# Norm of linear functional

1. Jun 30, 2011

### zok_peltek

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

2. Jun 30, 2011

### zok_peltek

3. Jun 30, 2011

### tiny-tim

welcome to pf!

hi zok! welcome to pf!
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)| ?

4. Jun 30, 2011

### zok_peltek

5. Jun 30, 2011

### Fredrik

Staff Emeritus
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).

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