Suppose \(f\in C^1([0,2])\) and \[ f(0)=f'(0)=f(2)=f'(2)=0. \] Prove that for every \(\varepsilon>0\) there exists \(T_\varepsilon\) such that for all \(t>T_\varepsilon\), \[ \left|\int_0^2 f(x)e^{itx}\,dx\right|\le \frac{\varepsilon}{t}. \]
Solution
Since \(f(0)=f(2)=0\), integration by parts gives, for \(t\ne0\), \[ \int_0^2 f(x)e^{itx}\,dx =-\frac1{it}\int_0^2 f'(x)e^{itx}\,dx. \] Because \(f'\in C([0,2])\subset L^1([0,2])\), the Riemann–Lebesgue lemma yields \[ \int_0^2 f'(x)e^{itx}\,dx\longrightarrow0 \qquad(t\to\infty). \] Hence for the given \(\varepsilon>0\) there exists \(T_\varepsilon\) such that \[ \left|\int_0^2 f'(x)e^{itx}\,dx\right|<\varepsilon \qquad(t>T_\varepsilon). \] Therefore, for \(t>T_\varepsilon\), \[ \left|\int_0^2 f(x)e^{itx}\,dx\right| =\frac1t\left|\int_0^2 f'(x)e^{itx}\,dx\right| <\frac{\varepsilon}{t}. \] This proves the claim.