Abstract Nonsense

Crushing one theorem at a time

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 - Posted by | Analysis | , , ,

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