Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 4 / 22F / Solution

If a linear map satisfies $\Vert Tx\Vert \le M\Vert x\Vert$, then $\Vert Tx-T{x}^{\prime }\Vert \le M\Vert x-x^{\prime }\Vert$, so it is continuous. Conversely continuity at zero gives $\delta >0$ such that $\Vert x\Vert <\delta$ implies $\Vert Tx\Vert <1$. Apply this to $\delta x/(2\Vert x\Vert )$ for $x\ne 0$ to obtain $\Vert Tx\Vert \le 2\Vert x\Vert /\delta$. Hence a linear map is continuous exactly when it is bounded.

For the separately continuous bilinear map, put $T_x(y)=F(x,y)$ for $\Vert x\Vert \le 1$. Every $T_x:Y\to Z$ is a bounded linear map by the first result. For each fixed $y$, continuity in the other variable gives $\Vert F(x,y)\Vert \le C_y\Vert x\Vert$, so ${\sup }_{\Vert x\Vert \le 1}\Vert T_{x}y\Vert <\infty$. The Uniform boundedness principle therefore gives ${\sup }_{\Vert x\Vert \le 1}\Vert T_x\Vert =M<\infty$, and bilinearity then yields

To spell out the uniform-boundedness step, define the closed subsets $E_m=\{y:{\sup }_{\Vert x\Vert \le 1}\Vert T_{x}y\Vert \le m\}$ of the Banach space $Y$. They cover $Y$. The Baire category theorem gives an $E_m$ containing a ball $B(y_0,r)$. For $\Vert h\Vert <r$, both $y_0+h$ and $y_0$ lie in $E_m$, so ${\sup }_x\Vert T_{x}h\Vert \le 2m$. Scaling $h=(r/2)y$ for $\Vert y\Vert \le 1$ bounds all the operator norms by $4m/r$. This proves the claimed uniform estimate without presuming joint continuity.