Let $\Lambda:X\rightarrow Y$ be a linear operator, where $X$ and $Y$ are two normed vector spaces. From the definition of an operator's norm, it is straightforward that $$ \|\Lambda(x)\|=\|\Lambda x\|\le \|\Lambda \|\cdot \|x\|_{X}. \tag 1 $$

Now consider two linear operators $\Lambda_1:X\rightarrow Y$ and $\Lambda_2:Y\rightarrow Z$. Is it true that

$$ \|\Lambda_2(\Lambda_1(x))\|\le \|\Lambda_2 \|\cdot \|\Lambda_1 \|\cdot \|x\|_X? $$

It seems to me that it is, but I am not sure about the fact that $ \|\Lambda_2\Lambda_1\|\le \|\Lambda_2 \|\cdot \|\Lambda_1 \| $. How can one prove this (if it is true of course)?

My idea is to prove that the space of linear operators is a normed vector space, so the inequality follows by equation (1).


Presumably $\Lambda_2: Y\to Z$ or something, for $\Lambda_2(\Lambda_1(x))$ to make sense.

It is true: $\|\Lambda_2(\Lambda_1(x))\| \leq \| \Lambda_2 \| \| \Lambda_1 (x) \| \leq \| \Lambda_2 \| \| \Lambda_1\| \|x\|$ by applying the definition of the operator norm twice. So, $\|\Lambda_2 \Lambda_1\| \leq \|\Lambda_2\| \| \Lambda_1\|$.

  • 3
    $\begingroup$ Notice the typographical difference between $\displaystyle ||\Lambda_2|| || \Lambda_1||$ and $\displaystyle \|\Lambda_2\| \| \Lambda_1\|$. I changed the former to the latter, which is standard usage. ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 4 '15 at 19:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.