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.
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 $\|f\|_1 \le \|g\|_2 \|\phi \|_2$.
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.
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 $p>1/2$.
