To each function $f \in E$, we associate the endomorphism $U$ of $E$ defined for all $x > 0$ by $$U ( f ) ( x ) = \int _ { 0 } ^ { + \infty } \left( \mathrm { e } ^ { \min ( x , t ) } - 1 \right) f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$ It has been shown that $U(f)$ satisfies $y'' - y' = -f(x)/x$ on $\mathbb{R}_+^*$. Let $\lambda \in \mathbb { R } ^ { * }$. We assume that $\lambda$ is an eigenvalue of $U$. Let $f$ be an eigenvector associated with it. Show that $f$ is a solution of the differential equation $(E_{1/\lambda}) : x(y'' - y') + \frac{1}{\lambda} y = 0$.
To each function $f \in E$, we associate the endomorphism $U$ of $E$ defined for all $x > 0$ by
$$U ( f ) ( x ) = \int _ { 0 } ^ { + \infty } \left( \mathrm { e } ^ { \min ( x , t ) } - 1 \right) f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$
It has been shown that $U(f)$ satisfies $y'' - y' = -f(x)/x$ on $\mathbb{R}_+^*$. Let $\lambda \in \mathbb { R } ^ { * }$. We assume that $\lambda$ is an eigenvalue of $U$. Let $f$ be an eigenvector associated with it. Show that $f$ is a solution of the differential equation $(E_{1/\lambda}) : x(y'' - y') + \frac{1}{\lambda} y = 0$.