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Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. It is well known that if $S\in \mathcal{B}(F)$, then $$\|S\|:=\sup_{\substack{x\in F\\ x\not=0}}\frac{\|Sx\|}{\|x\|}$$

Why $$\|S\|=\sup_{\|x\| \leq 1}\|Sx\|?$$

Schüler
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    Show they are both equivalent to $ \sup_{||x||=1} ||Sx||$. Hint: Use the linearity of S. The solution isn't enlightening, its more important you understand why you can restrict to the unit sphere. – Luke Peachey Dec 26 '17 at 14:46
  • @LukePeachey Thank you. Is what I write true $$|S|:=\sup_{\substack{x\in F\ x\not=0}}\left|S(\frac{x}{|x|})\right|=\sup_{|x| =1}|Sx|?$$ – Schüler Dec 26 '17 at 15:06
  • Yes its correct. There are quite a few equivalent definitions of the operator norm like this, just get comfortable using them all and make sure you understand why it only depends on the sphere. – Luke Peachey Dec 26 '17 at 15:10
  • @LukePeachey Thank you very much. Why it only depends on the sphere? – Schüler Dec 26 '17 at 15:15
  • Notice that $\frac{||S(x)||}{||x||}$ is constant on a line passing through the origin. Therefore this function is completely determind by its values on any central sphere, in particular the unit sphere. – Luke Peachey Dec 26 '17 at 15:21

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A very standard exercise in functional analysis. One can show $$ A:=\sup_{\substack{x\in F\\ x\not=0}}\frac{\|Sx\|}{\|x\|}=\sup_{\|x\| \leq 1}\|Sx\|=:B $$ by showing $A\leq B$ and $A\geq B$. One direction is almost trivial.

  • Note that $\left\|\frac{x}{\|x\|}\right\|=1$ for $x\neq 0$ and observe that the norm is by definition absolutely homogeneous: $\|ax\|=|a|\|x\|$. Show that $A\leq B$.
  • To show $A\geq B$, it suffices to show that $A\geq \|Sx\|$ for all $\|x\|\leq 1$. There are two cases.
    • If $x=0$, then this is trivially true.
    • If $0<\|x\|\leq 1$, then $\frac{\|Sx\|}{\|x\|}\geq\|Sx\|$ implies that $A\geq \|Sx\|$.

(The proof above is also true when $F$ is only a Banach space.)