Equivalent Definitions of the Operator Norm

Question

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

Answer 1

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 $S_3 = I$.

Answer 2

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 $a=b=c=d$.

Answer 3

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...

Answer 4

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.

Answer 5

Another equivalent definition, which I can rarely see: $$\lVert A\rVert=\sup_{\lVert x \rVert < 1} \lVert Ax \rVert$$ It's easy to see that $$\sup_{\lVert x \rVert < 1} \lVert Ax \rVert \leqslant \sup_{\lVert x \rVert \leqslant 1} \lVert Ax \rVert$$ For the other direction, let $(r_n)$ be a sequence of real numbers, with $0 \leqslant r_n < 1$ and $r_n \to 1$, and $x \in V$ with $\lVert x \rVert \leqslant 1$. Then, we have that $\lVert r_n x\rVert<1$, which implies that $\lVert A(r_n x) \rVert \leqslant \sup_{\lVert x \rVert < 1} \lVert Ax \rVert$, but $\lVert A(r_n x) \rVert \to \lVert A x\rVert$, which implies that $\lVert Ax \rVert \leqslant \sup_{\lVert x \rVert < 1} \lVert Ax \rVert$.

Answer 6

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 $c\geq b.$

Answer 7

@xxxg: I want to make some further details. In other words, I want to prove the following two things.

(i) The infimum of the first set, i.e., I, also satisfies the condition in the first set, which means that $$\|Av\| \leq I \|V\|, \quad \text{for all $v \in V$}. $$ Proof of (i): Let us assume that there exists a $v_0$ such that $$\frac{\|Av_0\|}{\|v_0\|} > I.$$ Then, by the definition of infimum, there exists $$c_0 \in \{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\}$$ such that $$c_0 < \left(\frac{\|Av_0\|}{\|v_0\|} + I\right) /2 < \frac{\|Av_0\|}{\|v_0\|},$$ which makes a contradiction. End of proof of (i).

(ii) Let $$ S_4 = \sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0, \|v\| \leq r\right\}, $$ $$ S_5 = \sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0, \|v\| = r\right\}. $$ Then $$I=S_1=S_2=S_3=S_4=S_5, \quad \text{for any $r>0$}.$$ Proof of (ii): For any $v\in V$, $\|v\|=1$, there exists $g = rv \in V$, $\|g\|=r$,
it holds $$\frac{\|Ag\|}{\|g\|} = \|Av\|.$$ So $S_2 \leq S_5$. Then we have $$S_2 \leq S_5 \leq S_4 \leq S_3.$$ End of proof of (ii).

Answer 8

It is of course possible to prove direct equivalence of any two of the definitions as well. Here is one example.
Let $$ \begin{align*} a &= \inf \left\{ k > 0: \left\lVert L v \right\rVert \leq k \left\lVert v \right\rVert \text{ for all } v \in V \right\} \text{ and} \\ b &= \sup \left\{ \left\lVert L v \right\rVert \text{ for all } v \in V \text{ with } \left\lVert v \right\rVert \leq 1 \right\} . \end{align*} $$ Now we prove that $a = b$ by proving that $b \leq a$ and $a \leq b$.

First, we have that for any $v \in V$ $$ \left\lVert L v \right\rVert \leq a \left\lVert v \right\rVert , $$ since $a$ was the infimum of all numbers that satisfies this equation. In particular, for any $v$ with $ \left\lVert v \right\rVert \leq 1$ we have $$ \left\lVert L v \right\rVert \leq a \left\lVert v \right\rVert \leq a . $$ This tells us that $a$ is an upper bound for $\left\lVert L v \right\rVert$ when $\left\lVert v \right\rVert \leq 1$, and hence an upper bound for the set in the definition of $b$. Since $b$ is the supremum (smallest upper bound), we have that $$ b \leq a . $$ The opposite direction requires a simple trick. Let $v \in V$ be any vector and choose an $\epsilon < 1$. Then the vector defined by $$ u = \frac{v}{\left\lVert v \right\rVert + \epsilon} , $$ fulfils $$ \left\lVert u \right\rVert = \frac{\left\lVert v \right\rVert}{\left\lVert v \right\rVert + \epsilon} < 1 . $$ Hence, we see that it is in the set that defines $b$, and we must have $$ \left\lVert L u \right\rVert \leq b. $$ Inserting the relation between $u$ and $v$ we get $$ \left\lVert L v \right\rVert \leq b \left(\left\lVert v \right\rVert + \epsilon \right) , $$ which in the limit $\epsilon \to 0$ becomes $$ \left\lVert L v \right\rVert \leq b \left\lVert v \right\rVert . $$ Since $a$, being the infimum of all numbers satisfying this relation, we must have $$ a \leq b , $$ and we are done