# Norm of product of linear operators in normed vector space

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, $\|\Lambda_2 \Lambda_1\| \leq \|\Lambda_2\| \| \Lambda_1\|$.
• 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. ${}\qquad{}$ – Michael Hardy Nov 4 '15 at 19:57