Function defined by integral - continuity, integrability, differentiability

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Let $f: [0,+\infty) \to \mathbb{R}$ be Lebesgue integrable and define $$F(x) := \int^{+\infty}_0 e^{-tx}\cos(tx)f(t)dt $$

Determine if:

a) F is defferentiable on $(0,+\infty)$

b) F is continuous on $[0,+\infty)$

c) F is integrable on $[0,+\infty)$

I have a problem with c) but I would also welcome solutions to a) and b)


My attempt in c):

Integrability would require $$\int^{+\infty}_0\int^{+\infty}_0 |e^{-tx}\cos(tx)f(t)|dtdx < +\infty$$ so I would like to use Fubini's Theorem to calculate $\int^{+\infty}_0\int^{+\infty}_0 e^{-tx}\cos(tx)f(t)dtdx = \int^{+\infty}_0 f(t)\int^{+\infty}_0 e^{-tx}\cos(tx)dxdt = \int^{+\infty}_0 \frac{f(t)}{2t} dt \stackrel{?}{=} +\infty$

and now I'm not sure about two things:

1) While $\frac{1}{2t}$ is of course non-integrable is it true that $\frac{f(t)}{2t}$ is also non-integrable?

2) Can I Use Fubini's Theorem to show non-integrability? That is the if last integral turns out to be divergent is correct to say that the original integral is also divergent?

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Take $f(t) = \frac{1}{t^\epsilon}\mathbb 1_{[1, \infty[} + t^{1-\epsilon}\mathbb 1_{]0, 1]}$ for $\epsilon > 0$ as a counter example for 1).

According to the Fubini-Tonelli theorem, we can deduce that if $$\int_X \int_Y |f(x,y)| \lambda(dy)\lambda( dx) = \infty$$ then $$f \notin L^1(X\times Y)$$