Let \(I\) denote the interval \((0,1)\). Suppose that \(f : I \to \mathbb{R}\) with \(\int_0^1 |f(t)| \, dt < +\infty\). Define \(g : I \to \mathbb{R}\) by
\[g(x) = \int_x^1 \frac{f(t)}{t} \, dt.\]
Show that \(g \in L^1(I)\).
Solution
1. Estimate the \(L^1\) norm by Tonelli’s theorem.
For \(x\in(0,1)\), \[ |g(x)| \le \int_x^1\frac{|f(t)|}{t}\,dt. \] Therefore, using Tonelli for the nonnegative integrand, \[ \begin{aligned} \int_0^1|g(x)|\,dx &\le \int_0^1\int_x^1\frac{|f(t)|}{t}\,dt\,dx\\ &=\int_0^1\frac{|f(t)|}{t} \left(\int_0^t dx\right)dt\\ &=\int_0^1|f(t)|\,dt. \end{aligned} \] The right-hand side is finite by hypothesis. Hence \[ \boxed{g\in L^1(0,1),\qquad \|g\|_1\le\|f\|_1.} \]