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

Question

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.

Answer 1

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.

Answer 2

$\int_{0}^{\infty}\int_{t}^{\infty}... dx dt=\int_{0}^{\infty}\int_{0}^{x}... dt dx$. 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.