= Solution
A <linear map> $T:X\to Y$ between normed spaces is a <bounded linear operator> if $\|Tx\|\leq C\|x\|$ for one finite $C$ and every $x$. This implies $\|Tx-Ty\|\leq C\|x-y\|$, hence continuity. Conversely, continuity at zero supplies $\delta>0$ such that $\|x\|<\delta$ implies $\|Tx\|<1$. Apply this to $\delta x/(2\|x\|)$ to obtain $\|Tx\|\leq2\|x\|/\delta$ for nonzero $x$. Thus \b[boundedness and continuity are equivalent for <linear maps>].
The <operator norm> is $\|T\|=\sup_{\|x\|\leq1}\|Tx\|$, equivalently $\sup_{x\ne0}\|Tx\|/\|x\|$. The bounded operators form a <vector space>, since $\|(S+T)x\|\leq(\|S\|+\|T\|)\|x\|$ and scalar multiples remain bounded. Taking suprema proves the <triangle inequality> and absolute homogeneity for the <operator norm>; $\|T\|=0$ forces $Tx=0$ for every $x$. Thus $B(X,Y)$ is a normed space, whether or not $X,Y$ are complete.
Back to article page
