Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 2 / 22F (Linear Analysis)

The Uniform boundedness principle states that a pointwise bounded family $\mathcal{T}$ of bounded linear maps from a Banach space $X$ to a normed space $Y$ is uniformly bounded in operator norm. Define closed sets $E_n=\{x:{\sup }_{T\in \mathcal{T}}\Vert Tx\Vert \le n\}$. Pointwise boundedness makes their union $X$. The Baire category theorem gives a ball $B(x_0,r)\subset E_n$ for some $n$. For $\Vert v\Vert <r$, both $x_0+v$ and $x_0$ lie in $E_n$, so $\Vert Tv\Vert \le 2n$ for every $T$. Scaling yields ${\sup }_T\Vert T\Vert \le 2n/r$, proving the principle.

For the separately continuous bilinear map, consider $T_x(y)=F(x,y)$ with $\Vert x\Vert \le 1$. Each $T_x$ is bounded and linear. For fixed $y$, continuity of the other linear map gives ${\sup }_{\Vert x\Vert \le 1}\Vert F(x,y)\Vert <\infty$. Apply the Uniform boundedness principle to this family to getBilinearity then gives $F(x,y)-F(x_0,y_0)=F(x-x_0,y)+F(x_0,y-y_0)$; the bound proves joint continuity.

For the trilinear map, fix $w$. The just-proved bilinear result bounds $G(\cdot ,\cdot ,w)$. Thus the bounded linear maps $T_{x,y}:w\mapsto G(x,y,w)$, for $\Vert x\Vert ,\Vert y\Vert \le 1$, are pointwise bounded. Apply the Uniform boundedness principle once more to obtain $\Vert G(x,y,w)\Vert \le C\Vert x\Vert \Vert y\Vert \Vert w\Vert$. Expansion of the difference in three slots proves joint continuity. Separate continuity suffices for these multilinear maps.

Without linearity it does not suffice. On ${\mathbb{R}}^2$ set $H(x,y)=xy/(x^2+y^2)$ away from the origin and $H(0,0)=0$. Each coordinate section is continuous, but $H(t,t)=1/2$ for $t\ne 0$, so $H$ is not continuous at the origin.