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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.

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What have you tried ? –  Klaus Feb 4 '13 at 23:40
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I tried with Holder's inequality (because of $f\in L^p(0,\infty)$). –  user55529 Feb 4 '13 at 23:52
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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

1 Answer 1

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 $t^{1/p}$.

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(+1) Nice answer. –  Mhenni Benghorbal Jul 26 '13 at 18:40

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