Operator Norm

Proof. Let $(A,||\cdot |{|}_a),(B,\left|\right|\cdot \left|{|}_b\right)$ over a field $𝔽$ be normed vector spaces and $L:(A,||\cdot |{|}_a)\to (B,\left|\right|\cdot \left|{|}_b\right)$ be a linear map, that is,

$$L\left(\alpha a_1\right)=\mathit{\alpha L}\left(a_1\right)\text{and}L(a_1+a_2)=L\left(a_1\right)+L\left(a_2\right)\text{for}{a}_1,a_2\in A\text{with}\alpha \in 𝔽$$

$\Rightarrow$: Suppose $L$ is bounded, then for all vectors $a,h\in A$ with $h\ne 0$,

$$0\le \left|\right|L(a+h)-L\left(a\right)|{|}_b=|\left|L\right(h\left)\right|{|}_b\le M\left|\right|h|{|}_a\text{for some constant }M$$

As $h$ goes to 0,

$$\underset{h\to 0}{\lim <!--\; FUNCTION\; APPLICATION\; -->}0\le \underset{h\to 0}{\lim <!--\; FUNCTION\; APPLICATION\; -->}\left|\right|L(a+h)-L\left(a\right)|{|}_b=\underset{h\to 0}{\lim <!--\; FUNCTION\; APPLICATION\; -->}|\left|L\right(h\left)\right|{|}_b\le \underset{h\to 0}{\lim <!--\; FUNCTION\; APPLICATION\; -->}M\left|\right|h|{|}_a.$$

Since $\underset{h\to 0}{\lim <!--\; FUNCTION\; APPLICATION\; -->}=0$ and $\underset{h\to 0}{\lim <!--\; FUNCTION\; APPLICATION\; -->}M\left|\right|h|{|}_a=0$, applying squeeze theorem gives

$$\underset{h\to 0}{\lim <!--\; FUNCTION\; APPLICATION\; -->}\left|\right|L(a+h)-L\left(a\right)|{|}_b=0$$

Hence $L$ is continuous.

$\Leftarrow$: By definition of continuous, for all $𝜀>0$, there exists $\delta >0$ such that

$$\left|\right|h-0|{|}_a<\delta \text{implies}|\left|L\right(h)-L(0\left)\right|{|}_b<𝜀$$

for all $h\in A$. Since $L$ is a linear map, $\left|\right|L\left(h\right)-L\left(0\right)|{|}_b=|\left|L\right(h\left)\right|{|}_b$. Now taking $𝜀=1$, then for all nonzero $\alpha \in A$, we have

$$\begin{array}{llll}\hfill \left|\right|L\left(\alpha \right)|{|}_b& =\left|\right|\frac{2}{\delta }L\left(\frac{\delta }{2}\alpha \right)|{|}_b=|\left|\frac{2\left|\right|\alpha |{|}_a}{\delta }L\right(\frac{\delta }{2}\frac{\alpha }{\left|\right|\alpha |{|}_a}\left)\right|{|}_b=\frac{2\left|\right|\alpha |{|}_a}{\delta }\left|\right|L\left(\frac{\delta }{2}\frac{\alpha }{\left|\right|\alpha |{|}_a}\right)|{|}_b& \hfill & \\ \hfill & <\frac{2\left|\right|\alpha |{|}_a}{\delta }=\frac{2}{\delta }\left|\right|\alpha |{|}_a& \hfill & \end{array}$$

Hence $L$ is bounded.

Proof. Let

$$\begin{array}{llll}\hfill & I=\inf <!--\; FUNCTION\; APPLICATION\; -->\{c\ge 0:|\left|\mathit{Ax}\right|{|}_Y\le c\left|\right|x|{|}_X\text{for all}x\in X\}& \hfill & \\ \hfill & S_1=\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\right|\left|\mathit{Ax}\right|{|}_Y:\left|\right|x|{|}_X\le 1\text{and}x\in X\}& \hfill & \\ \hfill & S_2=\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\right|\left|\mathit{Ax}\right|{|}_Y:\left|\right|x|{|}_X=1\text{and}x\in X\}& \hfill & \\ \hfill & S_3=\sup <!--\; FUNCTION\; APPLICATION\; -->\{\frac{\left|\right|\mathit{Ax}|{|}_Y}{\left|\right|x|{|}_X}:x\ne 0\text{and}x\in X\}& \hfill & \end{array}$$

First, we show $S_1=S_2=S_3$.
Notice that

$$\left\{\right|\left|\mathit{Ax}\right|{|}_Y:\left|\right|x|{|}_X=1\text{and}x\in X\}\subseteq \left\{\right|\left|\mathit{Ax}\right|{|}_Y:\left|\right|x|{|}_X\le 1\text{and}x\in X\}$$

The supremum of $\left\{\right|\left|\mathit{Ax}\right|{|}_Y:\left|\right|x|{|}_X=1\text{and}x\in X\}$ must not be greater then the supremum of $\left\{\right|\left|\mathit{Ax}\right|{|}_Y:\left|\right|x|{|}_X\le 1\text{and}x\in X\}$, that is

$$S_2\le S_1$$

On the other hand, since $A$ is a linear operator, the elements in $\{\frac{\left|\right|\mathit{Ax}|{|}_Y}{\left|\right|x|{|}_X}:x\ne 0\text{and}x\in X\}$ can be written in the form

$$\frac{\left|\right|\mathit{Ax}|{|}_Y}{\left|\right|x|{|}_X}=\left|\right|A\frac{x}{\left|\right|x|{|}_X}|{|}_Y,\text{and}|\left|\frac{x}{\left|\right|x|{|}_X}\right|{|}_X=1$$

So

$$\{\frac{\left|\right|\mathit{Ax}|{|}_Y}{\left|\right|x|{|}_X}:x\ne 0\text{and}x\in X\}\subseteq \left\{\right|\left|\mathit{Ax}\right|{|}_Y:\left|\right|x|{|}_X=1\text{and}x\in X\}$$

One may check that the two sets are actually identical, but for convenience only subset relation will be taken here and $S_3\le S_2$ can be obtained by the relation.

Also, for all $\left|\right|x|{|}_X\le 1$, the inequality

$$\left|\right|\mathit{Ax}|{|}_Y\le \frac{1}{\left|\right|x|{|}_X}|\left|\mathit{Ax}\right|{|}_Y$$

is sufficient to give $S_1\le S_3$.
Hence we conclude $S_1=S_2=S_3$.

Now, we show $I=S_3$ and therefore $S_1=S_2=S_3=I$.
Since $S_3=\sup <!--\; FUNCTION\; APPLICATION\; -->\{\frac{\left|\right|\mathit{Ax}|{|}_Y}{\left|\right|x|{|}_X}:x\ne 0\text{and}x\in X\}$, for all $x\in X$ the inequality $\left|\right|\mathit{Ax}|{|}_Y\le S_3|\left|x\right|{|}_X$ must be satisfied, so $I\le S_3$.
By definition of supremum, there is $x_n\in X$ with

$$I\ge \frac{\left|\right|A{x}_n|{|}_Y}{\left|\right|x_n|{|}_X}\ge S_3-\frac{1}{n}\text{for all}n\in \mathbb{N}$$

this gives $I\ge S_3$.
Since $I\le S_3$ and $I\ge S_3$, equality must hold and therefore $I=S_3$.

Note that we have left the case $\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\right|\left|\mathit{Ax}\right|{|}_Y:\left|\right|x|{|}_X\in \{0,1\left\}\right\}$. As $\left|\right|\mathit{Ax}|{|}_Y$ must be nonnegative and

$$\left|\right|x|{|}_X=0\Rightarrow x=0\Rightarrow \mathit{Ax}=0\Rightarrow |\left|\mathit{Ax}\right|{|}_Y=0$$

Therefore, $\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\right|\left|\mathit{Ax}\right|{|}_Y:\left|\right|x|{|}_X\in \{0,1\left\}\right\}$ is exactly the same as $S_2$.