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The question is as follows:

Find the spectrum of the operator $A : L_2[-1,1] \to L_2[-1,1]$ defined as $$(Ax)(t) = \int_{-1}^{1} t^2s x(s) ds$$.

$\textbf{Some definition and observation:}$

We have that the spectrum of $A$ is $\sigma(A) = \{ \lambda \in \mathbb{C} \mid A -\lambda I \text{ is not invertible} \}.$

I think $A$ is compact. Because $k(t,s)=t^2 s \in L_2[-1,1] \times L_2[-1,1]$ and $\int_{-1}^{1} \int_{-1}^{1} \left| t^2 s \right|^2 dt ds = 0 $.

But I do not know how to find its spectrum.

Can someone give me an idea on how to find it?

Thanks!

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Suppose that $Ax = \lambda x$ for some $0 \neq x \in L_2[-1,1]$. Then \begin{equation} \tag{1} \lambda x(t) = t^2 \int^1_{-1}sx(s)ds. \end{equation} Multiplying by $t$ and integrating gives $$\lambda \int^1_{-1} tx(t) dt = 0.$$ This equation is satisfied iff $$\lambda = 0 \,\,\,\,\,\, \text{ or } \,\,\,\,\,\, \int^1_{-1} tx(t) dt = 0.$$ However, $\lambda = 0$ in turn implies $$\int^1_{-1} tx(t) dt = 0$$ by $(1)$ and likewise if $$\int^1_{-1} tx(t) dt = 0$$ (and $x(t) \neq 0$), then $\lambda = 0$ by $(1)$.

Thus it seems to me that $\sigma(A) = \{0\}$ and that the set of eigenvectors corresponding to the eigenvalue $0$ is the set $$E = \left\{ x \in L_2[-1,1] \,\,:\,\, \int^1_{-1} tx(t) dt = 0 \right\}.$$ This set contains all even functions in $L_2[-1,1]$, for example.

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  • $\begingroup$ Many thanks! Clever idea! $\endgroup$ – user510716 Dec 12 '17 at 8:20

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