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The spectrum of an operator, in this case an unbounded operator, greatly depends on the domain of definition and the surrounding Banach or Hilbert space. For example on the space $C^2[0,1]$ of twice differentiable functions with the usual Banach norm, the Laplacian is a bounded operator, and hence its spectrum is bounded. On the other hand, if we take the $... 2 In the finite dimensional case, we get a similar argument to work: the max is continuous (as a function of$V$), and the space of$k$-dimensional vector spaces is compact, so the minimum is reached. In the infinite dimensional case, the main difference is that$\mathcal{G} (k, \infty)$is no longer compact. It is still a metric space, and the application$...

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For the first part: If $V$ has dimension $k$ then we have $$\max_{\substack{x \in V \\ x \neq 0}} \frac{\langle \,x , Ax \rangle}{\langle \, x, x \rangle} = \max_{\substack{x \in V \\ x \neq 0}} \langle \, \frac{x}{\Vert x \Vert} , A \frac{x}{\Vert x \Vert} \rangle = \max_{\substack{x \in V \\ \Vert x \Vert = 1}} \langle \,x , Ax \rangle.$$ This is a maximum ...

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Suppose that $T-\lambda I$ is invertible, then it has trivial kernel and is bounded. Particularly you can solve the equation $$(T-\lambda I)x = y$$ for any $y\in\ell^p$. Writing $x = (x_m)$ and $y = (y_m)$, we see that $$(\alpha_m-\lambda)x_m = y_m,$$ i.e. $$x_m = \frac{1}{\alpha_m-\lambda}y_m.$$ Note that $\lambda=\alpha_m$ is a serious problem here. ...

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It is clear that $0$ is in the spectrum of $K$, since $K$ is compact. Let us address if it is an eigenvalue: if $$Kf=0,$$ this means we have $$\tag{1} 0=t\int_t^1f(s)\,ds+\int_0^ts\,f(s)\,ds.$$ Differentiating (via Lebesgue's Differentiation Theorem), $$\tag{2} 0=\int_t^1 f(s)\,ds-tf(t)+tf(t)=\int_t^1f(s)\,ds,\ \ \ \text{a.e.}.$$ Then, for any $v,t\in[0,... 1 The connection with the spectral measure$P$is $$P(E) = \chi_{E}(A).$$ So, for example, $$P[a,b] = \chi_{[a,b]}(A), \;\; P(a,b) = \chi_{(a,b)}(A) \\ P[a,b] = P(a,b) + P\{a\}+P\{b\}$$ The spectral measure$P\$ is regular in the strong topology, which gives you \begin{align} P[a,b]x & = \lim_{\...

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