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Let $\phi(t,s)$ be a real-valued function smooth away from the diagonal, and equal to 0 on the diagonal. Assume that $0\le \phi(t,s)\le |t-s|$ for $t,s\in \mathbb{R}$. Let $$T_\lambda f(t)=\int \frac{\sin(\lambda\phi(t,s))}{|t-s|}f(s)ds,\ \lambda\gg1$$ Then do we have the uniform bound $$||T_\lambda||_{L^2[0,1]\to L^2[0,1]}\le C,$$ where $C$ is a constant independent of $\lambda$?

For example, if $\phi(t,s)=|t-s|$, it is well known that the answer is YES, by the multiplier theorem. Indeed, $\frac{\sin(\lambda(t-s))}{t-s}$ is essentially the Dirichlet kernel for the partial summation of one-dimensional Fourier series. But I don't know whether this simple generalization has been considered before.

Any comments are welcome;) I will be grateful if you can point out some related references to me. Thanks!

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    $\begingroup$ Isn't it false if $\phi$ is constant away from the diagonal ? $\endgroup$
    – Simon Henry
    Commented Nov 16, 2016 at 16:26
  • $\begingroup$ @SimonHenry "Assume that $0\le \phi(t,s)\le |t-s|$" $\endgroup$
    – Mr.right
    Commented Nov 16, 2016 at 16:29
  • $\begingroup$ That does not prevent $\phi$ from being constant 'away from the diagonal', like $\phi(t,s)= min(1,|t-s|)$. $\endgroup$
    – Simon Henry
    Commented Nov 16, 2016 at 16:31
  • $\begingroup$ @SimonHenry I am not sure... We can only consider the operator on $L^2[0,1]$, for simplicity. $\endgroup$
    – Mr.right
    Commented Nov 16, 2016 at 16:44
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    $\begingroup$ what if $\varphi(t,s)=\max(t-s,0)$? $\endgroup$
    – Fedor Petrov
    Commented Nov 16, 2016 at 16:54

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