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$T_kf=\int K(x,y)f(y)dy$

where $K(x,y)=\frac{\phi(x)\phi(y)}{|x-y|^{n-\alpha}}$

$\phi(x)$ is a smooth function on a compact support. $f$ is defined on $R^n$ and $K$ is defined on $R^n\times R^n$

Show that $T_k$ is compact from $L^p$ to $L^q$ when $p and $\frac1r = \frac1q - \frac\alpha n$.

I believe this exercise is an application of Hardy's fractional inequality and Ascoli-Arzela Theorem

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    I don't see how this question has anything to do with Hilbert Spaces.2012-12-07

1 Answers 1

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I am really confused by the $|n-\alpha|$ constraint, so I hope its ok if I just assume that $K(x,y) = \frac{\phi(x)\phi(y)}{|x-y|^{\beta}}$.

If I'm not mistaken then for $p,q$ with $1/p + 1/q = 1/r$ and for $\beta < 1/q$ one has that $T_K$ given by $ f\mapsto \int K(x,y) f(y) dy $ is a well defined operator from $L^p$ to $L^r$ and in operator norm bounded by $||K||_q$. Now let $\chi_n(x,y)$ be a sequence of simple functions that converge to $K(x,y)$ in $L^q$. Then $T_{\chi_n} $ given by
$ f\mapsto \int \chi_n(x,y) f(y) dy $ converges to $T_K$ w.r.t. the operator norm. If we could show that all the $T_{\chi_n} $ are compact operators then it would follow that $T_K$ is compact since the set of compact operators is closed in the operator norm topology. But since $\chi_n$ is just a linear combination of indicator functions, $T_{\chi_n}$ is just a linear combination of functionals that look like $ f\mapsto \int_A f(y) dy. $ This shows that $T_K$ is compact.