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I am studying for qualifying exams and I found the following problem from University of Michigan May 2013 analysis qualifying exam.

Let $f$ be a measurable function on $(0, \infty)$. Let $p > 1/2$ and define $g(x) = (x^p + x^{−p})f(x)$. Show that if $g ∈ L^2(0,\infty)$ then $f ∈ L^1(0,\infty)$.

I used Holder and now I am not sure how to show that the integral $(x^p + x^{−p})^2$ from $0$ to infinity is finite.

I am new to this website so I apologize for not having it in LaTex.

Gamma Function
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Lucy Lewis
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    It doesn't. I'm guessing you mean $(x^p+x^{-p})^{-2}$ (which is what you get when you try and apply Holders to $f$. This is true by rewriting it as $\displaystyle \frac{x^{2p}}{1+2x^{2p}+x^{4p}}$. The denominator is non-vanishing at $0$, so there are no convergence issues there, and as $x\to \infty$ you have that the function is $O(x^{-2p})$ which converges. – Alex Youcis Aug 29 '13 at 06:12
  • yeah sorry typo. – Lucy Lewis Sep 26 '13 at 21:40

3 Answers3

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Let $\phi(x) = \frac{1}{x^{-p}+x^p}$. Then $|\phi(x)| \le 1 $ for $x \in (0,1]$, and $|\phi(x)| \le x^{-p}$ for $x >1$. Then, since $2p>1$, we have $\phi \in L^2$.

We see that $f(x) = g(x) \phi(x)$, so Hölder gives $\|f\|_1 \le \|g\|_2 \|\phi \|_2$.

copper.hat
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Indeed, as you say, one has for all $0<a<b$ $$\int_a^b|f(x)|dx \leq \sqrt{\int_a^b\left(f(x)(x^p+x^{-p})\right)^2}dx + \sqrt{\int_a^b\frac{dx}{(x^p+x^{-p})^2}}.$$

Now, the first integral is converging when $a\to 0,b\to\infty$. The second can be simplified $$\int_a^b\frac{x^{2p}}{x^{4p}+2x^{2p}+1}dx$$ So there are no problems at $a=0$, and by the comparison test with $\int_1^\infty\frac{dx}{x^{2p}}$, which due to $2p>1$ is known to converge, this improper integral converges as well.

Jonathan Y.
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By Holder's inequality, we have $$\int_0^\infty|f(x)|dx=\int_0^\infty|g(x) (x^p + x^{−p})^{-1}|dx\leq \left(\int_0^\infty|g(x)|^2dx\right)^{\frac{1}{2}}\left( \int_0^\infty(x^p + x^{−p})^{-2}dx\right)^{\frac{1}{2}} \tag{1}.$$ Since $g\in L^2(0,\infty)$, if we can show that $\int_0^\infty(x^p + x^{−p})^{-2}dx<\infty$, then $f\in L^1(0,\infty)$ by $(1)$.

Note that $$\int_0^\infty(x^p + x^{−p})^{-2}dx= \int_0^\infty\frac{1}{x^{2p} + x^{−2p}+2}dx$$ $$ =\int_0^1\frac{1}{x^{2p} + x^{−2p}+2}dx+\int_0^\infty\frac{1}{x^{2p} + x^{−2p}+2}dx $$ $$\leq \int_0^1\frac{1}{2}dx+\int_1^\infty\frac{1}{x^{2p}}dx=\frac{1}{2}-\frac{1}{2p-1}\frac{1}{x^{2p-1}}\big|_1^\infty=\frac{1}{2}+\frac{1}{2p-1}<\infty$$ where we have used the assumption $p>1/2$.

Paul
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