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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.

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    It might be nicer to look at $L^2$-functions. Let us for simplicity take $L^2[-\pi,\pi]$. A nice basis for this space is given by the set of functions $\{\sin(nx)\}_{n\in\mathbb{N}} \cup \{\cos(nx)\}_{n\in\mathbb{N}} \cup \{1\}$. Here all the trigonometric functions will map to zero, 1 will go to something non-zero, and you have a basis for the kernel.2011-02-12
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    @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

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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).

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    I have just started studying this part of linear algebra, so I am not sure what co-dimension is. So from what I understand, the kernel has infinitely many elements and not one (clearly), what about the basis for the kernel? Is it infinite, and can we write it out?2011-02-12
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    @InterestedQuest: The kernel is infinite dimensional. It is as hard to write down as basis for the kernel as it is to write down as basis for $C[0,1]$. Co-dimension one means that you would need to add one more vector to a basis of the kernel to get a basis of $C[0,1]$.2011-02-12
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    @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 kernel2011-02-12