I have the following SL problem:
$$ (x^{2} f')'+ \lambda f = 0 $$
where $ \lvert f(x) \rvert $ is bounded as $ x \rightarrow 0 $ and $ f(1) = 0 $
I have to show that the above problem has NO eigenvalues.
I know that this is a Cauchy-Euler ODE and set $ f = x^m $ and plugged it in.
I solved for $ m $ using the quadratic equation in terms of $ \lambda $ but I am not sure how to prove there are no eigenvalues.
There are solution $x^m$ of $(x^2 f')'+\lambda f=0$, where $$ m(m+1)+\lambda=0 \\ m^2+m+\lambda = 0 \\ (m+1/2)^2=1/4-\lambda \\ m=-1/2\pm\sqrt{1/4-\lambda}. $$ For convenience, normalize this solution at $x=1$ by imposing $f(1)=0$, $f'(1)=1$, which gives $$ f(x)=\frac{x^{-1/2+\sqrt{1/4-\lambda}}-x^{-1/2-\sqrt{1/4-\lambda}}}{2\sqrt{1/4-\lambda}} \\ = \frac{x^{\sqrt{1/4-\lambda}}-x^{-\sqrt{1/4-\lambda}}}{2\sqrt{x}\sqrt{1/4-\lambda}} \\ = \frac{e^{i\ln(x)\sqrt{\lambda-1/4}}-e^{-i\ln(x)\sqrt{\lambda-1/4}}}{2i\sqrt{x}\sqrt{\lambda-1/4}} \\ = \frac{\sin(\ln(x)\sqrt{\lambda-1/4})}{\ln(x)\sqrt{\lambda-1/4}}\cdot\frac{\ln(x)}{\sqrt{x}} $$ This function is not bounded near $x=0$, regardless of the value of $\lambda$.