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I need to show that $\int_{-\infty}^{\infty}|x-y|^{-p}f(y)\text{d}y<\infty$ for a.e. $x$ when $f\in L^1(\mathbb{R})$ and $p\in (0,1)$. Is the convolution still finite a.e. if instead of $|\cdot|^{-p}$ we use an arbitrary function in $L^1_{loc}(\mathbb{R})$?

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2 Answers 2

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Hint: Show that $g(x) = \int_{x-1}^{x+1} |x - y|^{-p}f(y)\,dy$ is finite $a.e.$ by showing that $g(x)$ is in $L^1$. Showing the rest of $\int_{-\infty}^{\infty} |x - y|^{-p}f(y)\,dy$ is finite for all $x$ can be done by directly bounding the integrand.

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I think this is an elegant method. –  timur Aug 16 '12 at 15:38

Supposing that $f$ is continuous at $0$, the reason for the integral $\int_{-\infty}^{\infty}|y|^{-p}f(y)\text{d}y$ being finite is that $|y|^{-p}$ is integrable near $y=0$, and that $f$ is globally integrable. So the proof must somehow separate the integral into two integrals.

By the Lebesgue theorem, $$ F(x) = \int_{-\infty}^xf(t)\mathrm{d}t, $$ is differentiable almost everywhere, with $F'(x)=f(x)$. We pick such an $x$, and without loss of generality, let us assume $x=0$. Also, let $\delta>0$. Then $$ \int_{-\infty}^\infty|y|^{-p}f(y)\mathrm{d}y = \int_{-\delta}^\delta|y|^{-p}f(y)\mathrm{d}y + \int_{|x|>\delta}|y|^{-p}f(y)\mathrm{d}y. $$ The second integral in the right hand side can be estimated easily as $$ \int_{|x|>\delta}|y|^{-p}f(y)\mathrm{d}y \leq \delta^{-p}\int_{|x|>\delta}|f(y)|\mathrm{d}y \leq \delta^{-p}\|f\|_{L^1}. $$ For the first integral, we have $$ \int_{-\delta}^\delta\frac{F'(y)}{|y|^{p}}\mathrm{d}y = \left.\frac{F(y)}{|y|^{p}}\right|_{-\delta}^{\delta} + p\int_{-\delta}^\delta\frac{yF(y)}{|y|^{p+2}}\mathrm{d}y. \qquad\qquad(*) $$ The first term in the right hand side is obviously finite. Note that by definition $$ F(y)=F(0)+yF'(0)+o(|y|), $$ and that we can take $F(0)=0$ since the left hand side of ($*$) depends only on $F'$. Taking this into account, we infer $$ \int_{-\delta}^\delta\frac{yF(y)}{|y|^{p+2}}\mathrm{d}y = \int_{-\delta}^\delta\frac{y^2F'(0)+o(|y|^2)}{|y|^{p+2}}\mathrm{d}y =O(\delta^{1-p}), $$ which completes the demonstration.

The answer to the second question is no. You can take $f$ to be a Gaussian, and convolve it with a fast growing function.

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