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Take an $x\in H\setminus \{0\}$. You need to show that $\langle x,Tx\rangle > 0$. Can you see that a certain projection with one-dimensional range could be helpful?


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Yes, let $X=L^{2}[0,1]$, and let $(Tx)(t)=tx(t)$, i.e., multiplication by $t$. This is the prototypical example in many regards. You don't have any eigenfunctions because $Tx=\lambda x$ would require $(t-\lambda)x(t)=0$ for a.e. $t\in[0,1]$ which means that $x=0$ a.e.. However, you have approximate eigenfunctions. For example, if $\lambda\in(0,1)$ and ...


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Let $H$ be a separable Hilbert space as desired. For simplicity assuming we are working over the real field. Then being an integral operator, we have $$ K: H\rightarrow H, (Kf)(x)=\int K(x,y)f(y)dy $$ Since $K$ is a symmetric compact operator, it is normal and can be diagonalized. Let us write its eigenvectors as $\phi_{j}$ with eigenvalue $\lambda_{j}$. ...


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You can always define an unbounded operator on the whole space $X$, as long as $X$ is infinite dimensional. Simply take any unbounded linear functional $\varphi : X \to \Bbb{K}$ (with $\Bbb{K} \in \{\Bbb{R}, \Bbb{C}\}$) and some $x_0 \in X \setminus \{0\}$ and define $T : X \to X, x \mapsto \varphi(x) \cdot x_0$. For the existence of $\varphi$, see On ...


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You are given that $$ (-2\pi i)Tf = \int_{-\pi}^{\pi}K(x-y)f(y)\,dy \\ =\int_{-\pi}^{\pi}(\pi\,\mbox{sgn}(x-y)-(x-y))f(y)\,dy\\ =\int_{-\pi}^{x}(\pi\,-(x-y))f(y)\,dy\\ +\int_{x}^{\pi}(-\pi-(x-y))f(y)\,dy\\ =(\pi-x)\int_{-\pi}^{x}f(y)\,dy+\int_{-\pi}^{x}yf(y)\,dy \\ -(\pi+x)\int_{x}^{\pi}f(y)\,dy+\int_{x}^{\pi}yf(y)\,dy. $$ ...


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First observation: For all $u,v\in\mathcal H$ and $j\ne k$ $$ \langle T_ju,T_k v\rangle=\langle T_j^*u,T_k^* v\rangle=0. $$ Thus the linear subspaces $T_j^*\mathcal H$, $j\in\mathbb Z$, are perpendicular to each other, and so are their closures. Set $$ Y=\bigoplus_{j\in\mathbb Z}\overline{T_j^*\mathcal H}. $$ This infinite direct sum contains elements of ...


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Let $P_{k}$ be the orthogonal projection onto the closure of the range of $T_{k}$. Then $P_{k}P_{k'}=P_{k'}P_{k}=0$ for $k\ne k'$ because $(T_{k}x,T_{k'}y)=(T_{k'}^{\star}T_{k}x,y)=0$ for all $x,y \in H$. Similarly, if $Q_{k}$ is the orthogonal projection onto the closure of the range of $T_{k}^{\star}$, then $Q_{k}Q_{k'}=0$ for $k\ne k'$. Furthermore, $$ ...


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See if you can extend your operator to some $L^2$ space containing your functions. Maybe your operator is even self adjoint and still compact. Use the result for compact self adjoint operators on Hilbert spaces to find a basis of eigenvectors in that $L^2$ space. Show that in fact your eigenfunctions are already in your original space. Show that any ...


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I think you are thinking about it backwards. Every Banach space is a vector space. Every vector space has a basis. If your eigenfunctions are: (1) linearly independent under the inner product (2) have the same dimension as the space. They must be an equivalent basis. However, neither (1) nor (2) are necessary conditions of eigenfunctions as far as I know. ...


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Compact operators on Banach spaces can in general be upper-triangularized---generalizing the Jordan canonical form. This means that the space is Schauder-spanned by the generalized eigenvectors. So what you would need to show is that, given any non-zero $\lambda \in \sigma(L)$, the finite dimensional subspace $\mbox{ker}(L - \lambda)$ is spanned by ...


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The following is a solution to part (c). One direction was proved in part (a). Therefore, I will only prove the other direction: assuming $\sum_{n = 1}^\infty |\lambda_n|^2 < \infty$, I will show that $T$ is a Hilbert-Schmidt operator (where $\lambda_n$ are the eigenvalues corresponding to some orthonormal basis of eigenvectors, such that $T(f) = \sum_{n ...


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There is no any relation between stability and connectivity. Take two non intersecting invariant sets ( it might be any 2 trajectories of the system, which in regular autonomous case do not intersect). The union of these trajectories is also an invariant set, which is not connected. I assume you question is more relevant when the invariant set is minimal. ...


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Since everything is taking place in the orthogonal complement of the nullspace of $A$, we can assume that the Hilbert space is seperable. Pick then an orthonormal basis of the range of $P$, say $\varphi_n$, such that $$ Px=\sum_n \langle x,\varphi_n\rangle\varphi_n. $$ From this form of $P$, it is not hard to see that $$ |\langle Px,y\rangle|\leq|\langle ...



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