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How do you prove that these four definitions of the operator norm are equivalent? $$\begin{align*} \lVert A\rVert_{\mathrm{op}} &= \inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\}\\ &=\sup\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert\leq 1\}\\ &=\sup\{\lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert = 1 \}\\ &=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}. \end{align*}$$

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Let $$\begin{align*} I &= \inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\}\\ S_1&=\sup\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert\leq 1\}\\ S_2&=\sup\{\lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert = 1 \}\\ S_3&=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}. \end{align*}$$ Notice that $S_2 \le S_1$ and as $\|Av\| /\|v\| = \| A(v / \|v\|)\|$ we have $S_3 \le S_2$. Now if $\|v\|\le 1$ we have $\|Av\| \le \|Av\| /\|v\|$. Then $S_1 \le S_3$ and $$ S_1=S_2=S_3.$$ Now note that $$ \|Av\| \le S_3 \|v\| \quad \forall v \in V.$$ Then $I \le S_3$ and by definition of $\sup$ we have $$ I \ge \|Av_n\| /\|v_n\| \ge S_3 - 1/n \quad \forall n.$$ Then $S_3 = I$.

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  • $\begingroup$ @ user29999 $S_1 \nleq S_3$. You have proved the inequality for $||v||\leq 1$ but $S_3$ is supremm over $X-${0} $\endgroup$ – Infinite Aug 25 '17 at 6:51
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    $\begingroup$ Why $ I \ge \|Av_n\| /\|v_n\|$ ? And thank you $\endgroup$ – Schüler Jan 23 '18 at 16:17
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Remember that if $s=\sup X$ and $x\leq t$ for all $x\in X$, then $s\leq t$. Also, if $s=\inf X$ and $t\in S$, then $s\leq t$. Now, let us write $$\begin{align*} a&= \inf\{ k\;\colon\; \lVert Av\rVert\leq k\lVert v\rVert \text{ for all }v\in V\},\\ \\ b&=\sup\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert\leq 1\},\\ \\ c&=\sup\{\lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert = 1 \},\\\\ d&=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}. \end{align*}$$

Note that:

  • $\|Av\|\leq a\|v\|$ for all $v\in V$. Taking $v\in V$ with $\|v\|\leq 1$ we obtain $\|Av\|\leq a$ and thus $b\leq a$;

  • $\|Av\|\leq b$ for all $v\in V$ with $\|v\|\leq 1$. Take $v\in V$ with $\|v\|=1$ and define $v_n=(1-1/n)v$. Since $\|v_n\|=1-1/n\leq 1$, we conclude that $\|Av_n\|\leq b$ for all $n\in\mathbb{N}$. Taking the limit we obtain $\|Av\|\leq b$ and thus $c\leq b$;

  • $\|Av\|\leq c$ for all $v\in V$ with $\|v\|= 1$. Taking $v\in V$ with $v\neq 0$, we obtain $\left\|\frac{v}{\|v\|}\right\|=1$. Hence $\frac{\|Av\|}{\|v\|}=\left\|A\left(\frac{v}{\|v\|}\right)\right\|\leq c$ and thus $d\leq c$;

  • $\frac{\|Av\|}{\|v\|}\leq d$ for all $v\in V$ with $v\neq 0$. Hence $\|Av\|\leq d\|v\|$ for all $v\in V$ and thus $a\leq d$.

This shows that $a\leq d\leq c\leq b\leq a$ and thus $a=b=c=d$.

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  • $\begingroup$ Why $b\leq a$?Thanks $\endgroup$ – Student Jan 23 '18 at 16:22
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    $\begingroup$ @Student As explained in the first item, we have $\|Av\|\leq a$ for all $v\in V$ such that $\|v\|\leq 1$. This shows that $a$ is an upper bound of $\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert\leq 1\}$. Since $b$ is the least upper bound (supremum) of this set, we obtain $b\leq a$. $\endgroup$ – Pedro Jan 23 '18 at 17:02
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    $\begingroup$ Thank you very much but my problem: why $\|Av\|\leq a\|v\|$ for all $v\in V$? $\endgroup$ – Student Jan 23 '18 at 21:05
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    $\begingroup$ @Student You are right, $a$ does not necessarily belong to the said set. However, there exists a sequence $(k_n)$ in the set such that $k_n\overset{n\to\infty}{\longrightarrow} a$ (see this or this). We have $\|Av\|\leq k_n\|v\|$ for all $n$ and thus, taking the limit as $n\to\infty$, we obtain $\|Av\|\leq a\|v\|$. $\endgroup$ – Pedro Jan 23 '18 at 23:25
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    $\begingroup$ @Student We know that $\inf A=\sup B=\sup C=\inf D$ and you are asking if $\inf A=\inf E$, where $A,B,...,E$ are the appropriate sets. Note that $A\subset E$ and thus $e=\inf E\leq \inf A=a$. As explained above (with sequences), $a\in A$. A similar argumment (with sequences) shows that $e\in E$ and thus $e$ is an upper bound of $D$. This implies that $d=\sup D\leq e$. Therefore $e\leq a=d\leq e$ which implies $a=e$. $\endgroup$ – Pedro Jan 25 '18 at 13:01
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I'll give you part of one to give you an idea of the flavor, but you should really do them yourself.

Let $w\neq 0$. Then $\frac{1}{\lVert w\rVert}$ makes sense. Now notice that $$A\left(\frac{1}{\lVert w\rVert}w\right) = \frac{1}{\lVert w\rVert}A(w).$$ Therefore, $$\left\lVert A\left(\frac{w}{\lVert w\rVert}\right)\right\rVert = \frac{\lVert Aw\rVert}{\lVert w\rVert}.$$ But $\frac{w}{\lVert w\rVert}$ is a vector of norm $1$, so...

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The four definitions are equivalent only if $V\neq\{0\}$ (which, admittedly, is usually the case).

Reason: When $V=\{0\}$, there is no $v\in V$ such that $\|v\|=1$. So $\sup\varnothing = -\infty$, but we clearly want $\|A\|=0$ if $A$ is the zero operator. This (small but still) mistake can be found in many textbooks.

In summary, only the first two expressions make sense.

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    $\begingroup$ It is not uncommon to take the supremum of an empty set of non-negative real numbers as $0$, generally if $L$ is a lattice with smallest element $s$, $\sup \varnothing = s$ when looking at subsets of $L$. Of course that should be mentioned. Which hardly any text does. And in the first definition, the constraint $c \geqslant 0$ is missing. Without that, $V = \{0\}$ leads to $\lVert A\rVert = -\infty$ for $A \colon V \to W$. $\endgroup$ – Daniel Fischer Jul 15 '18 at 8:59
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Let $a=\inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\},b=\sup\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert\leq 1\},c=\sup\{\lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert = 1 \},d=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}.$

$a=\inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\}= \inf\{ c\;\colon\;\frac{\lVert Av\rVert}{\lVert v\rVert}\leq c \text{ for all }v\neq 0\}$ (easily verified), which is the infimum of the set of upper bounds for the set $\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}$, i.e., the least upper bound. So $a=d$. Next see that $\frac{\lVert Av\rVert}{\lVert v\rVert}=\lVert A(\frac{v}{\lVert v\rVert})\rVert$, and $\frac{v}{\lVert v\rVert}$ has norm $1$. And if $v$ has norm $1$, then we can write $v =\frac{v}{\lVert v\rVert}.$ This shows that $c=d$. Finally I'll show that $b=c$ (because I find this more interesting). $b\geq c$ is easily seen. For each $v$ with $\lVert v\rVert\leq 1$, there is $\tilde v=\frac{v}{\lVert v\rVert}$ having norm $1$, with $\lVert A\tilde v\rVert=\frac{\lVert Av\rVert}{\lVert v\rVert}\geq \lVert Av\rVert.$ This proves $c\geq b.$

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