Norms of compositions of bounded operators between different spaces

  • #1
Suppose I have [itex]B: X\to Y[/itex] and [itex]A: Y\to Z[/itex], where [itex]X,Y,Z[/itex] are Banach spaces and [itex]B\in \mathcal L(X,Y)[/itex] and [itex]A\in \mathcal L(Y,Z)[/itex]; that is, both of these operators are bounded. Does it follow that [itex]AB \in \mathcal L(X,Z)[/itex] and
[tex]
\| AB \|_{\mathcal L(X,Z)} \leq \|A\|_{\mathcal L(Y,Z)} \|B\|_{\mathcal L(X,Y)}
[/tex]
It seems like this should be the case, but any time I try to prove a functional analytic result like this, I always get mired in uncertainty about the details...
 

Answers and Replies

  • #2
Hmmm...okay, well, I think I've found my answer in Eqn. 3.3 of this PDF. Does this suffice for a proof of this result:
[tex]
\| AB \|_{\mathcal L(X,Z)} = \sup_{x\in X, \|x\|_X = 1} \|ABx\|_Z \leq \sup_{x\in X, \|x\|_X = 1} \|A\|_{\mathcal L(Y,Z)} \|Bx\|_Y = \|A\|_{\mathcal L(Y,Z)} \sup_{x\in X, \|x\|_X = 1} \|Bx\|_Y = \|A\|_{\mathcal L(Y,Z)} \|B\|_{\mathcal L(X,Y)}.
[/tex]
 
  • #3
WannabeNewton
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This is ok. Another way of proving it is to note that ##\|A\|## is the smallest number ##C## such that [tex]\|Ax\|\leq \|A\|\|x\|[/tex] for each ##x## in the domain. It follows that for each ##x\in X##: [tex]\|ABx\|\leq \|A\|\|Bx\|\leq \|A\|\|B\|\|x\|[/tex] So ##\|A\|\|B\|## is some number ##C## such that ##\|ABx\|\leq C\|x\|##. It follows that ##AB## is bounded and that ##\|AB\|\leq \|A\|\|B\|##, since ##\|AB\|## is by definition the smallest ##C## such that ##\|ABx\|\leq C\|x\|##.
 

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