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0

We will show that $T:L^{2}[1,\infty)\rightarrow L^{2}[1,\infty)$ is compact if and only if $-\infty<\alpha<0$. Observe that the integral kernel $K$ of $T$ is given by \begin{align*} K(x,y)=\dfrac{x^{\alpha}}{y}\chi_{[1\leq x\leq y<\infty)}, \tag{1} \end{align*} where $\chi$ is the characteristic function of the set $\left\{(x,y)\in [1,\infty)^{2} : ...


1

By Lax-Milgram there exists only one $\theta\in W_0^{1,2}$ such that \begin{equation} \int_\Omega\nabla\theta\cdot\nabla\phi dx=\int_\Omega f\phi dx \end{equation} for all $\phi\in W_0^{1,2}$. So in fact $T$ is well defined, linear and continuous from $W_0^{1,2}$ to $W_0^{1,2}$ with norm $|\theta|_{W_0^{1,2}}$. Obviously it is compact, since the range of $T$ ...


0

No. Since $\lambda_k\to 0$ you have a norm-convergent expansion. The $*$-operation is norm-continuous.


2

To show noncompactness, let $a> 0$ and set $f_a(x)= (a/2)^{-1/2}\chi_{(a/2,a)}.$ Then $f$ is a unit vector in $L^2.$ It is straightforward to see $$T(f_a)(x) = \begin{cases} 0,\, 0 < x < a/2 \\ (a/2)^{1/2}/x,\, x> a.\end{cases}.$$ Suppose $a<b/2.$ Because $T(f_b) = 0$ on $(0,b/2),$ $$T(f_a)(x)-T(f_b)(x) = (a/2)^{1/2}/x, a<x<b/2.$$ ...


6

I thought I would just offer an alternate solution using Minkowski's integral inequality applied to: $$ \| Tf \|_2 = \left\{ \int_0^{\infty} \left( \int_0^x x^{-1} f(t)\,dt \right)^{2}\,dx \right\}^{1/2} $$ But first we do a variable substitution $t \rightarrow xt$ so the inner integral is integrating over a fixed space: $$ \| Tf \|_2 = \left\{ ...


6

For the boundedness, observe that \begin{align*} \| Tf \|_2^2 &= \int_{0}^{\infty} |Tf(x)|^2 \, dx \\ &\leq \int_{0}^{\infty} \frac{1}{x^2} \int_{0}^{x} \int_{0}^{x} |f(u)||f(v)| \, dudvdx \\ &= \int_{0}^{\infty}\int_{0}^{\infty} |f(u)||f(v)| \left( \int_{u \vee v}^{\infty} \frac{dx}{x^2} \right) \, dudv \\ &= ...


3

Doing only 3). For each $n \in \mathbb{N}$ we define the operators $T_n: \ell^2 \to \ell^2$ as follows, for each $x=\{ x_j\}_j \in \ell^2$ put $$ T_n(x) = \left( \sum_{k=1}^\infty a_{1k} x_k, \cdots, \sum_{k=1}^\infty a_{nk} x_k, 0 ,\cdots \right) $$ By Hölder, for any $j\in \mathbb{N}$ $$ \left|\sum_{k=1}^\infty a_{jk} x_k\right| \leq \left( ...



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