- #1
laonious
- 9
- 0
Hi everyone,
I am studying past analysis prelim exams to take in the fall and have run into one which really has me stumped:
Let f be a real-valued Lebesgue integral function on [0,\infty).
Define
F(x)=\int_{0}^{\infty}f(t)\cos(xt)\,dt.
Show that F is defined on R and is continuous on R.
Show that \lim_{\rightarrow \infty}F(x)=0.
That F is defined is straightforward I think, the part I am struggling with is showing that it is continuous.
Going the route of a direct proof:
F(x)-F(y)$=\int_{0}^{\infty}f(t)(\cos(xt)-\cos(yt))\,dt,
but it isn't clear to me how to show that this is enough to make the integral small.
Any ideas would be greatly appreciated, thanks!
I am studying past analysis prelim exams to take in the fall and have run into one which really has me stumped:
Let f be a real-valued Lebesgue integral function on [0,\infty).
Define
F(x)=\int_{0}^{\infty}f(t)\cos(xt)\,dt.
Show that F is defined on R and is continuous on R.
Show that \lim_{\rightarrow \infty}F(x)=0.
That F is defined is straightforward I think, the part I am struggling with is showing that it is continuous.
Going the route of a direct proof:
F(x)-F(y)$=\int_{0}^{\infty}f(t)(\cos(xt)-\cos(yt))\,dt,
but it isn't clear to me how to show that this is enough to make the integral small.
Any ideas would be greatly appreciated, thanks!