Some questions on the Laplace transform of a periodic function

Question

$f$ is a periodic continuous function of period $T > 0 $.

1. Let $ a > 0$, prove that: $ \int_{0}^{+ \infty} f(t)e^{-at}dt $ is absolutely convergent.

2. Find a positive constant $ \lambda $ dependent only on $a$ and $T$ such that: $$ \int_{0}^{+ \infty} f(t)e^{-at}dt = \lambda \int_{0}^{T}f(t)e^{-at}dt $$

3. Find the expression of $ \mu $ that is dependent only on $ a$ and $T$ such that:

$$\left|\int_{0}^{T}f(t)e^{-at}\,dt\right| \leq \mu\sqrt{\int_{0}^{T}f^{2}(t)e^{-at}\,dt} $$

4. Find $ \lim_{a \to + \infty } f(t) e^{-at} dt$.

---

1. $f$ is continuous on the closed interval $[0, T]$, there is an $M \in \mathbb{R}$ such that:

   $|f(t)| < M \implies |f(t) e^{-at}| < Me^{-at} $

$ \int_{0}^{+ \infty}Me^{-at}$ is convergent, by comparison test, so does $\int_{0}^{+ \infty}|f(t) e^{-at}|$, which proves that $ \int_{0}^{+ \infty} f(t)e^{-at}dt $ is absolutely convergent. Is this correct?

I also need help with question 2, 3 and 4.

Thank you for your help.

Answer 1

Hint 2:

Using the substitution $t = u + (k-1)T$,

$$\int_0^\infty f(t)e^{-at} \, dt = \sum_{k=1}^\infty\int_{(k-1)T}^{kT}f(t)e^{-at} \, dt = \sum_{k=1}^\infty \int_{0}^{T}f(u+(k-1)T)e^{-au}e^{-a(k-1)T} \, du \\ = \sum_{k=1}^\infty e^{-a(k-1)T}\int_{0}^{T}f(u)e^{-au} \, du $$

Hint 3:

Use the Cauchy-Scwharz inequality on the RHS of

$$\int_0^\infty f(t)e^{-at} \, dt = \int_0^\infty f(t)e^{-at/2}e^{-at/2} \, dt $$

Answer 2

$4$. Since $f$ is a bounded function the given limit equals $0$ for any $t>0$;

$3$. By the Cauchy-Schwarz inequality

$$ \left|\int_{0}^{T}f(t)\,e^{-at}\,dt\right|\leq \sqrt{\int_{0}^{T}e^{-at}\,dt\int_{0}^{T}f(t)^2 e^{-at}\,dt} $$ hence we may take $\mu$ as $\sqrt{\frac{1-e^{-aT}}{a}}$;

$2$. $$\int_{0}^{+\infty}f(t)e^{-at}\,dt = \int_{0}^{T}f(t)\left(e^{-at}+e^{-a(t+T)}+\ldots\right)\,dt $$ hence we may take $\lambda$ as $\frac{1}{1-e^{-aT}}$.