Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f\in L^p(0,\infty)$, $p>1$. Show that $\int_0^\infty f(x)\frac{\sin xy}{x} dx$ converges uniformly in $y$ in every finite interval. Show also that $|g(t+y)-g(y)|\leqslant M|t|^{\frac{1}{p}}$. Thanks in advance.

share|cite|improve this question
I tried with Holder's inequality (because of $f\in L^p(0,\infty)$). – user55529 Feb 4 '13 at 23:52
Then I tried to put $t=x y$, but I don't know how should I conclude that the convergence is uniform. Any help is welcome... – user55529 Feb 4 '13 at 23:57

For the integral $\int_0^1 f(x)\frac{\sin xy}{x}\,dx$, use $|\sin xy|\le xy$, canceling $x$ in the denominator.

For the integral $\int_1^\infty f(x)\frac{\sin xy}{x}\,dx$, use $|\sin xy|\le 1$ and then Hölder's inequality: $$ \int_1^\infty f(x)\frac{1}{x}\,dy \le \|f\|_{L^p} \left(\int_1^\infty \frac{1}{x^q}\,dx\right)^{1/q} $$ where $q$ is the conjugate exponent.

For the last part, estimate $\left|\sin [x(y+t)]-\sin xy\right|$ by $\min(tx,2)$. Using Hölder's inequality again, you end up estimating the $L^q$ norm of $\min(tx,2)/x$. This is of order $t^{1/p}$.

share|cite|improve this answer
(+1) Nice answer. – Mhenni Benghorbal Jul 26 '13 at 18:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.