Laplace Transform

[Tsakanikas Nickos]

Let \( \displaystyle{ f : \left[0,+\infty \right) \rightarrow \mathbb{R} , \, f(t) = \sin ( {e}^{t^2} ) } \)

(i) Is \( \displaystyle{f} \) of exponential order?

(ii) Prove that \(f^\prime \) is not of exponential order, but its Laplace transform exists.

[Tsakanikas Nickos]

Let me give it a try.

(i) \( \displaystyle f \) is clearly of exponential order (and specifically of exponential order \( \displaystyle 0 \)), since it is bounded.

(ii) We have that \( \displaystyle f^\prime (t) = 2t{e}^{t^2}\cos( {e}^{t^2} ) \, , \, t \geq 0 \) . Suppose that \( \displaystyle f^\prime \) is of exponential order.
Then there exist \( \displaystyle M>0, a \in \mathbb{R} \) and \( t_{0} \geq 0 \) such that \[ \displaystyle \left| f^\prime(t) \right| \leq M{e}^{at} \, , \, \forall \, t \geq t_{0} . \] Thus,
\[ 2t{e}^{t^2-at} \left| \cos( {e}^{t^2} ) \right| \leq M \, , \, t \geq t_{0} , \] which implies that the function
\( \displaystyle g(t) = 2t{e}^{t^2-at} \left| \cos( {e}^{t^2} ) \right| \, , \, t \geq t_{0} \), is bounded. Consider now the sequence \( \displaystyle \left( \sqrt{ \log{ ( n\pi ) } } \right) _ {n \in \mathbb{N} } \) and note that
\[ \displaystyle \left| \cos( {e}^{\left( \sqrt{ \log{ ( n\pi ) } } \right) ^{2} } ) \right| = \left| \cos( n\pi ) \right| = 1 \, , \, n \in \mathbb{N} . \] Furthermore, from the archimedean property of \( \mathbb{R} \) there exists \( \displaystyle n_{0} \in \mathbb{N} \) such that \( n_{0}>t_{0} \). Additionally, note that \( \displaystyle \lim_{ t \to \infty } ( 2t{e}^{t^2-at} ) = \infty \) and \( \displaystyle \lim_{n \to \infty} \sqrt{ \log{ ( n\pi ) } } = \infty \) .
Consequently, for sufficiently large n ( obviously for \( n>n_{0} \)),
\[ \displaystyle g( \sqrt{ \log{ (n\pi) }} ) = 2\sqrt{ \log{ (n\pi) }} {e}^{ \left( \sqrt{ \log{ (n\pi) }} \right)\left( \sqrt{ \log{ (n\pi) }} \, - a \right) } \] becomes arbitrarily large. However, this contradicts the fact that \( \displaystyle g \) is bounded, which is equivalent to the hypothesis that \( \displaystyle f^{\prime} \) is of exponential order. Therefore,\(\, \displaystyle f^{\prime} \) is not of exponential order.
Its Laplace Transform, though, exists. Indeed; Since \( \displaystyle f \) is continuous and of exponential order 0, its Laplace Transform
\( \displaystyle \mathcal{L} \{f(t)\} \) exists for \( \displaystyle s>0 \). Now, by the definition of Laplace Transform of \( \displaystyle f^{\prime} \), we can easily see that \( \displaystyle \mathcal{L} \{f^{\prime}(t)\} = s\mathcal{L} \{f(t)\} \, - \, sin(1) \, , \, s>0 \).