$f''(x)\ge m>0$, show that $ |\int_0^{+\infty}\exp(if(t))\mathrm{d}t | \le 8/\sqrt{m}$

Question

How can this inequality be proven? I thought of writing $e^{if(t)}$ as $ \frac{1}{f'(t)} \cdot f'(t) e^{if(t)}$ to integrate by parts but I can't finish.

Answer 1

This inequality seems to be a consequence of the Van der Corput lemma, but the constants given by this lemma give $\frac{10}{\sqrt{m}}$ instead of $\frac{8}{\sqrt{m}}$. To get this better inequality, we shall adapt the proof of Van Der Corput lemma.

Step 1 : Let $a<b$, $f:[a,b]\to\mathbb{R}$ be continuously differentiable with $f''\ge 0$ such that $|f'| \ge \lambda>0$, then $$ \left|\int_a^b e^{if}\right| \le \dfrac{3}{\lambda}$$

Proof : Because $|f'|>0$ we can write : $$ \int_a^b e^{if} = \int_a^b \dfrac{1}{if'(t)}\cdot if'(t)e^{if(t)}\mathrm{d}t= \dfrac{e^{if(b)}}{if'(b)} - \dfrac{e^{if(a)}}{if'(a)} + \int_a^b \dfrac{f''(t)}{f'(t)^2}e^{if(t)}\mathrm{d}t$$ Taking module on both sides, and applying triangular inequality we get : $$ \left|\int_a^b e^{if}\right|\le \frac{1}{|f'(b)|} + \frac{1}{|f'(a)|} + \int_a^b \dfrac{f''(t)}{f'(t)^2}\mathrm{d}t$$ $$ \le \dfrac{2}{\lambda} + \dfrac{1}{f'(a)} - \dfrac{1}{f'(b)}\le \dfrac{3}{\lambda}$$ Which concludes the proof of step 1.

Step 2 :Let $a < b$, $f:[a,b]\to\mathbb{R}$ be twice continuously differentiable such that $f''\ge m >0$, And $f'(t)\neq 0$ for all $t$ then $$ \left|\int_a^b e^{if}\right| \le \dfrac{4}{\sqrt{m}}$$

Proof : Because $f'(t)\neq 0$ for all $t$, and that $f'$ is continous, then $f'$ is of constant sign on $[a,b]$. Suppose first that $f'>0$. Let $\alpha >0$, with $\alpha < b-a$, then if $x\in [a+\alpha, b]$: $$ f'(x) = f'(a) + \int_a^{x} f''(t)\mathrm{d}t\ge \alpha m$$ Then, by applying step 1 we get : $$ \left|\int_a^b e^{if(t)}\mathrm{d}t\right| \le \int_a^{a+\alpha} 1 + \dfrac{3}{\alpha m} = \alpha + \dfrac{3}{\alpha\cdot m}$$ If $\alpha \ge b-a$, then the inequality still holds as the integral is less than $b-a$.

Taking $\alpha = \frac{1}{\sqrt{m}}$ we get : $$ \left| \int_a^b e^{if}\right| \le \dfrac{4}{\sqrt{m}}$$ In the case where $f'<0$, the proof is the same but instead, we split the integral between $a$ and $b-\alpha$, and $b-\alpha, b$. So the proof of step 2 is done.

Step 3 : let $f:\mathbb{R}_+\to\mathbb{R}$ be twice continuously differentiable such that $f''\ge m>0$, then $$ \left|\int_0^{\infty} e^{if(t)}\mathrm{d}t\right|\le \dfrac{8}{\sqrt{m}}$$

Proof : first, it is needed to show that the integral converges. For $x$ big enough, $f'\ge 1$, so we can integrate by parts like in step 1 and get a converging integral. Now, one of two things can happen : Either $f'(t)\neq 0$ for all $t$, then we can apply step 2 directly by taking $a=0$ and letting $b\to+\infty$ to get the even better inequality $$ \left|\int_0^{\infty} e^{if(t)}\mathrm{d}t \right| \le \dfrac{4}{\sqrt{m}}$$

Or, their exists $c\ge 0$ such that $f'(c)=0$, because $f''>0$, this $c$ is unique, if $c=0$, we can just let $a\to 0$ and $b\to+\infty$ in step 2 to get the same inequality as before. Otherwise, let $\varepsilon > 0$, then : $$ \left| \int_0^{\infty} e^{if}\right| \le \left| \int_0^{c-\varepsilon} e^{if}\right| + 2\varepsilon + \left|\int_{c+\varepsilon}^{+\infty} e^{if}\right| \le \dfrac{8}{\sqrt{m}} + 2\varepsilon$$ Letting $\varepsilon \to 0$, we get what we want.