Exercise: Linear operator on Hilbert space $H=L^2(-\pi,\pi)$

Question

On Hilbert space $H=L^2(-\pi,\pi)$ let be defined the operator \begin{equation} [Tf](x)=\cos^2 x\int_{-\pi}^{\pi}dy\space \sin(y)\space f(y). \end{equation} Find kernel, image, eigenvalues and eigenvectors of $T$. Verify if $T$ is bounded and, if it's true, determine the $T$ norm.

I've proceeded in this way:

KERNEL:

$f\in Ker(T) \iff [Tf](x)=0$, so from the T definition I've \begin{equation} I=\int_{-\pi}^{\pi}dy\space \sin(y)\space f(y)=0. \end{equation} The integration range is symmetric, so I need an even function to make the integral $I=0$. I've decided to use trigonometrical polynomials base to define a generic $f(x)$: \begin{equation} f(x)=\frac{1}{\sqrt{2\pi}}+\sum_{n=1}^\infty \frac{cos\space nx}{\sqrt\pi}+\sum_{n=1}^\infty\frac{sin\space nx}{\sqrt\pi}=c_0+\sum_{n=1}^\infty(c_n+s_n) \end{equation} At this point I've chosen this particular linear combination of $s_n$ and $c_n$ \begin{equation} f(x)=\frac{1}{\sqrt{2\pi}}+\sum_{n=1}^\infty \frac{\cos\space nx}{\sqrt\pi} \end{equation} because it makes $I=0$. Using the integral definition of scalar product I've rewritten $(Tf)$ as \begin{equation} (Tf)=\pi \frac{cos^2 x}{\sqrt\pi}\int_{-\pi}^{\pi}dy \frac{siny}{\sqrt\pi}f(y)=\pi \frac{cos^2 x}{\sqrt\pi}<s_1,f> \end{equation} so $<s_1,f>=0$ that means they are orthogonal. I've concluded that

$f(x) \in \ker(T) \iff f(x)=c_0+\sum_{n=1}^\infty c_n+\sum_{n=2}^\infty s_n$

IMAGE:

$g\in Im(T) \iff (Tf)=g$

so I've used again the relationship $I=<s_1,f>$, but this time must be nonzero and it's true just for $<s_1,s_1>$, so \begin{equation} g\in Im(T) \quad \text{if} \quad g=\sqrt\pi \cos^2 x \quad \text{or in general} \quad g=(costant)\cdot \cos^2 x \end{equation}

EIGENVALUES AND EIGENVETORS:

$\lambda$ is an eigenvalue if $(Tf)=\lambda f$, so

$\lambda=0$ if $(Tf)=0f$ so the eigenvectors are all the functions of the Kernel of T;

for $\lambda\neq 0$ I've considered the action of T and as we can see $(Tf)= costant \cdot <s_1,f>$ where is nonzero if $<s_1,f>=<s_1,s_1>$, but I want an $f(x)=cos^2 x$. I've concluded that there aren't nonzero eigenvalues because if I choose $f(x)=\cos^2 x$, the integral goes to zero.

T NORM:

$T$ is bounded if $\exists R>0$ with $R\in\mathbb{R}\ :\ \dfrac{\|Tf\|}{\|f\|}\le R$, at this point I think it's better to consider $\|Tf\|^2=<Tf,Tf>$ and $\|f\|^2=<f,f>$ using $f=s_1$, but I'm not sure about it and i don't know how to complete the exercise. Can you help me, please?

Another question: can you tell me if exists something like a .pdf or a book where can I find these kind of exercises? Thank you so much!

Answer 1

The operator may be written as the one-dimensional operator $$ (Tf)(x) = \langle f,\sin\rangle \cos^2(x) $$ where $\langle h,k\rangle=\int_{-\pi}^{\pi}h(x)\overline{k(x)}dx$ is the usual inner product on $L^2(-\pi,\pi)$.

• The kernel is every function in $L^2(-\pi,\pi)$ that is orthogonal to $\sin$.
• The image is the one-dimensional space $[\{ \cos^2 \}]$ that is spanned by the function $\cos^2$.
• The only eigenvector is a scalar multiple of $\cos^2$ because the range of $T$ consists of all scalar multiples of $\cos^2$.
• The only eigenvalue is $\lambda$ such that $$ T(\cos^2)=\langle \cos^2,\sin\rangle\cos^2=\lambda\cos^2 \\ \implies \lambda = \langle \cos^2,\sin\rangle = \int_{-\pi}^{\pi}\cos^2(x)\sin(x)dx=0. $$
• The operator is bounded because $$ \|Tf\|=|\langle f,\sin\rangle|\|\cos^2\| \le (\|\sin\|\|\cos^2\|)\|f\| \\ \implies \|T\| \le \|\sin\|\|\cos^2\| $$ And, $$ \|T\sin\|=\|\sin\|^2\|\cos^2\|\\\implies \|T\|=\|\sin\|\|\cos^2\|. $$

Answer 2

Preliminar:

It is well known that an orthonormal basis of $\mathcal{L}^2(0,1)$ is $\{e^{2\pi inx}\}_{n\in\mathbb{Z}}$, so by using a simple change of variables it follows that an orthonormal basis for $\mathcal{L}^2(-\pi,\pi)$ is $\{e^{in(x+\pi)}\}_{n\in\mathbb{Z}}$. Observing that $e^{in(x+\pi)}=\cos(n(x+\pi))+i\sin(n(x+\pi))$ and that the functions $\cos(n(x+\pi)),\sin(m(x+\pi))$ are orthonormal for any $n,m\in\mathbb{Z}$, we can introduce the orthonormal basis $\{\cos(n(x+\pi))\}_{n\in\mathbb{Z}}\cup\{\sin(m(x+\pi))\}_{m\in\mathbb{Z}}$, but by simple computations one can see that $$\cos(n(x+\pi))=(-1)^n cos(nx)\quad\text{ and }\quad\sin(m(x+\pi))=(-1)^m cos(mx),$$ so the set $$\mathcal{B}=\{\cos(nx)\}_{n\in\mathbb{Z}}\cup\{\sin(mx)\}_{m\in\mathbb{Z}}$$ is an orthonormal basis of $\mathcal{L}^2(-\pi,\pi)$.

Kernel:

Since $T=\langle \sin(x) , \cdot \rangle \cos^2(x)$ and $\sin(x)\in \mathcal{B}$, it follows that $$\ker T= \text{span}\{\sin(x)\}^\perp=\text{span}(\mathcal{B}\backslash\{\sin(x)\}).$$

Image:

Since $T=\langle \sin(x) , \cdot \rangle \cos^2(x)$, $$\text{Im}(T)=\text{span}\{\cos^2(x)\}.$$

Eigenvalues and eigenvectors:

Since $Tf=\langle \sin(x) , f \cdot \rangle \cos^2(x)=\lambda f$ can be satisfied just if $\lambda=0$ or $f\in \text{span}\{\cos^2(x)\}$, it follows that the eigenvalues are $\{0,\langle\sin(x),\cos^2(x)\rangle\}$, however $\langle\sin(x),\cos^2(x)\rangle=0$ so there is just one eigenvalue and its eigenspace is $$E_0=\ker T.$$