Normed linear space

Question

In Walter Rudin's Complex Analysis, it states that by definition$$\|\Lambda\|=\text{sup}\{\|\Lambda x\|: x\in X, \|x\|\leq1\}$$ and then later he shows that $\|\Lambda x\|\leq \|\Lambda\|\|x\|.$ But, the only thing I know is $\|\Lambda\|\geq\|\Lambda x\|$.

How to show that indeed $\|\Lambda x\|\leq \|\Lambda\|\|x\|.$

Answer 1

You can prove that $$ \|\Lambda\|=\sup_{\substack{x\in X\\x\ne0}}\frac{\|\Lambda x\|}{\|x\|} $$ by noticing that $$ \left\|\frac{1}{\|x\|}x\,\right\|=1 $$

Answer 2

If $x=0$, that inequality is trivial.

In the other cases, $x=\lVert x\rVert\times\frac x{\lVert x\rVert}$. Since, $\left\lVert\frac x{\lVert x\rVert}\right\rVert=1$ and $\Lambda$ is linear,$$\lvert\Lambda x\rvert=\lVert x\rVert\left\lvert\Lambda\frac x{\lVert x\rVert}\right\rvert\leqslant\lVert x\rVert\lVert\Lambda\rVert.$$