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In mathematics, the '''Malgrange preparation theorem''' is an analogue of the [[Weierstrass preparation theorem]] for [[smooth function]]s. It was conjectured by [[René Thom]] and proved by {{harvs|txt=yes|first=B. |last=Malgrange|authorlink=Bernard Malgrange|year1=1962–1963|year2=1964|year3=1967}}.
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==Statement of Malgrange preparation theorem==
Suppose that ''f''(''t'',''x'') is a smooth  complex function of ''t''∈'''R''' and ''x''∈'''R'''<sup>''n''</sup> near the origin, and let ''k'' be the smallest integer such that
:<math>f(0,0)=0, {\partial f\over \partial t}(0,0)=0, \dots, {\partial^{k-1} f\over \partial t^{k-1}}(0,0)=0, {\partial^{k} f\over \partial t^{k}}(0,0)\ne0.</math>
Then one form of the preparation theorem states that near the origin ''f'' can be written as the product of a smooth function ''c'' that is nonzero at the origin and a smooth function that as a function of ''t'' is a polynomial of degree ''k''. In other words,  
:<math>f(t,x) = c(t,x)\left(t^k+a_{k-1}(x)t^{k-1}+\cdots+a_0(x) \right)</math>
where the functions ''c'' and ''a'' are smooth and ''c'' is nonzero at the origin.
 
A second form of the theorem, occasionally called the '''Mather division theorem''', is a sort of "division with remainder" theorem: it says that if ''f'' and ''k''  satisfy the conditions above and ''g'' is a smooth function near the origin, then we can write
:<math>g=qf+r\ </math>
where ''q'' and ''r'' are smooth, and as a function of ''t'', ''r'' is a polynomial of degree less than ''k''. This means that  
:<math>r(x)=\sum_{0\le j<k}t^jr_j(x)</math>
for some smooth functions ''r''<sub>''j''</sub>(''x'').
 
The two forms of the theorem easily imply each other: the first form is the special case of the "division with remainder"  form where ''g'' is ''t''<sup>''k''</sup>, and the division with remainder form follows from the first form of the theorem as we may assume that ''f'' as a function of ''t'' is a polynomial of degree ''k''.
 
If the functions ''f'' and ''g'' are real, then the functions ''c'', ''a'', ''q'', and ''r'' can also be taken to be real. In the case of the Weierstrass preparation theorem these functions are uniquely determined by ''f'' and ''g'', but uniqueness no longer holds for the Malgrange preparation theorem.
 
==Proof of Malgrange preparation theorem==
The Malgrange preparation theorem can be deduced from the Weierstrass preparation theorem. The obvious way of doing this does not work: although smooth functions have a formal power series expansion at the origin, and the Weierstrass preparation theorem applies to formal power series, the formal power series will not usually converge to smooth functions near the origin. Instead one can use the idea of decomposing a smooth function as a sum of analytic functions by applying a partition of unity to its Fourier transform.  
For a proof along these lines see {{harv|Mather|1968}} or {{harv|Hörmander|1983a|loc=section 7.5}}
 
==Algebraic version of the Malgrange preparation theorem==
The Malgrange preparation theorem can be restated as a theorem about [[module (mathematics)|modules]] over [[ring (mathematics)|rings]] of smooth, real-valued [[sheaf (mathematics)|germs]]. If ''X'' is a [[manifold]], with ''p''∈''X'', let ''C''<sup>∞</sup><sub>''p''</sub>(''X'') denote the ring of real-valued germs of smooth functions at ''p'' on ''X''. Let ''M''<sub>''p''</sub>(''X'') denote the unique [[maximal ideal]] of ''C''<sup>∞</sup><sub>''p''</sub>(''X''), consisting of germs which vanish at p.&nbsp;Let ''A'' be a ''C''<sup>∞</sup><sub>''p''</sub>(''X'')-module, and let ''f'':''X''&nbsp;→&nbsp;''Y'' be a smooth function between manifolds. Let ''q'' = ''f''(''p'').  ''f'' induces a ring homomorphism ''f''<sup>*</sup>:''C''<sup>∞</sup><sub>''q''</sub>(Y)&nbsp;→&nbsp;''C''<sup>∞</sup><sub>''p''</sub>(''X'') by composition on the right with ''f''. Thus we can view ''A'' as a ''C''<sup>∞</sup><sub>''q''</sub>(''Y'')-module. Then the Malgrange preparation theorem says that if ''A'' is a finitely-generated ''C''<sup>∞</sup><sub>''p''</sub>(''X'')-module, then ''A'' is a finitely-generated ''C''<sup>∞</sup><sub>''q''</sub>(''Y'')-module if and only if ''A''/''M''<sub>''q''</sub>(''Y'')A is a finite-dimensional real vector space.
 
==References==
*{{Citation |title=Stable Mappings and Their Singularities|last=Golubitsky |first=Martin|authorlink=Marty Golubitsky |coauthors=Victor Guillemin |year=1973| publisher = [[Springer-Verlag]]|isbn= 0-387-90073-X | series = Graduate Texts in mathematics 14 }}
*{{citation|first=L.|last= Hörmander|title=The analysis of linear partial differential operators I|series= Grundl. Math. Wissenschaft. |volume= 256 |publisher= Springer  |year=1983a|isbn= 978-3-540-00662-6}}
*{{citation|mr=0160234 |last= Malgrange|first= Bernard |title=Le théorème de préparation en géométrie différentiable I–IV |year= 1962–1963  |series=Séminaire Henri Cartan, 1962/63 |volume= 11–14|publisher= Secrétariat mathématique, Paris}}
*{{citation|mr=0182695
|last=Malgrange|first= Bernard
|title=The preparation theorem for differentiable functions. 1964 Differential Analysis, Bombay Colloq.|year= 1964 |pages= 203–208|publisher= Oxford Univ. Press|publication-place= London }}
*{{citation|mr=0212575
|last=Malgrange|first= B.
|title=Ideals of differentiable functions
|series=Tata Institute of Fundamental Research Studies in Mathematics|volume= 3 |publisher=Oxford University Press|publication-place= London|year= 1967 |pages=vii+106|unused_data=Tata Institute of Fundamental Research, Bombay;}}
*{{citation|mr=0232401
|last=Mather|first= John N.
|title=Stability of ''C''<sup>&infin;</sup> mappings. I. The division theorem.
|journal=Ann. Of Math. (2)|volume=87|year= 1968 |pages=89–104
|doi=10.2307/1970595|issue=1|publisher=The Annals of Mathematics, Vol. 87, No. 1|jstor=1970595
}}
 
[[Category:Singularity theory]]
[[Category:Theorems in analysis]]

Latest revision as of 23:44, 23 August 2014

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When an increased profile attorney might be the commitment of success you dream of, they might be also too hectic to deal with your situation. This could sometimes make you in the hands of a less seasoned legal professional with the business, or by itself occasionally when you need your attorney. Seek out an individual having a balance of energy and expertise.

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