Suppose that \(f : (0, 1) \to \mathbf{R}\) is a continuous function with \(\int_0^1 |f(t)| \, dt < \infty\). Define \(g : (0, 1) \to \mathbf{R}\) by
\[ g(x) = \int_x^1 \frac{f(t)}{t} \, dt. \]
Show that \(\int_0^1 |g(x)| \, dx < \infty\).
Solution
1. For every \(x\in(0,1)\), the defining integral for \(g(x)\) isabsolutely convergent.
On the interval \([x,1]\) one has \(1/t\le1/x\). Therefore \[ \int_x^1\left|\frac{f(t)}t\right|\,dt \le \frac1x\int_x^1|f(t)|\,dt \le \frac1x\int_0^1|f(t)|\,dt <\infty. \] Thus \(g(x)\) is well-defined.
2. For every \(x\in(0,1)\),\[ |g(x)| \le \int_x^1\frac{|f(t)|}{t}\,dt. \]
This is the triangle inequality for the absolutely convergent integral from step 1.
3. One has the estimate\[ \int_0^1 |g(x)|\,dx \le \int_0^1 |f(t)|\,dt. \]
By step 2 and Tonelli’s theorem for the nonnegative integrand, \[ \begin{aligned} \int_0^1|g(x)|\,dx &\le \int_0^1\int_x^1\frac{|f(t)|}{t}\,dt\,dx\\ &= \int_0^1\int_0^t\frac{|f(t)|}{t}\,dx\,dt\\ &= \int_0^1 \frac{|f(t)|}{t}\,t\,dt\\ &= \int_0^1|f(t)|\,dt. \end{aligned} \] The right-hand side is finite by hypothesis.
4. Consequently,\[ \boxed{ \int_0^1|g(x)|\,dx<\infty. } \]
This follows immediately from step 3 and the assumed integrability of \(f\).
5. Q.E.D.
Step 4 is the required conclusion.