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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 ...


1

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.


2

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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