Convolution operator

Question

I'm trying to find the spectrum of the convolution operator and understand $$T : L_2 \left[ -\pi, \pi \right] \longrightarrow L_2\left[ -\pi, \pi \right],$$ $$ f(t) \longmapsto \int_{-\pi}^\pi \sin^2(t-s)f(s) \, \mathrm{d}s. $$

I think the best option would be to go over to the Fourier transform $F$ due to this answer enter link description here. But the Fourier transformation is defined in space $L_2 (\mathbb{R}) $ (as a limit on the Schwarz class) ($F : L_2 (\mathbb{R}) \longrightarrow L_2 (\mathbb{R}) $). Then we know that $L_2\left[ -\pi, \pi \right] \subset L_2 (\mathbb{R}) $ and $F \circ T \circ F^{-1} : F \left( L_2\left[ -\pi, \pi \right] \right) \longrightarrow F \left( L_2\left[ -\pi, \pi \right] \right)$.

In this case we get that the operator $F \circ T \circ F^{-1} $ is the multiplication operator $F \circ T \circ F^{-1} = M_{g}$, where $$g(x) = \int_{-\pi}^\pi \sin^2 (t) e^{-ixt } \, \mathrm{d}t$$

We know also that $\sigma(M_g) = \sigma \left(F \circ T \circ F^{-1} \right) =\sigma(T)$. But then we want to define spectrum of $T$. We know that in $ B \left(L_2 \left( \mathbb{R} \right) \right)$ spectrum of $M_g$ is essential range of $g$ (in this case is simply range).

But our space is not all $L_2\left( \mathbb{R} \right)$, we only have a part $ \operatorname{Im}(F) = F \left( L_2\left[ -\pi, \pi \right] \right) $ and i don't know how to show that in $B(\operatorname{Im}(F))$ operator $M_g$ has the same spectrum.

Sorry for my english and thank you very much!

Answer 1

Using standard trig identities: \begin{align} \cos(2(t-s)) & = \cos^2(t-s)-\sin^2(t-s) \\ & = 1-2\sin^2(t-s)\\ \sin^2( t -s) & = \frac{1-\cos(2(t-s))}{2} \\ & = \frac{1-\cos(2t)\cos(2s)-\sin(2t)\sin(2s)}{2}. \end{align} Therefore, $T$ has a simple form: $$ Tf = \frac{1}{2}\langle f,1\rangle 1 - \frac{1}{2}\langle f,\sin(2s)\rangle\sin(2t) - \frac{1}{2}\langle f,\cos(2s)\rangle\cos(2t). $$ Your operator $T$ reduces to a matrix operator on the subspace $M$ spanned by the orthonormal basis $\left\{\frac{1}{\sqrt{\pi}},\sqrt{\frac{2}{\pi}}\cos(2t),\sqrt{\frac{2}{\pi}}\sin(2t)\right\}$, and $T$ is $0$ on the orthogonal complement of $M$. These basis elements are eigenvectors of $T$. So the operator $T$ is diagonalized on this orthonormal basis. The spectrum is the union of these eigenvalues and $0$. A finite rank operator such as this one is always compact.

Answer 2

(I don't have time to check it carefully, but this is an idea of an approach)

Because the kernel of $T$ is in $L^2([0,1]\times[0,1])$, $T$ is compact. So its spectrum consists of eigenvalues (in general, it could be the case that $0$ is not an eigenavlue, but it is in this case). If $$ f(t)=\lambda \int_{-\pi}^{\pi} \sin^2 \left( t-s \right)f\left( s \right) \mathrm{d}s. $$ Since the convolution with a $C^\infty$ function is $C^\infty$, we deduce that $f\in C^\infty[0,1]$; in particular, it is bounded. Now \begin{align} \lambda^2 f(t)&=T^2f(t)=\int_{-\pi}^{\pi} \sin^2 \left( t-s \right)\,Tf\left( s \right) {d}s\\ \ \\ &=\int_{-\pi}^{\pi}\int_{-\pi}^{\pi} \sin^2 \left( t-s \right)\sin^2(s-r)\,f\left( r \right)\,dr\, {d}s\\ \ \\ &=\int_{-\pi}^{\pi}\,f\left( r \right)\int_{-\pi}^{\pi} \sin^2 \left( t-s \right)\sin^2(s-r)\,ds\, {d}r\\ \ \\ &=\frac\pi4\,\int_{-\pi}^{\pi}\,f\left( r \right)(3-2\sin^2(t-r))\, {d}r\\ \ \\ &=\frac{3\pi}4\int_{-\pi}^\pi f(r)\,dr-\frac\pi2\,Tf(t)\\ \ \\ &=\frac{3\pi}4\int_{-\pi}^\pi f(r)\,dr-\frac{\lambda\pi}2f(t). \end{align} So $$ \left(\lambda^2+\frac{\lambda\pi}2\right)f(t)=\frac{3\pi}4\int_{-\pi}^\pi f(r)\,dr. $$ If $f$ is not constant, the above forces $\lambda^2+\frac{\lambda\pi}2=0$, which gives eigenvalues $\lambda=0$ and $\lambda=-\pi/2$.

If $f$ is constant, we may assume that $f=1$, and then $$ \lambda=\frac1{ \int_{-\pi}^{\pi} \sin^2 \left( t-s \right)\,ds}=\frac1\pi. $$