Estimate norm of convolution operator

Question

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.

Answer 1

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 $v_0=1/2$.

Answer 2

Too long for a comment, but not a full answer.

This post suggests that the norm of a convolution operator $f\mapsto f*g$ on $\mathbb R$ is $\|g\|_{L^1}$ if $g\geq 0$ (in general, it is the $L^\infty$ norm of the Fourier transform of $g$). The main intuition behind it is that, by scaling symmetry, you can think of $g$ as being vey close to a Dirac delta multiplied by $\|g\|_{L^1}$. Most likely, the same is not true here, due to the boundedness of the domain which does not allow for arbitrary scaling of functions.

One estimate I can think of on the operator norm of $T$ is the following. By Riesz-Thorin theorem, one has the bound $$ \|T\|_{L^2\to L^2}\leq \|T\|_{L^1\to L^1}^{1/2}\|T\|_{L^\infty\to L^\infty}^{1/2}, $$ where $\|T\|_{L^p\to L^p}$ is the operator norm of the operator on $L^p(0,1)$. The norm $\|T\|_{L^1\to L^1}$ is $\|g\|_{L^1(-1/2,1/2)}$, where $g(x):=|\sin(x)|^{-\alpha}$. The same is true for the norm $\|T\|_{L^\infty\to L^\infty}$. So, by the above bound, one has $$ \|T\|_{L^2\to L^2}\leq \int_{-1/2}^{1/2}|\sin(x)|^{-\alpha}dx, $$ which is quite close to what you have done to show the boundedness of $T$, if not the same. The above estimate is likely not optimal, and I don't know if there is an easy way of finding a better estimate or compute the exact operator norm.

Another way of estimating the norm $\|T\|_{L^2\to L^2}$ is by using Schur's test, but the standard one with $q=p\equiv 1$ gives the exact same bound I wrote above.

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Edit 1. This operator is an integral Hilbert-Schmidt operator if $\alpha<1/2$. For this kind of operator, one also has the bound $$ \|T\|_{L^2\to L^2}\leq \int_{[0,1]\times[0,1]}|\sin(x-y)|^{-2\alpha}dxdy, $$

see e.g. this pdf, Theorem 8.2.1.

The first estimate I gave sounds better, but maybe the second one is better for some values of $\alpha$.

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Edit 2. The operator $T$ is a self-adjoint, non-negative compact operator on $L^2(0,1)$ for any $0<\alpha<1$. By general facts, this implies that all the eigenvalues of $T$ are non-negative real numbers, and $\|T\|_{L^2\to L^2}$ coincides with the largest eigenvalue of $T$. Therefore, if one has a way of finding or estimating the eigenvalues of $T$, this in principle gives an estimate on the operator norm. I don't see how to find the eigenvalues at the moment, but this has better hope of giving the sharp constant.

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Edit 3. I don't have time to write a proof of $\|T\|_{L^1\to L^1}=\|g\|_{L^1(-1/2,1/2)}$. I can give some heuristics though. For a function $f$ with $\|f\|_{L^1(0,1)}=1$, it is evident that in order to maximize $\|Tf\|_{L^1(0,1)}$, we can assume $f\geq 0$ a.e.. Now think of $f$ as an infinite linear combinations of Dirac deltas $\delta_{x_0}$ centered at $x_0\in [0,1]$. The operator $T$ applied to $\delta_{x_0}$ formally gives $$ T\delta_{x_0}(x)=|\sin(x-x_0)|^{-\alpha} $$ The norm $\|T\delta_{x_0}\|_{L^1(0,1)}$ is maximized by choosing $x_0=1/2$, and in that case we have $$ \|T\delta_{1/2}\|_{L^1(0,1)}=\|g\|_{L^1(-1/2,1/2)} $$ by a simple change of variables.

Coming back to $f$, since we want to maximize $\|Tf\|_{L^1(0,1)}$, the optimal way to do that is to choose $f$ in whose linear expansion in terms of Deltas, $\delta_{1/2}$ gives the main contribution (that is, $f$ approximates $\delta_{1/2}$ in some sense). This is because all the contributions coming from $\delta_{x_0}$ with $x_0$ far from $1/2$ will give a sub-optimal contribution to $\|Tf\|_{L^1}$. Then, by choosing $f_n(x)=\chi_{[1/2-1/n,1/2+1/n]}$, one verifies that $$ \|Tf_n\|_{L^1(0,1)}\to \|g\|_{L^1(-1/2,1/2)} $$ as $n\to\infty$. This suggests that my claim is correct.