1. ## Determine Eigenfunctions?

if g = T(f) and g(x) = integral(f, -infinity, infinity) then prove that every possible lambda is an eigenvalue for T and determine the eigenfunctions corresponding to lambda.

When I set up the integral = (lambda)(f), i guess I need to take the derivative? How would I take the derivative of an improper integral?

2. Originally Posted by hashshashin715
if g = T(f) and g(x) = integral(f, -infinity, infinity) then prove that every possible lambda is an eigenvalue for T and determine the eigenfunctions corresponding to lambda.

When I set up the integral = (lambda)(f), i guess I need to take the derivative? How would I take the derivative of an improper integral?
I am a little confused so let me see if I understand.

You are given a linear Transformation
$T:X \to \mathbb{R}$ where $X$ is some space of integrable functions on all of $\mathbb{R}$ and for $f \in X$
$\displaystyle T(f)=\int_{-\infty}^{\infty}f(x)dx$

Here is my point of confusion for every integrable $f$(if it is a function of 1 variable) this integral is a constant

$\displaystyle \int_{-\infty}^{\infty}f(x)dx$

But the only constant function that is integrable on all of $\mathbb{R}$ is the zero function $f(x) = 0$. This is the zero vector of your space so it cannot be an eigenvector!

3. What you have written is impossible. For any f(x), such that the integral converges, $\int_{-\infty}^\infty f(x)dx$ is a number so there are two ways to interpret this:
(1) that T(f) is a linear transformation from the set of functions integrable on $(-\infty, \infty)$ to the real numbers. But, in order to have eigenvalues and eigenvectors, a linear transformation must be from some vector space V to itself not from one vector space to another.
(2) that T(f) is a inear transfromation from the set of functions integrable on $(-\infty, \infty)$ to itself. But then T(f) is a constant function and a non-zero constant function is not in that set.

Did you mean $T(f)= \int_{-\infty}^x f(t)dt$? In that case, you have $\int_{-\infty}^x f(t)dt= \lambda f$, an "integral equation". But, yes, you can differentiate both sides, using the "fundamental theorem of Calculus" on the left to get the differential equation $f(x)= \lambda \frac{df}{dx}$ or $\frac{df}{dx}= \frac{1}{\lambda} f$. What are the solutions to that equation?

4. Ok, thanks all I wanted to know was if I can use the fundamental theorem of Calculus on the equation I gave in the OP, because I can get the right answer that way.

I didn't give the specifics because I really didn't think they mattered. But it seems they do (I never use them).