One-dimensional quadratic oscillatory amplitude bound

Rescale $u=Rt$ by [F3], put $\mathrm{\Lambda }=\lambda R^2$, and define $\tilde{H}(t)=H(Rt)$. If $\mathrm{\Lambda }\le 4$, [F4] bounds the integral by $2R\mathrm{\parallel }H{\mathrm{\parallel }}_{\mathrm{\infty }}\le 4\mathrm{\parallel }H{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}\sqrt{\lambda }$, so the claim holds. Assume $\mathrm{\Lambda }>4$, put $\delta =2{\mathrm{\Lambda }}^{-1\mathrm{/}2}<1$, and split $[-1,1]$ into $[-\delta ,\delta ]$, $[\delta ,1]$, and $[-1,-\delta ]$. By [F5], the integral is the sum over these subintervals; the central integral in the rescaled variable is at most $2\delta \mathrm{\parallel }H{\mathrm{\parallel }}_{\mathrm{\infty }}=4\mathrm{\parallel }H{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}\sqrt{\mathrm{\Lambda }}$ by [F4].

On $[\delta ,1]$, [F1] and [F2] integrate by parts using $(i\sigma \mathrm{\Lambda }t{)}^{-1}$ and $\tilde{H}$; the boundary at $1$ vanishes because $\tilde{H}$ is compactly supported in $(-1,1)$. The remaining boundary is at most $\mathrm{\parallel }H{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}(\mathrm{\Lambda }\delta )$. The integral term is at most $\frac{1}{\mathrm{\Lambda }}(R\mathrm{\parallel }{H}^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{\delta }^1{t}^{-1}dt+\mathrm{\parallel }H{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{\delta }^1{t}^{-2}dt)\le \frac{R\mathrm{\parallel }{H}^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }H{\mathrm{\parallel }}_{\mathrm{\infty }}}{\mathrm{\Lambda }\delta }$, since both integrals are at most $1\mathrm{/}\delta$. [F4] justifies the complex integral norm bounds; the negative interval has the same estimates. With $\delta =2{\mathrm{\Lambda }}^{-1\mathrm{/}2}$, each boundary is at most $\mathrm{\parallel }H{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}(2\sqrt{\mathrm{\Lambda }})$ and each integral is at most $(R\mathrm{\parallel }{H}^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }H{\mathrm{\parallel }}_{\mathrm{\infty }})\mathrm{/}(2\sqrt{\mathrm{\Lambda }})$. By [F5], adding the central piece and two tails gives $(6\mathrm{\parallel }H{\mathrm{\parallel }}_{\mathrm{\infty }}+R\mathrm{\parallel }{H}^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }})\mathrm{/}\sqrt{\mathrm{\Lambda }}$ in the rescaled integral. Multiplication by $R$ and $\mathrm{\Lambda }=\lambda R^2$ give the claimed bound.