A bilinear estimate with a simple one-dimensional oscillatory integral kernel

Question

Let $f\in L^{p}(\mathbb{R})$, $1\leq p\leq 2$. I am trying to show that $$\int_{\mathbb{R}}\int_{\mathbb{R}} \,K(y,z)\, \frac{f(y)f(z)}{y^{\frac{1}{2\,p^{\prime}}}\,z^{\frac{1}{2\,p^{\prime}}}}\,dy\,dz\leq C\,\epsilon\, \|f\|_{L^{p}(\mathbb{R})}^{2},\tag{$*$}$$ for some $1\leq p \leq 2$, where $$K(y,z):=\int_{1}^{1+\epsilon}\,e^{i (y-z) t}\,t^{-\frac{1}{2}}\,dt.$$

Please note that I don't know if the estimate $(*)$ is true for any such values of $p$.

By Holder's inequality, the trivial bound $|k(y,z)|\leq C \epsilon$ is enough to obtain $(*)$ with the integral on $\mathbb{R}\times \mathbb{R}$ replaced by a bounded cube, say, $[-N,N]\times [-N,N]$, with $N>1$ fixed.

We may then consider the problem: $$\int_{|y|\gg1,\,|z|\gg1} \,K(y,z)\, \frac{f(y)f(z)}{y^{\frac{1}{2\,p^{\prime}}}\,z^{\frac{1}{2\,p^{\prime}}}}\,dy\,dz\leq C\,\epsilon\, \|f\|_{L^{p}(\mathbb{R})}^{2}\tag{$**$}.$$

Let us look a bit at the kernel: Let $\lambda>1$ be large and $\epsilon>0$ be small, and look at the oscillatory integral

$$I(\lambda):=\int_{1}^{1+\epsilon}\,e^{i \lambda t}\,t^{-\frac{1}{2}}\,dt.$$

Does $I$ satisfy an estimate of the form $$|I(\lambda)|\leq C \frac{\epsilon}{\lambda^{\beta}}$$ for some $\beta>0$ ?

Integration by parts is not useful if the amplitude is not of compact support in $(1,1+\epsilon)$ and I don't know how to go around that if at all possible. The best I can get is the estimate $$|I(\lambda)|\leq C\frac{|\sin{(\epsilon \lambda)}|}{\lambda}\leq C\min\{\epsilon, \, \frac{1}{\lambda}\},\tag1$$ for large enough $\lambda$. Are the first and second parts of $(1)$ the best one can do ? Does $(1)$ give the estimate $(**)$ ?

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A different approach: Using the fact that $\widehat{x^{-\frac{1}{2\,p^{\prime}}}}(\xi)=C\,\xi^{\frac{1}{2\,p^{\prime}}-1}$ in the sense of distributions, one could apply Fubini's theorem to work on the Fourier side as follows

$$\int_{\mathbb{R}}\int_{\mathbb{R}} \,K(y,z)\, \frac{f(y)f(z)}{y^{\frac{1}{2\,p^{\prime}}}\,z^{\frac{1}{2\,p^{\prime}}}}\,dy\,dz$$ $$=\int_{1}^{1+\epsilon}t^{-\frac{1}{2}}\,\int_{\mathbb{R}}\,e^{i t y}\frac{f(y)}{y^{\frac{1}{2\,p^{\prime}}}}dy\,\int_{\mathbb{R}}\, e^{-i t z}\frac{f(z)}{z^{\frac{1}{2\,p^{\prime}}}}dz \,dt$$ $$=C\int_{1}^{1+\epsilon}t^{-\frac{1}{2}}\,\int_{\mathbb{R}}\frac{\widehat{f}(x)}{|t-x|^{1-\frac{1}{2p^{\prime}}}}dx\, \,\int_{\mathbb{R}}\frac{{f}^{\vee}(y)}{|t-y|^{1-\frac{1}{2p^{\prime}}}}dy\,\,dt$$ $$=C \,\int_{\mathbb{R}}\,\int_{\mathbb{R}}\, \widehat{f}(x)\,{f}^{\vee}(y)\, H(x,y)\,dx\,dy,\tag2$$ where $$H(x,y):=\int_{1}^{1+\epsilon}t^{-\frac{1}{2}}\,\frac{1}{|t-x|^{1-\frac{1}{2p^{\prime}}}}\, \frac{1}{|t-y|^{1-\frac{1}{2p^{\prime}}}}\,dt$$ $$\approx\,\int_{1}^{1+\epsilon}\,\frac{1}{|t-x|^{1-\frac{1}{2p^{\prime}}}}\, \frac{1}{|t-y|^{1-\frac{1}{2p^{\prime}}}}\,dt.$$

Now, if ($|x|>2$, or $|x|<1/2$) and ($|y|>2$, or $|y|<1/2$), then $H$ is easy to estimate and $(*)$ can be obtained applying Holder's inequality to the right side of $(2)$ then the Hausdorff-Young inequality. We are only left with the case $1/2<|x|<2$ or $1/2<|y|<2$.

Let us start with the worst case, $1/2<|x|<2$ and $1/2<|y|<2$: By the triangle inequality, we have $$|x-y|\leq \,2\max\{|x-t|,|y-t|\}.$$ Integrating then using the mean-value theorem, we may estimate $$H(x,y)\leq\,\frac{1}{2}\, |x-y|^{{\frac{1}{2p^{\prime}}}-1} \int_{1}^{1+\epsilon}\, |t-y|^{{\frac{1}{2p^{\prime}}}-1}\,dt$$ $$\lesssim \epsilon\;|x-y|^{{\frac{1}{2p^{\prime}}}-1} \, |1-y|^{{\frac{1}{2p^{\prime}}}-1}.$$ Thus, by linearity, it suffices then to show that $$\int_{1/2<x<1}\int_{1/2<y<1} g(x)g(y)|x-y|^{{\frac{1}{2p^{\prime}}}-1} \, |1-y|^{{\frac{1}{2p^{\prime}}}-1}\,\lesssim\,1,$$ for $g\in L^{p^{\prime}}(\mathbb{R})$ with $\int_{1/2<x<1} |g(x)|^{\prime} dx=1$. But this last estimate is FALSE.

Have I been too sloppy with estimating/using the Kernel $H$ ?

What about the estimateS $(1)$ ? Are these optimal ? Then, is it possible to show that $$\int_{|y|\gg1,\,|z|\gg1} \frac{f(y)f(z)}{y^{\frac{1}{2\,p^{\prime}}}\,z^{\frac{1}{2\,p^{\prime}}}} \frac{|\sin{(\epsilon (y-z))}|}{|y-z|}\,dy\,dz\leq C\,\epsilon,\,$$ for $f\in L^{p}(\mathbb{R})$ with $\int_{|x|\gg1}|f(x)|^p \, dx=1$ ? This is probably false as well, but I can't find a simple counterexample here.

Answer 1

We can make more concrete what I suggested in my comment. Take $f(x)=x^{1/2p'}e^{-ix}$ on $1\le x\le 1/(10\epsilon)$ and $f=0$ otherwise. Then the LHS of (*) equals $$ \int_1^{1+\epsilon} |G(t)|^2\, \frac{dt}{\sqrt{t}}, \quad G(t)=\int_1^{1/(10\epsilon)} e^{iy(t-1)}\, dy . $$ Since $|G(t)|\gtrsim 1/\epsilon$ for $1\le t\le 1+\epsilon$, we have $\int_1^{1+\epsilon}|G|^2\gtrsim 1/\epsilon$.

On the other hand, $$ \|f\|^2_p \simeq \epsilon^{-1/p'-2/p} = \epsilon^{-1-1/p} , $$ so $\epsilon\|f\|^2_p\simeq \epsilon^{-1/p}$. This shows that the inequality does not hold for $p>1$.

For $p=1$, if we interpret $y^{-1/2p'}, z^{-1/2p'}=1$ (since $p'=\infty$ now), then the inequality is trivially true, since we have the bound $|K|\lesssim \epsilon$, as you already observed yourself.