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Let $\lambda>1$, $x,y\in [0,1]$, and $K_\lambda(x,y)=(1+\lambda|x-y|)^{-2}e^{i\lambda|x-y|}$. Consider the operator $$T_\lambda f(x)=\int_0^1 K_\lambda(x,y)f(y)dy.$$ We want to precisely estimate the norm $\|T_\lambda\|_{L^\infty([0,1])\to L^1([0,1])}$.

First, it is clear that $\|T_\lambda\|_{L^\infty([0,1])\to L^1([0,1])}\le \int_0^1\int_0^1|K_\lambda(x,y)|dxdy\approx \lambda^{-1}$.

Is this bound optimal? It looks doubtful, as it does not exploit the rapid oscillations in the kernel. Any comments are welcome!

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  • $\begingroup$ Consider the characteristic function $f$ of the set of $x$ for which $\cos \lambda x>0.99$. Then $\langle T_\lambda f, f\rangle$ seems to be of order $1/\lambda$ $\endgroup$
    – Fedor Petrov
    Commented Sep 2, 2023 at 10:46
  • $\begingroup$ Oscillation of the form $e^{i( \phi(x) + \psi(y)}$ ( which your example morally has the form of) does not actually influence operator norms no matter how rapidly oscillating as it can be absorbed into the input or output function. One needs genuinely bilinear phases to get cancellation (which occurs for instance in higher dimensions in which yoyr integral is essentially a Bochner-Riesz operator). $\endgroup$
    – Terry Tao
    Commented Sep 2, 2023 at 15:35

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