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capitalone
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Let $\lambda>1$, $x,y\in [0,1]$, and $K_\lambda(x,y)=(1+\lambda|x-y|)^{-2}e^{i\lambda|x-y|}$, $a\in C_0^\infty([0,1])$. 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!

Let $\lambda>1$, $x,y\in [0,1]$, and $K_\lambda(x,y)=(1+\lambda|x-y|)^{-2}e^{i\lambda|x-y|}$, $a\in C_0^\infty([0,1])$. 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!

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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capitalone
capitalone
  • 31
  • 2

$L^\infty-L^1$ norm of an oscillatory integral operatorsoperator

Let $\lambda>1$, $x,y\in [0,1]$, and $K_\lambda(x,y)=(1+\lambda|x-y|)^{-2}e^{i\lambda|x-y|}$, $a\in C_0^\infty([0,1])$. 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!

$L^\infty-L^1$ norm of oscillatory integral operators

Let $\lambda>1$, $x,y\in [0,1]$, and $K_\lambda(x,y)=(1+\lambda|x-y|)^{-2}e^{i\lambda|x-y|}$, $a\in C_0^\infty([0,1])$. 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.

$L^\infty-L^1$ norm of an oscillatory integral operator

Let $\lambda>1$, $x,y\in [0,1]$, and $K_\lambda(x,y)=(1+\lambda|x-y|)^{-2}e^{i\lambda|x-y|}$, $a\in C_0^\infty([0,1])$. 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!

Source Link
capitalone
capitalone
  • 31
  • 2

$L^\infty-L^1$ norm of oscillatory integral operators

Let $\lambda>1$, $x,y\in [0,1]$, and $K_\lambda(x,y)=(1+\lambda|x-y|)^{-2}e^{i\lambda|x-y|}$, $a\in C_0^\infty([0,1])$. 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.