Let $I(f)$ be defined as $\int_{0}^{1}f(x)dx$ from $C$[0, 1] to $\mathbb{R}$, where $C$[0, 1] is a linear space of all continuous functions from closed interval [0, 1] to $\mathbb{R}$. Is there a systematic way to find the kernel of $I$? Clearly, the zero function is there, as integral of $0$ from 1 to 0 will be equal to zero. Some trig functions might be in the kernel too (not sure if 1 in the integral is being interpreted as radian or degrees), but what would be an exhaustive way to find the basis of the kernel of $I$? Thanks.
What is the kernel of $I$ from $C$[0, 1] to $\mathbb{R}$?
4
$\begingroup$
linear-algebra
-
0@Dan Petersen: Note that this would not give a basis in the sense of linear algebra... And also there are obvious issues with the elements in $L^2$ not even being functions... However, the idea to take it as an illustration of how big the kernel is, is not bad, I guess! – 2011-02-13
1 Answers
4
Since you are looking at a surjective linear map from an infinite dimensional vector space to a one-dimensional vector space, the kernel is a subset in $C[0,1]$ of co-dimension one. Co-dimension one means that you would need to add one more vector to a basis of the kernel to get a basis of $C[0,1]$. This means that the kernel is very large.
For every function $f\in C[0,1]$ you can find a number $c\in\mathbb R$, such that $f-c\cdot 1$ is in the kernel (here $1$ denotes the constance function with value 1).
-
0@InterestedQuest: Wikipedia on [codimension](http://en.wikipedia.org/wiki/Codimension) and [cokernels](http://en.wikipedia.org/wiki/Coker_%28mathematics%29). Codimension and cokernel are useful terms when talking about infinite dimensional vector spaces because it's possible to have continuous linear maps with infinite-dimensional image that nonetheless have only a "small amount of stuff" outside the kernel – 2011-02-12