To each function $f \in E$, we associate the function $U ( f )$ with derivative $U(f)^\prime(x) = \mathrm { e } ^ { x } \int _ { x } ^ { + \infty } f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$. Show that for all $f \in E$ and for all $x > 0$, $$\left| U ( f ) ^ { \prime } ( x ) \right| \leqslant \mathrm { e } ^ { x } \| f \| \left( \int _ { x } ^ { + \infty } \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t \right) ^ { 1 / 2 } \leqslant \| f \| \frac { \mathrm { e } ^ { x / 2 } } { \sqrt { x } }$$
To each function $f \in E$, we associate the function $U ( f )$ with derivative $U(f)^\prime(x) = \mathrm { e } ^ { x } \int _ { x } ^ { + \infty } f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$. Show that for all $f \in E$ and for all $x > 0$,
$$\left| U ( f ) ^ { \prime } ( x ) \right| \leqslant \mathrm { e } ^ { x } \| f \| \left( \int _ { x } ^ { + \infty } \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t \right) ^ { 1 / 2 } \leqslant \| f \| \frac { \mathrm { e } ^ { x / 2 } } { \sqrt { x } }$$