Abstract Nonsense

Crushing one theorem at a time

Linear Operators and the Operator Norm (Pt. II)


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

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Operator Norm

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We saw in the proof of the last theorem that given a bounded linear operator T:\mathscr{V}\to\mathscr{W} there are many choices for the bounding constant. Of course a natural question is to ask what is the minimum such constant? This leads us naturally to the notion of the operator norm which is defined to be the infimum of all such constants, and it turns out (as the name suggests) that this forms a norm on the vector space (it’s obvious that it is one under pointwise scalar multiplication and addition, by use of the triangle inequality, etc.) of all bounded linear operators \mathscr{V}\to\mathscr{W} denoted \mathcal{B}\left(\mathscr{V},\mathscr{W}\right). To be more explicit if \left(\mathscr{V},\|\cdot\|_\mathscr{V}\right) and \left(\mathscr{W},\|\cdot\|_\mathscr{W}\right) are normed vector spaces we can define the function \|\cdot\|_\text{op}:\mathcal{B}\left(\mathscr{V},\mathscr{W}\right)\to\mathbb{R} given by

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\left\|T\right\|_\text{op}=\inf\left\{c\in\mathbb{R}:\|T(v)\|_\mathscr{W} \leqslant c\|v\|_\mathscr{V}\text{ for all }v\in\mathscr{V}\right\}

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We know that this function makes sense since the set on the right is non-empty (by assumption of boundedness) and bounded below by 0. We call \|\cdot\|_\text{op} the operator norm on \mathcal{B}\left(\mathscr{V},\mathscr{W}\right). Before we actually show it’s a norm we find a few alternate definitions of this, as for now, function.

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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 for any T\in\mathcal{B}\left(\mathscr{V},\mathscr{W}\right) one has 

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\displaystyle \begin{aligned}\|T\|_\text{op} &=\sup_{v\ne\bold{0}}\frac{\|T(v)\|_\mathscr{W}}{\|v\|_\mathscr{V}}\\ & \leqslant \sup_{\|v\|_\mathscr{V}\leqslant 1}\|T(v)\|_\mathscr{W}\\ &=\sup_{\|v\|_\mathscr{V}=1}\|T(v)\|_\mathscr{W}\end{aligned}

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Proof: To prove the first equality we note by definition that for any c which is an upper bound for T (in the sense of the definition of boundendess) and any v\in\mathscr{V}-\{\bold{0}\} one has

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\displaystyle \frac{\|T(v)\|_\mathscr{W}}{\|v\|_\mathscr{V}}\leqslant c

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since this is true for all upper bounds c and all v\in\mathscr{V}-\{\bold{0}\} we may conclude that

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\displaystyle \sup_{v\ne\bold{0}}\frac{\|T(v)\|_\mathscr{W}}{\|v\|_\mathscr{V}}\leqslant\inf\left\{c\in\mathbb{R}:\left\|T(v)\right\|_\mathscr{W}\leqslant c\|v\|_\mathscr{V}\right\}=\left\|T\right\|_\text{op}

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That said, it’s evidently true that for every x\ne\bold{0} one has

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

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and so evidently for every x\in\mathscr{V}

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

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and so

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\displaystyle \|T\|_\text{op}=\inf\left\{c\in\mathbb{R}:\|T(v)\|_\mathscr{W}\leqslant c\|v\|_\mathscr{V}\text{ for all }v\in\mathscr{V}\right\}\leqslant \sup_{v\ne\bold{0}}\frac{\|T(v)\|_\mathscr{W}}{\|v\|_\mathscr{V}}

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from where the first equality follows. To prove the second equality we note that evidently

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\displaystyle \sup_{\|v\|_\mathscr{V}\leqslant 1}\|T(v)\|_\mathscr{W}\leqslant\sup_{v\ne\bold{0}}\frac{\|T(v)\|_\mathscr{W}}{\|v\|_\mathscr{V}}

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since for 0<\|v\|_\mathscr{V}\leqslant 1 one has that \displaystyle \|T(v)\|_\mathscr{W}\leqslant \frac{\|T(v)\|_\mathscr{W}}{\|v\|_\mathscr{V}}. That said

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\displaystyle \begin{aligned}\left\{\frac{\|T(v)\|_\mathscr{W}}{\|v\|_\mathscr{V}}:v\ne\bold{0}\right\} &=\left\{\left\|T\left(\frac{v}{\|v\|_\mathscr{V}}\right)\right\|_\mathscr{W}:v\ne\bold{0}\right\}\\ &=\left\{\|T(v)\|_\mathscr{W}:\|v\|_\mathscr{V}=1\right\}\\ &\subseteq\left\{\|T(v)\|_\mathscr{W}:\|v\|_\mathscr{V}\leqslant 1\right\}\end{aligned}

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from where the reverse inequality readily follows and thus the entire second equality. Lastly, to prove the last equality we note that it’s evident that \displaystyle \sup_{\|v\|_\mathscr{V}=1}\|T(v)\|_\mathscr{W}\leqslant\sup_{\|v\|_\mathscr{V}\leqslant 1}\|T(v)\|_\mathscr{W}. That said, let \|x\|_\mathscr{V}\leqslant 1 and note that

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\displaystyle \|T(x)\|_\mathscr{W}=\|x\|_\mathscr{V} \left\|T\left(\frac{x}{\|x\|_\mathscr{V}}\right)\right\|_\mathscr{W}\leqslant \left\|T\left(\frac{x}{\|x\|_\mathscr{V}}\right)\right\|_\mathscr{W}\leqslant \sup_{\|v\|_\mathscr{V}=1}\|T(v)\|_\mathscr{W}

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and so the last equality holds. \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 | , , , ,

3 Comments »

  1. […] Point of Post: This post is a continuation of this one. […]

    Pingback by Linear Operators and the Operator Norm (Pt. III) « Abstract Nonsense | May 24, 2011 | Reply

  2. […] is the operator norm. Taking the limit as of both sides gives from where the conclusion […]

    Pingback by Differentiability Implies Continuity « Abstract Nonsense | May 24, 2011 | Reply

  3. […] is the operator norm. To show the other limit is zero we note that for any there exists some such that if . But, we […]

    Pingback by The Chain Rule « Abstract Nonsense | May 24, 2011 | Reply


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