Find the eigenvalues and the eigenvectors of the linear operator
$T:C^\infty(0, 1)\to C^\infty(0, 1)$
$T(f)(x) = \frac{f'(x)}{x}, x \in (0,1) $
Using the definition :
$TF = \lambda F \iff \frac{f'(x)}{x} = \lambda f(x) $
At this point, though, I'm stuck.
Can anyone help ?
Best Answer
suppose $\lambda$ is an eigenvalue of $T.$ we will first show that $\lambda$ must be real.
$\lambda$ is an eigenvalue of $T$ means there is an $f \neq 0$ such that $\dfrac{f^\prime(x)}{x} = Tf(x) = \lambda f(x)$ since $f$ is not identically zero, there is an $x_0 \in (0,1)$ such that $f(x_0) \neq 0$ that is $\lambda =\dfrac{f^\prime(x_0)}{x_0f(x_0)}$ which is a real number.
now will solve for $f$ the equation $\dfrac{f^\prime(x)}{x} = \lambda f(x)$ separating the variables, we find $$\dfrac{df}{f} = \lambda x$$ on integrating gives you $$ f = e^{\lambda x^2/2}\text{ is an eigenfunction corresponding to the eigenvalue } \lambda, \text{ where $\lambda$ is any real number.}$$