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Suppose $f$ is Lebesgue measurable on $[0,\infty)\times [0,\infty)$ and $g\in L^1([0,\infty))$. If $|xf(x,y)|\leq g(x)$ for all $y\in [0,\infty)$ prove that

$$\sup_{y>0}\left|\int_0^\infty \int_t^\infty f(x,y) \cos\left(\dfrac{t}{y}\right)dx\,\,dt\right|<\infty. $$

I believe since $f(x,y)$ is Lebesgue mesurable and $[0,\infty)$ is $\sigma$-finite with respect to Lebesgue measure, we should get in somehow a nonnegative function $h$ and hence apply Tonelli's theorem (which requires $\sigma$-finite product space and a nonnegative measurable function). At the end we should get something like $$\sup_{y>0}\int_0^\infty |xf(x,y)|\,dx\leq \sup_{y>0}\int_0^\infty g(x)\,dx <\infty,\quad\text{ since } g\in L^1$$.

This is an old prelim problem I trying to solve as a preparation for a prelim exam in January. Any help is appreciated.

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The presence of the $\cos$ term leads one to think that there might be some cancellation in the integral and that the triangle inequality is not the right way to start. For a prelim question that is a bit misleading, in my opinion. For any $y$ you have $$ \left| \int_0^\infty \int_t^\infty f(x,y) \cos \left( \frac ty \right) \, dx dt \right| \le \int_0^\infty \int_t^\infty |f(x,y)| \, dx dt.$$ Introduce an indicator function to get the variable out of the limits of integration: $$ \int_0^\infty \int_t^\infty |f(x,y)| \, dx dt = \int_0^\infty \int_0^\infty |f(x,y)| \chi_{x \ge t} \, dxdt.$$ Apply the Tonelli theorem: $$ \int_0^\infty \int_0^\infty |f(x,y)| \chi_{x \ge t} \, dxdt = \int_0^\infty \int_0^\infty |f(x,y)| \chi_{x \ge t} \, dtdx = \int_0^\infty |f(x,y)| \int_0^\infty \chi_{x \ge t} \, dtdx$$ where $$\int_0^\infty \chi_{x \ge t} \, dt = \int_0^x \, dt = x.$$ Thus $$\left| \int_0^\infty \int_t^\infty f(x,y) \cos \left( \frac ty \right) \, dx dt \right| \le \int_0^\infty x|f(x,y)| \, dx$$ which is easily bounded by your remark above.

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  • $\begingroup$ First thanks for commenting. For your solution there a tiny mistake in the line after you say "Apply the Tonelli theorem:" last integral should be $dt dx$ and hence the next line is not true. Anyway if $\int_0^\infty\chi_{x\geq t}dt= x$ then we are done. Now do you think this this true? $\endgroup$ – Ruzayqat Dec 19 '14 at 16:01
  • $\begingroup$ Sorry about the typo. I fixed it. $\endgroup$ – Umberto P. Dec 19 '14 at 16:06
  • $\begingroup$ Could you elaborate on $$\int_0^\infty \chi_{x \ge t} \, dt = \int_0^x \, dt = x.$$ $\endgroup$ – Ruzayqat Dec 19 '14 at 16:12
  • $\begingroup$ $\chi_{x \ge t}$ equals $1$ if $x \ge t$ and $0$ if $x < t$. $$\int_0^\infty \chi_{x \ge t} \, dt = \int_0^x \chi_{x \ge t}\, dt + \int_x^\infty \chi_{x \ge t} \, dt = \int_0^x 1 \, dt + \int_x^\infty 0 \, dt.$$ $\endgroup$ – Umberto P. Dec 19 '14 at 16:13
  • $\begingroup$ Got it .. no need .. thanks a lot! Actually I started thinking of this problem as you did by getting rid of cos.. but i couldnt continue because i needed x in somehow. $\endgroup$ – Ruzayqat Dec 19 '14 at 16:15
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$\int_{0}^{\infty}\int_{t}^{\infty}... dx dt=\int_{0}^{\infty}\int_{0}^{x}... dt dx$. Thus $\int_{0}^{\infty}\int_{0}^{x} f(x,y) \cos(\frac {t}{y}) dt dx=\int_{0}^{\infty} f(x,y) y\sin(\frac{x}{y})dx$.

Because $\sin(x)\leq x$ for $ x\geq 0$ we have $\int_{0}^{\infty} f(x,y) y\sin(\frac{x}{y})dx\leq \int_{0}^{\infty} xf(x,y) dx$. now you can put the absolute values and it's done.

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  • $\begingroup$ How you did that: $$\int_{0}^{\infty}\int_{t}^{\infty}... dx dt=\int_{0}^{\infty}\int_{0}^{x}... dt dx$$??? $\endgroup$ – Ruzayqat Dec 19 '14 at 16:03
  • $\begingroup$ You need to be a bit more careful in the first step justifying the change of order of integration. $\endgroup$ – Umberto P. Dec 19 '14 at 16:07
  • $\begingroup$ @Leo $0<t\leq x< \infty$. $\endgroup$ – Haha Dec 19 '14 at 16:15
  • $\begingroup$ But you changed the order of $dx dt$ to be $dt dx$. This can't happen because we don't know if $f(x,y)\cos(t/y)\ge 0$, so we can't apply Tonelli's. Also, we don't know if $f(x,y)\cos(t/y)$ is integrable so we can't apply Fubini. $\endgroup$ – Ruzayqat Dec 19 '14 at 16:22

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