The operator norm is defined as ($x \in X$ (a normed space))

$$\lVert T \rVert = \sup_{x \neq 0} \frac {\lVert Tx \rVert}{\lVert x \rVert} = \sup_{\lVert x \rVert=1} \lVert Tx \rVert = \sup_{\lVert x \rVert \leq1} \lVert Tx \rVert $$

What I don't understand is why is $\sup_{||x||=1} (\cdot) = \sup_{||x|| \leq1} (\cdot)$?


The supremum is achieved when $\lVert x \rVert = 1$. Note that if you had an $x$ with norm less than $1$ such that the supremum were achieved, note that using the unit vector $\frac{x}{||x||}$ has norm $|| T \frac{x}{||x||}|| = ||T x||/||x|| > ||Tx||$. So, the supremum is achieved for unit vectors.

  • $\begingroup$ Thank you for the explanation $\endgroup$ – Parinn Jun 25 '17 at 22:53

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