Let \(f:[0,2]\to\mathbb R\) be \(C^1\) and suppose \[ f(0)=f'(0)=f(2)=f'(2)=0. \] Prove that for every \(\varepsilon>0\) there exists \(t_\varepsilon>0\) such that \[ \left|\int_0^2 f(x)e^{itx}\,dx\right|\le\frac{\varepsilon}{t} \qquad(t\ge t_\varepsilon). \]
The argument uses only \(f(0)=f(2)=0\) and \(f'\in L^1([0,2])\).
Solution
For \(t>0\), integration by parts and the endpoint conditions on \(f\) give \[ \begin{aligned} \int_0^2 f(x)e^{itx}\,dx &=\left[\frac{f(x)e^{itx}}{it}\right]_0^2 -\frac1{it}\int_0^2 f'(x)e^{itx}\,dx\\ &=-\frac1{it}\int_0^2 f'(x)e^{itx}\,dx. \end{aligned} \] Since \(f'\) is continuous on \([0,2]\), it belongs to \(L^1([0,2])\). By the Riemann–Lebesgue lemma, \[ \int_0^2 f'(x)e^{itx}\,dx\longrightarrow0 \qquad(t\to\infty). \] Therefore, for the given \(\varepsilon>0\), there exists \(t_\varepsilon>0\) such that \[ \left|\int_0^2 f'(x)e^{itx}\,dx\right|\le\varepsilon \qquad(t\ge t_\varepsilon). \] Substituting into the integration-by-parts identity yields \[ \left|\int_0^2 f(x)e^{itx}\,dx\right| \le\frac{\varepsilon}{t} \qquad(t\ge t_\varepsilon). \]