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4

Take the Fourier transform. Then $\hat{Af}=\hat{f}\hat{h}$, and therefore we have turned convolution into multiplication. Now, for the multiplication operator, the spectrum is the range of $h$ (since $h$ is cont. range=essential range). As the Fourier transform is a unitary, we get that spectrum of $A$ is also the range of $h$. Edit: Let $M_h$ be the ...

1

Very rough outline: Suppose first that $A$ is self-adjoint. By the spectral theorem, we can write $A = \int \lambda dE$ where $E$ is the projection-valued spectral measure for $A$. Show that for sufficiently small $\epsilon$ we have that the projection $E([\epsilon, \infty))$ has infinite rank (if not, $A$ would be compact). Let $H_0$ be the image of this ...

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Maybe try working with the following: The class of compact operators is an ideal. In a separable Hilbert space, any ideal that contains a noncompact operator must equal the whole space.

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As it turns out, you do still have integral representation of the duality. If $1\leq p<\infty$, the dual of $L^p$ is isometrically isomorphic to $L^{p'}$ where $p'$ is the dual exponent, i.e. $\frac1p+\frac1{p'}=1$. (The dual exponent of $1$ is $\infty$.) The duality is given by $$\langle f,g\rangle=\int_0^1f(x)g(x)dx$$ for $f\in L^{p'}(0,1),g\in ... 2 Use Holder's inequality: $$\int_x^y |u(t)| \, dt \le \left( \int_x^y 1 \, dt \right)^{1/p'} \left( \int_x^y |u(t)|^p \, dt \right)^{1/p} \le |x-y|^{1/p'} \|u\|_p.$$ 5 Hint: If an operator can be approximated by finite rank operators, then it is compact. Try to show that$\|T-T_k\| \to 0$for a suitable chosen sequence$\{ T_k\}$of finite rank operators. Try with $$T_k: (x_1, x_2, \ldots)\mapsto (x_1, \frac{x_2}{2}, \ldots, \frac{x_k}{k}, 0, 0, \ldots ).$$ It is obvious that$rk(T_k)=k$, i.e., it is finite rank ... 1 Let's begin with showing that$g(x,y) := |x-y|^{a-1}$is in$L^2(\Omega)$where$\Omega := (0,1)\times(0,1)$. For this purpose we define for$\epsilon > 0$$$g_{\epsilon} := g(x,y) \cdot \mathbf{1}_{(0,x-\epsilon) \cup (x+\epsilon,1)}(y)$$ We calculate $$\int_{0}^{1} g_{\epsilon}^{2} dy =\int_{0}^{x-\epsilon} (x-y)^{2(a-1)} dy + \int_{x+\epsilon}^{1} ... 2 The Volterra integral operator V_K: L_2(0,1)\to L_2(0,1) which is given by$$ (V_K f)(x)=\int_{0}^{x}K(x,y) f(y)dy\qquad (f\in L_2(0,1))$$is of rank at most n if and only if the kernel is of the form$$ K(x,y)=g_1(x)\overline{h_1(y)}+\cdots+g_n(x)\overline{h_n(y)} \tag1$$for some functions g_j, h_j \in L_2(0,1) (1\leq j \leq n) such that$$ ... 0 I think that Proposition B.17 here might answer your question. To sum up: if$(e_n)_{n\geq 1}$is an orthonormal basis for your Hilbert space$H$, then for all trace class operators$T\in L^1(H)$, we consider the functional $$\varphi_T\colon S\mapsto\mathrm{tr}(TS)=\sum_{n=1}^\infty\langle TSe_n,e_n\rangle,\quad S\in B(H).$$ The functional$\mathrm{tr}$is ... 1 Note first that$w''^*w''$is a positive semidefinite operator such that$w''^*w''\leq \|w''^*w''\|$(on the right is a positive scalar times the identity operator). Since in$C^*$-algebra$\|w''^*w''\|=\|w''\|^2$we have$w''^*w''\leq \|w''\|^2$. Multiply this inequality from the left and the right side by a positive semidefinite$|u|\$ and you have the ...

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