What is the meaning of the norm of a linear functional?

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    Functional Linear Norm
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Discussion Overview

The discussion revolves around the meaning and equivalence of different definitions of the norm of a linear functional as presented in the book "Optimization by Vector Space Methods" by David Luenberger. Participants explore the implications of various formulations of the norm, particularly in relation to the conditions on the vector norms.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant, zok, presents confusion regarding the equivalence of the definitions of the norm of a linear functional, specifically questioning how ||f|| = sup(|f(x)|) for ||x|| <= 1 relates to other formulations.
  • Another participant suggests that the two definitions may be equivalent, citing the linearity of f and providing an example involving scalar multiplication.
  • A later reply discusses the relationship between the sets defined by the conditions ||x|| = 1 and ||x|| <= 1, arguing that the supremum of the first set is less than or equal to the supremum of the second set, but also noting that the supremum of the second set is the least upper bound.

Areas of Agreement / Disagreement

Participants express differing levels of understanding regarding the equivalence of the definitions, with some asserting that they are the same while others remain uncertain. The discussion does not reach a consensus on the clarity of the definitions.

Contextual Notes

There are unresolved assumptions regarding the properties of the linear functional and the implications of the definitions provided. The discussion also highlights the need for careful consideration of the conditions under which the supremum is taken.

zok_peltek
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Hi everyone,

I have been studying "Optimization by Vector Space Methods", written by David Luenberger and I am stuck in an obvious point at first glance. My problem is in page 105, where the norm of a linear functional is expressed in alternative ways. The definition for the norm of a linear functional f is (from now on ||y|| denotes the norm of y and |y| the absolute value of y):
||f|| = inf{M: |f(x)| <= M*||x|| for every x in X} where X is the vector space the functional is defined on.

now this definition can be modified and give:
||f|| = inf{M: |f(x/||x||)| <= M for every x in X and x not zero} since f is linear which is equivalent to:
||f|| = sup(|f(x)|/||x||) for x not zero or
||f|| = sup(|f(x)|) for ||x|| = 1

now in the book it states that also:
||f|| = sup(|f(x)|) for ||x|| <= 1
which I can't compehend given the previous definition.

any help is much appreciated,
zok
 
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welcome to pf!

hi zok! welcome to pf! :smile:
zok_peltek said:
||f|| = sup(|f(x)|) for ||x|| = 1

now in the book it states that also:
||f|| = sup(|f(x)|) for ||x|| <= 1
which I can't compehend given the previous definition.

aren't they the same?

eg if ||x|| = 1, then f(0.5x) = 0.5f(x) since f is linear, so |f(0.5x)| < |f(x)| ?
 
Note that \big\{|f(x)|\, \big| \,\|x\|=1\big\}\subset \big\{|f(x)|\,\big|\,\|x\|\leq 1\big\}. Let's call the first set A and the second set B. We have A\subset B. This means that \sup A\leq\sup B. On the other hand, you can easily show that an arbitrary member of B is smaller than sup A. This means that sup A is an upper bound of B, but sup B is the least upper bound, so \sup B\leq\sup A.

(Click the quote button if you want to see how I did the math).

You might also want to check out this thread.
 

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