Laplace expansion: Difference between revisions

From formulasearchengine
Jump to navigation Jump to search
No edit summary
Line 1: Line 1:
In mathematics, in the field of [[functional analysis]], a '''Minkowski functional''' is a function that recovers a notion of distance on a linear space.
The author's name is Andera and she [http://ltreme.com/index.php?do=/profile-127790/info/ telephone psychic] believes it sounds quite good. Office supervising is exactly where her primary income arrives  [http://www.weddingwall.com.au/groups/easy-advice-for-successful-personal-development-today/ psychic phone readings] from but she's already applied for another one. Mississippi is where her house is but her spouse wants them to move. My spouse doesn't like it the way I do but what I truly like doing is caving but I don't have the time recently.<br><br>My website: [http://www.article-galaxy.com/profile.php?a=143251 free tarot readings]
 
Let ''K'' be a symmetric convex body in a linear space ''V''. We define a function ''p'' on ''V'' as
 
:<math>p(x) = \inf \{ \lambda \in \mathbb{R}_{> 0} : x \in \lambda K \} </math>
 
if that [[infimum]] is well-defined.<ref>Thompson (1996) p.17</ref>
 
== Motivation ==
===Example 1===
 
Consider a [[normed vector space]] ''X'', with the norm ||·||. Let ''K'' be the unit sphere in ''X''. Define a function ''p : X →'' '''R''' by
:<math>p(x) = \inf \left\{r > 0: x \in r K \right\}. </math>
 
One can see that <math>p(x) = \|x\|</math>, i.e. ''p'' is just the norm on ''X''. The function ''p'' is a special case of a Minkowski functional.
 
=== Example 2===
 
Let ''X'' be a vector space without topology with underlying scalar field '''K'''. Take ''φ ∈  X' '', the algebraic dual of ''X'', i.e. ''φ : X →'' '''K''' is a linear functional on ''X''. Fix ''a > 0''. Let the set ''K'' be given by
 
:<math>K = \{ x \in X : | \phi(x) | \leq a \}. </math>
 
Again we define
 
:<math>p(x) = \inf \left\{r > 0: x \in r K \right\}. </math>
 
Then
 
:<math>p(x) = \frac{1}{a} | \phi(x) |.</math>
 
The function ''p''(''x'') is another instance of a Minkowski functional. It has the following properties:
 
#It is ''subadditive'': ''p''(''x'' + ''y'') ≤ ''p''(''x'') + ''p''(''y''),
#It is ''homogeneous'': for all ''α'' ∈ '''K''', ''p''(''α x'') = |''α''| ''p''(''x''),
#It is nonnegative.
 
Therefore ''p'' is a [[seminorm]] on ''X'', with an induced topology. This is characteristic of Minkowski functionals defined via "nice" sets. There is a one-to-one correspondence between seminorms and the Minkowski functional given by such sets. What is meant precisely by "nice" is discussed in the section below.
 
Notice that, in contrast to a stronger requirement for a norm, ''p(x) = 0'' need not imply ''x = 0''. In the above example, one can take a nonzero ''x'' from the kernel of ''φ''. Consequently, the resulting topology need not be [[Hausdorff space|Hausdorff]].
 
== Definition ==
 
The above examples suggest that, given a (complex or real) vector space ''X'' and a subset ''K'', one can define a corresponding Minkowski functional
 
:<math>p_K:X \rightarrow [0, \infty)</math>
 
by
 
:<math>p_K (x) = \inf \left\{r > 0: x \in r K \right\},</math>
 
which is often called the gauge of <math>K</math>.
 
It is implicitly assumed in this definition that 0 ∈ ''K'' and the set {''r'' > 0: ''x'' ∈ ''r K''} is nonempty. In order for ''p<sub>K</sub>'' to have the properties of a seminorm, additional restrictions must be imposed on ''K''. These conditions are listed below.
 
#The set ''K'' being [[convex set|convex]] implies the subadditivity of ''p<sub>K</sub>''.
#[[Homogeneous function|Homogeneity]], i.e. ''p<sub>K</sub>''(''α x'') = |''α''| ''p<sub>K</sub>''(''x'') for all ''α'', is ensured if ''K'' is ''balanced'', meaning ''α K'' ⊂ ''K'' for all |''α''| ≤ 1.
 
A set ''K'' with these properties is said to be [[absolutely convex set|absolutely convex]].
 
=== Convexity of ''K'' ===
 
A simple geometric argument that shows convexity of ''K'' implies subadditivity is as follows. Suppose for the moment that ''p<sub>K</sub>''(''x'') = ''p<sub>K</sub>''(''y'') = ''r''. Then for all ''ε'' > 0, we have ''x'', ''y'' ∈ (''r + ε'') ''K'' = '' K' ''. The assumption that ''K'' is convex means '' K' '' is also. Therefore ½ ''x'' + ½ ''y'' is in '' K' ''. By definition of the Minkowski functional ''p<sub>K</sub>'', one has
 
:<math>p_K\left( \frac{1}{2} x + \frac{1}{2} y\right) \le r + \epsilon = \frac{1}{2} p_K(x) + \frac{1}{2} p_K(y) + \epsilon .</math>
 
But the left hand side is ½ ''p<sub>K</sub>''(''x'' + ''y''), i.e. the above becomes
 
:<math>p_K(x + y) \le  p_K(x) + p_K(y) + \epsilon, \quad \mbox{for all} \quad \epsilon > 0.</math>
 
This is the desired inequality. The general case ''p<sub>K</sub>''(''x'') > ''p<sub>K</sub>''(''y'') is obtained after the obvious modification.
 
'''Note''' Convexity of ''K'', together with the initial assumption that the set {''r'' > 0: ''x'' ∈ ''r K''} is nonempty, implies that ''K'' is [[absorbing set|''absorbent'']].
 
=== Balancedness of ''K'' ===
 
Notice that ''K'' being balanced implies that
 
:<math>\lambda x \in r K \quad \mbox{if and only if} \quad x \in \frac{r}{|\lambda|} K.</math>
 
Therefore
 
:<math>p_K (\lambda x) = \inf \left\{r > 0:  \lambda x \in r K \right\}
=  \inf \left\{r > 0:  x \in \frac{r}{|\lambda|} K \right\}
= \inf \left\{ | \lambda | \frac{r}{ | \lambda | } > 0:  x \in \frac{r}{|\lambda|} K \right\}
= |\lambda| p_K(x).
</math>
 
== See also ==
 
* [[Hadwiger's theorem]]
* [[Hugo Hadwiger]]
* [[Morphological image processing]]
 
==Notes==
{{reflist}}
 
==References==
* {{cite book | title=Minkowski Geometry | series=Encyclopedia of Mathematics and Its Applications | first=Anthony C. | last=Thompson | publisher=[[Cambridge University Press]] | year=1996 | isbn=0-521-40472-X }}
 
[[Category:Functional analysis]]
[[Category:Convex analysis]]

Revision as of 19:20, 9 February 2014

The author's name is Andera and she telephone psychic believes it sounds quite good. Office supervising is exactly where her primary income arrives psychic phone readings from but she's already applied for another one. Mississippi is where her house is but her spouse wants them to move. My spouse doesn't like it the way I do but what I truly like doing is caving but I don't have the time recently.

My website: free tarot readings