Let $f\geq0$ and that $\int_{0}^{x}f(t)dt<e^x$ for all $0\leq x$, show $f(t)e^{-at}\in L^1[0,\infty)$, $a>1$

Question

Let $f\geq0$ and that $\int_{0}^{x}f(t)dt<e^x$ for all $0\leq x$, show $f(t)e^{-at}\in L^1[0,\infty)$, $a>1$

I have a proof by i am not sure if what I do is actually allowed. Define $h(x)=\int_{0}^{x}f(t)$, then we know that $h'=f$ a.e and so $$\int_{0}^{\infty} f(t)e^{-at}dt=\int_{0}^{\infty} h'(t)e^{-at}dt=[h(t)e^{-at}]_0^{\infty}-\int_{0}^{\infty} h(t)e^{-at}dt=-\int_{0}^{\infty} h(t)e^{-at}dt<-\int_{0}^{\infty} e^{1-at}<\infty$$

I used integration by parts but I am not sure I can do that. What are the conditions for Lebesgue integration to use integration by parts? Differentiable a.e is sufficient? And what is the actual solution here? We actually know $h$ is AC so perhaps this proof is fine after all.

Answer 1

Without IBP and any regularity assumption: you have \begin{align*} \int_0^{+ \infty} f(t) e^{-at} \: \mathrm{d} t & = \sum_{n \geq 0} \int_n^{n+1} f(t) e^{-at} \: \mathrm{d} t \\ & \leq \sum_{n \geq 0} e^{-an} \int_n^{n+1} f(t) \: \mathrm{d} t \\ & \leq \sum_{n \geq 0} e^{-an} \int_0^{n+1} f(t) \: \mathrm{d} t \\ & \leq \sum_{n \geq 0} e^{-an} e^{n+1} \\ & = e \sum_{n \geq 0}e^{(1-a)n} \end{align*} and this is finite since $a > 1$.

Answer 2

For integration by parts to hold on a closed and bounded interval, it suffices that the functions in question are continuously differentiable (here, that is your $h(x)$ and the exponential). In general, however, one must be careful if the interval is unbounded. For instance, taking $\frac{\sin x}{x}$, one can erroneously use IBP on an interval $(1,n]$ to "reduce" the problem to the integrability of $\frac{\cos x}{x^2}$ and take limits. However, the former is not Lebesgue integrable over $(1, \infty)$ (split into intervals that are integer multiples of $\pi$) whereas the latter is. The lesson is that integration by parts cannot always be used to infer the existence of one integral from another.

In this case, your integrand is non-negative, so you should be able to apply the monotone convergence theorem and get away with it.

Answer 3

Integrate over a finite interval $[0,x]$ using integration by parts. You get an upper bound $$\int^x_0f(t)e^{-at}\,dt=h(x)e^{-ax}+a\int^x _0 h(s)e^{-as}\,ds\leq e^{-x(a-1)}+a\int^x_0 e^{-s(a-1)}\,ds$$

where $h(x)=\int^x_0f$. Then let $x\rightarrow\infty$. You will require $f$ to be also locally integrable. Then your $h$ will satisfy the needed conditions for integration by parts (local finite variation)