I'm trying to find the operator norm for $T: L^2([0,1])\to L^2([0,1])$, defined as $Tf(x)=\int_{[0,1]}|\sin(x-y)|^{-\alpha}f(y)dy$, where $0<\alpha<1$. Using an upper bound on $|\sin(x)|\geq |x|/2$, I am able to find that for any $x\in[0,1]$, $\int |\sin(x-
y)|^{-\alpha} dy\leq \frac{2^\alpha}{1-\alpha}=C$, and then by Theorem 6.18 in Folland, it should follow that $$\lVert Tf\rVert_2 \leq C\lVert f\rVert_2$$
However, I'm currently stuck on finding or closely estimating the operator norm, so any help on this would be appreciated.
Best Answer
Too long to be a comment:
This is just an elementary upper bound.
We have \begin{align*} \Vert Tf \Vert_{2}^2 &= \int_0^1 \vert Tf(x)\vert^2 dx =\int_0^1 \left\vert \int_0^1 \vert \sin(x-y)\vert^{-\alpha} f(y) dy \right\vert^2 dx \\ &\leq \int_0^1 \left( \int_0^1 \vert \sin(x-y)\vert^{-\alpha} \vert f(y)\vert dy \right)^2 dx \\ &= \int_0^1 \left(\int_0^1 \vert\sin(x-y_1)\vert^{-\alpha} \vert f(y_1) \vert dy_1\right)\left( \int_0^1 \vert\sin(x-y_2)\vert^{-\alpha} \vert f(y_2) \vert dy_2 \right) dx \\ &= \int_0^1 \int_0^1 \int_0^1 \vert\sin(x-y_1)\vert^{-\alpha} \vert f(y_1) \vert \cdot \vert\sin(x-y_2)\vert^{-\alpha} \vert f(y_2) \vert dy_1 dy_2 dx. \end{align*} Using Cauchy-Schwarz and then Tonelli's theorem we get \begin{align*} \Vert Tf \Vert_{2}^2 &\leq \left( \int_0^1 \int_0^1 \int_0^1 \vert\sin(x-y_1)\vert^{-\alpha} \vert\sin(x-y_2)\vert^{-\alpha} \vert f(y_1) \vert^2 dy_1 dy_2 dx\right)^{1/2} \\ &\qquad \times \left( \int_0^1 \int_0^1 \int_0^1 \vert\sin(x-y_1)\vert^{-\alpha} \cdot \vert\sin(x-y_2)\vert^{-\alpha} \vert f(y_2) \vert^2 dy_1 dy_2 dx\right)^{1/2} \\ &=\left( \int_0^1 \int_0^1 \int_0^1 \vert\sin(x-y_1)\vert^{-\alpha} \vert\sin(x-y_2)\vert^{-\alpha} \vert f(y_1) \vert^2 dy_2 dx dy_1\right)^{1/2} \\ &\qquad \times \left( \int_0^1 \int_0^1 \int_0^1 \vert\sin(x-y_1)\vert^{-\alpha} \cdot \vert\sin(x-y_2)\vert^{-\alpha} \vert f(y_2) \vert^2 dy_1 dx dy_2\right)^{1/2} \\ &=\int_0^1 \int_0^1 \int_0^1 \vert\sin(x-s)\vert^{-\alpha} \cdot \vert\sin(x-t)\vert^{-\alpha} \vert f(t) \vert^2 ds dx dt. \end{align*} After the change of variables $w=s-x$ we get \begin{align*} \Vert Tf \Vert_{2}^2 &\leq \int_0^1 \int_0^1 \int_{-x}^{1-x} \vert \sin(w)\vert^{-\alpha} \cdot \vert \sin(x-t) \vert^{-\alpha} \cdot \vert f(t)\vert^2 dw dx dt \\ &\leq \left( \sup_{v\in [0,1]} \int_{-v}^{1-v} \vert \sin(w)\vert^{-\alpha} dw \right) \int_0^1 \int_0^1 \vert \sin(x-t) \vert^{-\alpha} \cdot \vert f(t)\vert^2 dx dt. \end{align*} Repeating this last step for the integration in $x$ we get \begin{align*} \Vert Tf \Vert_2^2 \leq \left( \sup_{v\in [0,1]} \int_{-v}^{1-v} \vert \sin(w)\vert^{-\alpha} dw \right)^2 \Vert f \Vert_2^2. \end{align*} Hence, we get \begin{align*} \Vert T f\Vert_2 \leq \left( \sup_{v\in [0,1]} \int_{-v}^{1-v} \vert \sin(w)\vert^{-\alpha} dw \right) \Vert f \Vert_2 \end{align*} and therefore \begin{align*} \Vert T \Vert \leq \sup_{v\in [0,1]} \int_{-v}^{1-v} \vert \sin(w)\vert^{-\alpha} dw. \end{align*} As the map \begin{align*} G:[0,1] \rightarrow \mathbb{R}, v \mapsto \int_{-v}^{1-v} \vert \sin(w)\vert^{-\alpha} dw \end{align*} is continuous and $[0,1]$ is compact, the maximum is attained at some point $v_0\in [0,1]$. In fact we have that the supremum is attained for $v_0=1/2$.