm{\parallel }h\mathrm{\parallel },\mathrm{\parallel }k\mathrm{\parallel }\}<\delta \le 1$, then $\mathrm{\parallel }B(h,k)\mathrm{\parallel }\le C{\delta }^2$, $\mathrm{\parallel }B(h,y_0)\mathrm{\parallel }\le C\delta \mathrm{\parallel }{y}_0\mathrm{\parallel }$, and $\mathrm{\parallel }B(x_0,k)\mathrm{\parallel }\le C\mathrm{\parallel }{x}_0\mathrm{\parallel }\delta$. Hence

The implication $(3)\Rightarrow (2)$ is immediate by specializing the point of continuity to $(0,0)$.

Assume $B$ is continuous at $(0,0)$. Applying [L3] with $\epsilon =1$ gives $\delta >0$ such that $\max \{\mathrm{\parallel }u\mathrm{\parallel },\mathrm{\parallel }v\mathrm{\parallel }\}<\delta$ implies $\mathrm{\parallel }B(u,v)\mathrm{\parallel }<1$.

Given $\epsilon >0$, choose $\delta \le 1$ so that the bound in step 1.1 is below $\epsilon$. Then [L3] shows that $B$ is continuous at $(x_0,y_0)$. Since $(x_0,y_0)$ was arbitrary, $(1)\Rightarrow (3)$.

If $x=0$ or $y=0$, bilinearity gives $B(x,y)=0$. Otherwise put $u:={\displaystyle \frac{\delta }{2\mathrm{\parallel }x\mathrm{\parallel }}}x$ and $v:={\displaystyle \frac{\delta }{2\mathrm{\parallel }y\mathrm{\parallel }}}y$. Then $\max \{\mathrm{\parallel }u\mathrm{\parallel },\mathrm{\parallel }v\mathrm{\parallel }\}=\delta \mathrm{/}2<\delta$, so $\mathrm{\parallel }B(u,v)\mathrm{\parallel }<1$ by step 1.3. By bilinearity, $B(u,v)={\displaystyle \frac{{\delta }^2}{4\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }}}B(x,y)$, hence $\mathrm{\parallel }B(x,y)\mathrm{\parallel }<{\displaystyle \frac{4}{{\delta }^2}}\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$. Therefore $B$ is bounded.

Steps 2.1, 1.2, and 2.2 prove $(1)\Rightarrow (3)\Rightarrow (2)\Rightarrow (1)$, so the three conditions are equivalent.