# Abstract Nonsense

## Linear Operators and the Operator Norm (Pt. III)

Point of Post: This post is a continuation of this one.

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With this we can now prove that the operator norm is, in fact, a norm:

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Theorem: Let $\left(\mathscr{V},\|\cdot\|_\mathscr{V}\right)$ and $\left(\mathscr{W},\|\cdot\|_\mathscr{W}\right)$ be two normed vector spaces, then the operator norm is a vector norm on $\mathcal{B}\left(\mathscr{V},\mathscr{W}\right)$. In particular for every $T,S\in\mathcal{B}\left(\mathscr{V},\mathscr{W}\right)$  and $c\in\mathbb{F}$ (where $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$) the following holds

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\begin{aligned}&\mathbf{(1)}\quad \left\|T+S\right\|_{\text{op}}\leqslant \left\|T\right\|_\text{op}+\left\|S\right\|_{\text{op}}\\ &\mathbf{(2)}\quad \left\|c T\right\|_\text{op}=|c|\left\|T\right\|_\text{op}\\ &\mathbf{(3)}\quad \left\|T\right\|_{\text{op}}=0\text{ implies }T=\bold{0}\end{aligned}

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Proof:

$\mathbf{(1)}$We merely note that for every $\|v\|_\mathscr{V}=1$ we have

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$\left\|(S+T)(v)\right\|_\mathscr{W}=\left\|S(v)+T(v)\right\|_\mathscr{W}\leqslant \|S(v)\|_\mathscr{W}+\|T(v)\|_\mathscr{W}\leqslant \|S\|_\text{op}+\|T\|_\text{op}$

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Since $\|v\|_\mathscr{V}=1$ was arbitrary we have that

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$\displaystyle \|S+T\|_\text{op}=\sup_{\|v\|_\mathscr{V}}\left\|(S+T)(v)\right\|_\mathscr{W}\leqslant \|S\|_\text{op}+\|T\|_\text{op}$

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$\mathbf{(2)}$: This is clear by definition since

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\begin{aligned}\displaystyle \sup_{\|v\|_\mathscr{V}=1}\|(cT)(v)\|_\mathscr{W} &=\sup_{\|v\|_\mathscr{V}=1}\|cT(v)\|_\mathscr{W}\\ &=\sup_{\|v\|_\mathscr{V}=1}|c|\|T(v)\|_\mathscr{W}\\ &=|c|\sup_{\|v\|_\mathscr{V}=1}\|T(v)\|_\mathscr{W}\\ &=|c|\|T\|_\text{op}\end{aligned}

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$\mathbf{(3)}$: For this we merely note that if $\|T\|_\text{op}=0$ then

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$\displaystyle \sup_{v\ne\bold{0}}\frac{\|T(v)\|_\mathscr{W}}{\|v\|_\mathscr{V}}=0$

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and so evidently $\|T(v)\|_\mathscr{W}=0$ for all $v\ne\bold{0}$ and so $T(v)=\bold{0}$ for all $v\ne\bold{0}$. The conclusion follows. $\blacksquare$

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One of the most important qualities about the operator norm is that it enables one to bound the actual value of a linear operator. More precisely:

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Theorem: Let $\left(V,\|\cdot\|_\mathscr{V}\right)$ and $\left(\mathscr{W},\|\cdot\|_\mathscr{W}\right)$ be normed vector spaces and $T\in\mathcal{B}\left(\mathscr{V},\mathscr{W}\right)$. Then, for every $v\in\mathscr{V}$ one has that $\|T(v)\|_\mathscr{W}\leqslant \|T\|_\text{op}\|v\|$

Proof: If $v=\bold{0}$ the result is clear, if not we note that by definition

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$\displaystyle \frac{\|T(v)\|_\mathscr{W}}{\|v\|_\mathscr{V}}\leqslant \sup_{x\ne\bold{0}}\frac{\|T(x)\|_\mathscr{W}}{\|x\|_\mathscr{V}}=\|T\|_\text{op}$

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from where the result follows immediately. $\blacksquare$

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From this we can easily see  that the operator norm is submultiplicative in the sense that:

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Theorem: Let $\left(\mathscr{V}_k,\|\cdot\|_k\right)$ be normed vector spaces for $k=1,2,3$ and let $\mathscr{V}_1\xrightarrow{T}\mathscr{V}_2\xrightarrow{S}\mathscr{V}_3$ be bounded linear operators. Then, $\|ST\|_\text{op}\leqslant \|T\|_\text{op}\|S\|_\text{op}$.

Proof: To prove this we merely note that for any $v\in\mathscr{V}_1$ then

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$\left\|S(T(v))\right\|_3\leqslant \|S\|_\text{op}\|T(v)\|_2\leqslant \|S\|_\text{op}\|T\|_\text{op}\|v\|_1$

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and so

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$\|S\|_\text{op}\|T\|_\text{op}\geqslant\inf\left\{c\in\mathbb{R}:\|S(T(v))\|_3\leqslant c\|v\|_1\text{ for all }v\in\mathscr{V}_1\right\}=\|ST\|_\text{op}$

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$\blacksquare$

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References:

1. Kreyszig, Erwin. Introductory Functional Analysis with Applications. New York: Wiley, 1978. Print.

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May 24, 2011 -

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