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

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    $\begingroup$ 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. $\endgroup$ – Luke Peachey Dec 26 '17 at 14:46
  • $\begingroup$ @LukePeachey Thank you. Is what I write true $$\|S\|:=\sup_{\substack{x\in F\\ x\not=0}}\left\|S(\frac{x}{\|x\|})\right\|=\sup_{\|x\| =1}\|Sx\|?$$ $\endgroup$ – Schüler Dec 26 '17 at 15:06
  • $\begingroup$ 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. $\endgroup$ – Luke Peachey Dec 26 '17 at 15:10
  • $\begingroup$ @LukePeachey Thank you very much. Why it only depends on the sphere? $\endgroup$ – Schüler Dec 26 '17 at 15:15
  • $\begingroup$ 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. $\endgroup$ – Luke Peachey Dec 26 '17 at 15:21

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.)

  • $\begingroup$ Thank you for your answer, but I don't understand very well why $A\leq B$ $\endgroup$ – Schüler Dec 26 '17 at 15:40
  • $\begingroup$ @Thierry: Note that by homogeneity of the norm, $\{\frac{\|Sx\|}{\|x\|}:x\neq 0\}$ is the same as $\{\|Sx\|:\|x\|=1|\}$, which is a subset of $\{\|Sx\|:\|x\leq1|\}$. $\endgroup$ – Jack Dec 26 '17 at 15:57
  • $\begingroup$ Thank you, now it is very clear. $\endgroup$ – Schüler Dec 26 '17 at 15:59

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