Lecture 9
Functions (Part II)
Limits of Sequences of functions
When we talk about limits of functions, we have to be careful. We must
define what we mean by
|| fn-f || -> 0
There are several obvious choices:
- || fn-f || = max x in D |fn(x)-f(x)|
- || fn-f || =
[
|fn(x)-f(x)| 2dx]
1/2
- || fn-f || =
[ |fn(x)-f(x)| pdx]
1/p, 1<=p
Each of these preserves slightly different properties (and smoothness) of
functions. They are called
norms, and rely on an underlying notion of
distance. Norms satisfy the following three properties
- || x || = 0 if and only if x = 0
- || k x || = |k| ||x||, for all real k
- || x + y || <= ||x|| + ||y|| for all x, y
Extensions of functions
We can define a "function of a function" (more precisely, an operator).
As an example consider the derivative,
D = d/dx
or the integral
f(x) dx
These are examples of linear operators.
Finite Dimensional Approximations -
A very important application of convergence of sequences of functions.
Suppose we want to solve L[u]=f, where u and f are functions, and
L is linear. Suppose we can find a solution of different problem,
L[un]=fn which is finite dimensional. Furthermore,
suppose that fn -> f as n increases. Does
un -> u, a solution of the original problem?
Examples - finite difference solutions, or fourier series, or finite
elements, or ...
Problems -
Moderately Hard Problem: Does the linear operator have the
property that L[f] = k f, where k is real or complex?
If so, it has an eigen-value and an eigen-function.
In the case of D, f is an eigen function if f'(x) = k f(x).
which clearly has solutions of the form f(x)=ekx
Unsolved Problem
Invariant Subspace Problem. Does a linear operator
L have an invariant suspace A, that is, is L[f] in A for
all f in A?.
In this problem, we ignore the trivial subpsaces of {0} and A itself.