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I have the following problem:

Determine the point spectrum and the continuous spectrum of the operator $$(A\psi )(x)=\theta (x)(\cos x)\psi (x)$$ on $L_2(\mathbb R,dx)$, where $\theta(x)=0$ for $x<0$, $\theta(x)=1$ for $x\geq 0$.

Now, for the point spectrum I guess I only have to find a $\psi(x)$ such that $(A\psi )(x)=\theta(x)(\cos x)\psi(x)=\lambda\psi (x)$ for some $\lambda\in\mathbb C$, but I don't know where to look. Or even if this is the way of dealing with problems like these. As for the continuous spectrum, I don't really know where to start.

Basically I've been trying to "guess" a good $\psi$ and looking for a clue in the Fourier transform/Fourier series of the equation, but I haven't come up with anything. I've only really dealt with eigenvalues/eigenvectors in the context of matrices and differential equations before.

Also, if anyone knows of any internet resources with problems like these (and preferably a few examples of how to solve them,) I'd be happy to know.

Thanks

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2 Answers

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Hint: Recall the spectrum $\sigma(A)$ is defined as $$\sigma(A)=\{\lambda:\ A-\lambda I\ \text{ is not invertible}\}.$$ So lets consider $A-\lambda I$, and find when this is non-invertible. Lets try an example first, what happens if $\lambda=2$. Can you find the inverse of $A-2I$? (yes) What is it an why did it work? Now find out what goes wrong with $|\lambda|\leq 1$, and conclude the spectrum is the closed interval $[-1,1]$. That is $$\sigma(A)=\{\lambda\in \mathbb{R}:\ |\lambda|\leq 1|\}.$$ The problem is that if $|\lambda|\leq 1$, there will exist $x$ such that $\left(A(\phi)\right)(x)=0$ for all $\phi$. Also, the $\theta (x)$ is of no importance.

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The problem isn't directly the points at which applying $(A-\lambda)$ involves multiplying by $0$, since the set of such points has measure $0$. The problem is all the nearby points where applying $(A-\lambda)$ involves multiplying by something small. –  Chris Eagle Mar 23 '11 at 16:14
    
Allright, I think I've got it. $(A-\lambda I)^{-1}=(\theta (x)\cos x-\lambda)^{-1}$ is undefined for some $x$ when $|\lambda|\leq 1$, so $\sigma(A)={\lambda\in\mathbb R:|\lambda|\leq 1}$. Also, since $A\psi(x)=0$ for $x<0$ for all $\psi(x)$ (because of $\theta (x)$,) the only possible eigenvalue is $0$, with eigenfunctions $\{\psi (x) : \psi(x)\neq 0 \Rightarrow x<0\}$. So the point spectrum consists of only $\lambda=0$, and the $0<|\lambda|\leq1$ are the continuous spectrum. Is that right? –  henrik Mar 23 '11 at 22:53
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I may be missing something, but what if you think about $\psi$ that are $0$ for positive $x$ and nonzero for negative $x$. Don't these work as eigenvectors for all $\lambda \in \mathbb{C}$ ?

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math.ucdavis.edu/~hunter/book/pdfbook.html This is a book that talks about spectral theory. It isn't the greatest, but it is free. –  Brian Mar 23 '11 at 15:49
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No, those are eigenvectors with eigenvalue 0. –  Chris Eagle Mar 23 '11 at 15:49
    
Oh... right. Sorry, I shouldn't have posted without thinking it through more. Well, there is one element of the point spectrum for you anyway podboq. –  Brian Mar 23 '11 at 15:53
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Thanks, Brian. The book you linked to is actually more verbose than what we use in my course, which is good for knuckleheads like me. –  henrik Mar 23 '11 at 22:58
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