Anal. Theory Appl., 32 (2016), pp. 396-404.
Published online: 2016-10
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Here we consider the following strongly singular integral $$T_{\Omega,\gamma,\alpha,\beta}f(x,t)=\int_{R^n} e^{i|y|^{-\beta}}\frac {\Omega(\frac{y}{|y|})}{|y|^{n+\alpha}}f(x-y,t-\gamma(|y|))dy,$$ where $\Omega\in L^p(S^{n-1}),$ $ p>1,$ $n>1,$ $\alpha>0$ and $\gamma$ is convex on $(0,\infty)$.
We prove that there exists $A(p,n)>0$ such that if $\beta>A(p,n)(1+\alpha)$, then $T_{\Omega,\gamma,\alpha,\beta}$ is bounded from $L^2(R^{n+1})$ to itself and the constant is independent of $\gamma$. Furthermore, when $\Omega\in C^\infty(S^{n-1})$, we will show that $T_{\Omega,\gamma,\alpha,\beta}$ is bounded from $L^2(R^{n+1})$ to itself only if $\beta>2\alpha$ and the constant is independent of $\gamma$.
Here we consider the following strongly singular integral $$T_{\Omega,\gamma,\alpha,\beta}f(x,t)=\int_{R^n} e^{i|y|^{-\beta}}\frac {\Omega(\frac{y}{|y|})}{|y|^{n+\alpha}}f(x-y,t-\gamma(|y|))dy,$$ where $\Omega\in L^p(S^{n-1}),$ $ p>1,$ $n>1,$ $\alpha>0$ and $\gamma$ is convex on $(0,\infty)$.
We prove that there exists $A(p,n)>0$ such that if $\beta>A(p,n)(1+\alpha)$, then $T_{\Omega,\gamma,\alpha,\beta}$ is bounded from $L^2(R^{n+1})$ to itself and the constant is independent of $\gamma$. Furthermore, when $\Omega\in C^\infty(S^{n-1})$, we will show that $T_{\Omega,\gamma,\alpha,\beta}$ is bounded from $L^2(R^{n+1})$ to itself only if $\beta>2\alpha$ and the constant is independent of $\gamma$.