Fredholm's alternative

In mathematics , Fredholm's alternative named after Ivar Fredholm is a result of Fredholm's theory . It can be expressed in different ways: as a theorem of linear algebra , as a theorem of integral equations, or as a theorem of Fredholm operators . In particular, it says that a complex number not equal to 0 in the spectrum of a compact operator is an eigenvalue .

Version of linear algebra

In a -dimensional vector space , exactly one of the following statements applies to a linear mapping : ${\displaystyle n}$ ${\displaystyle V}$ ${\displaystyle A\colon V\to V}$

1. For every vector in there is a vector in such that . In other words: is surjective. ${\displaystyle v}$ ${\displaystyle V}$ ${\displaystyle u}$ ${\displaystyle V}$ ${\displaystyle Au=v}$ ${\displaystyle A}$
2. There is an in with , that is: is not injective. ${\displaystyle u\neq 0}$ ${\displaystyle V}$ ${\displaystyle Au=0}$ ${\displaystyle A}$

Fredholm integral equations

Be an integral kernel . Consider the homogeneous Fredholm integral equation , ${\displaystyle K(x,y)}$

${\displaystyle \lambda \phi (x)-\int _{a}^{b}K(x,y)\phi (y)\,dy=0}$,

as well as the inhomogeneous equation

${\displaystyle \lambda \phi (x)-\int _{a}^{b}K(x,y)\phi (y)\,dy=f(x)}$.

Fredholm's alternative says that for a complex number , either the first equation has a nontrivial solution, or the second equation has a solution for any right-hand side . ${\displaystyle 0\neq \lambda \in \mathbb {C} }$ ${\displaystyle f(x)}$

A sufficient condition for this theorem to be true is the integrability of squares on the rectangle (where a and / or b may also be plus or minus infinity). ${\displaystyle K(x,y)}$ ${\displaystyle [a,b]\times [a,b]}$

Fredholm's alternative

statement

Be a compact operator on and be with . Then there is a Fredholm operator with Fredholm index 0. The Fredholm alternative is now: ${\displaystyle K\in K(X)}$ ${\displaystyle X}$ ${\displaystyle \lambda \in \mathbb {C} }$ ${\displaystyle \lambda \neq 0}$ ${\displaystyle Tx:=\lambda x-Kx}$

• Either both have the homogeneous equation

${\displaystyle \lambda x-Kx=0}$

as well as the adjoint equation

${\displaystyle \lambda x'-K'x'=0}$

only the trivial solution is zero and thus the equations are inhomogeneous

${\displaystyle \lambda x-Kx=y}$

and

${\displaystyle \lambda x'-K'x'=y'}$

clearly solvable,

• or the homogeneous equation

${\displaystyle \lambda x-Kx=0}$

and the adjoint equation

${\displaystyle \lambda x'-K'x'=0}$

have exactly linearly independent solutions (where the identical figure denotes) and thus the equation would be inhomogeneous ${\displaystyle n=\dim \ker(\lambda \operatorname {id} -K)<\infty }$ ${\displaystyle \operatorname {id} }$

${\displaystyle \lambda x-Kx=y}$

solvable if and only if . ${\displaystyle y\in (\ker(\lambda \operatorname {id} -K'))^{\bot }}$

In connection with the integral equations

Note that the delta distribution is the identity of the convolution . Let be a Banach space , for example, and be a Fredholm operator which through ${\displaystyle X}$ ${\displaystyle X=L^{2}([a,b])}$ ${\displaystyle T\colon X\to X}$

${\displaystyle T\phi (x)=\int _{a}^{b}\lambda \delta (x-y)\phi (y)-k(x,y)\phi (y)dy=\lambda \phi (x)-\int _{a}^{b}k(x,y)\phi (y)dy,}$

is defined, where must hold in order to obtain a Fredholm operator. Then there is a compact operator and you can see that this statement generalizes the statement about Fredholm's integral equations. ${\displaystyle k\in L^{2}([a,b])}$ ${\displaystyle \textstyle \int _{a}^{b}k(x,y)\phi (y)dy}$

Fredholm's alternative can then be formulated as follows: A is either an eigenvalue of or it is part of the resolvent set ${\displaystyle \lambda \neq 0}$ ${\displaystyle K}$

${\displaystyle \rho (K)=\{\lambda \in \mathbb {C} :(\lambda \operatorname {id} -K){\text{ beschränkt invertierbar}}\}}$.

literature

• Dirk Werner : functional analysis , Springer-Verlag, Berlin, ISBN 978-3-540-72533-6 .