Exercise 2.9

The answer is: No. To prove this, we only assume that $X$, $Y$, $Z$ are normed spaces. Since $B$ is continous at the origin, there exists a positive $r$ such that

Given nonzero $x,y$, let $s$ range over $]0,r[$, so that the folllowing bound

is effective. It is now obvious that

which achieves the proof.

As a concrete example, choose $X=Y=Z=C_c(R)$, topologized by the supremum norm. $C_c(R)$ is not complete (see 5.4.4 of Laurent Schwartz’ Analyse III (in French), Hermann, 1997.), nevertheless the bilinear product

is bounded (since $\parallel f\cdot g{\parallel }_{\infty }\le \parallel f{\parallel }_{\infty }\cdot \parallel g{\parallel }_{\infty }$), and continuous. To show this, pick a positive scalar $𝜖$ smaller than $1$, provided any $(f,g)$. Next, define

We now restrict $(u,v)$ to a particular neighborhood of $(f,g)$. More specifically,

Next, remark that $\parallel u{\parallel }_{\infty }\le r+\parallel f{\parallel }_{\infty }$ and so obtain (bear in mind that $r<1$)

Since $𝜖$ was arbitrary, it is now established that $B$ continuous at every $(f,g)$.