Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

could anybody please help me with the following task?

Consider the operator $$ Af(x):=\int\limits_{-\pi}^{\pi}\sin(x-y)f(y)\, dy, x\in [-\pi,\pi], f\in L_2(-\pi,\pi). $$ Show that the operator $A\in\mathcal{L}(L_2(-\pi,\pi))$ is normal. Determine the Singular Value Decomposition (SVD) of A.

In order to check if A is normal, I determined the adjoint operator with the result that $$ A^* f(x)=\int\limits_{-\pi}^{\pi}\overline{\sin(x-y)} f(y)\, dy. $$

Then I calculated $AA^*$ and $A^*A$. Here are my results: $$ AA^* f(x)=\int\limits_{-\pi}^{\pi}\sin(x-y) \int\limits_{-\pi}^{\pi}\overline{\sin(y-z)} f(z)\, dz\, dy $$ $$ A^*Af(x)=\int\limits_{-\pi}^{\pi}\overline{\sin(x-y)}\int\limits_{-\pi}^{\pi}\sin(y-z)f(z)\, dz\, dy $$ And this is identical because $\sin(x)=\overline{\sin(x)}$.

Could you please write me in a comment if it is okay until now?

Thanks a lot.

Edit:

Concerning the SVD:

Is it right, that I have to determine the eigenvalues (resp. eigenfunctions) of $AA^*$? Does one need the convolution theorem of the Fouriertransformation? Explicitly: To my opinion it is $$ AA^*f(x)=(\sin\star A^*f)(x) $$ and therefore $$ AA^* f(x)=\lambda f(x)\Leftrightarrow \mathcal{F}(\sin\star A^*f)=(2\pi)^{1/2}\mathcal{F}(\sin)\cdot\mathcal{F}(A^*f)=\lambda\mathcal{F}(f), $$ i.e. $$ (2\pi)^{1/2}\mathcal{F}(\sin)\cdot\mathcal{F}(A^*f)=\lambda\mathcal{F}(f). $$ Can one use that equation to determine now $\lambda$ resp. $f$?

Greetings

share|improve this question
    
Here is a related problem for normal matrices. –  Mhenni Benghorbal Feb 12 '13 at 11:03

2 Answers 2

Yes, Fourier transform is the way to go.

Recall the spectral theorem: a bounded operator on a Hilbert space is normal if and only it is unitarily equivalent to a multiplication operator. What you have here is a special case, with the implementing unitary being the Fourier transform $\mathcal{F}$.

As you noted, the convolution theorem says $\mathcal{F}$ maps convolution to multiplication. The Pontryagin dual in this case, $\mathbb{Z}$, is discrete. This means $A$ is diagonalizable, in particular normal. The $\mathcal{F}(h)$ for any $h \in L^1(\mathbb{T})$ is a sequence vanishing at infinity. So $A$ is compact and SVD makes sense.

In Fourier domain, your answer is very simple: $A$ is projection onto the basis element $\sin x$ (up to a scaling factor, possibly). So the eigenvalues of $A^*A$ is same as $A$: $1$ for the basis element projected on and $0$'s for every other element (again, you might need to adjust for a scaling factor that makes $\mathcal{F}$ unitary).

share|improve this answer

Note that, if $$ K \equiv \int_{a}^{b}k(x,t) dt $$ and

$$ H \equiv \int_{a}^{b}h(x,t) dt, $$ then

$$ KH \equiv \int_{a}^{b}k(x,z)h(z,t) dz.$$

Now, apply this result to your operators $A$ and $A^{*}.$

share|improve this answer
    
Why is KH that way? –  math12 Feb 12 '13 at 15:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.