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How to construct a function $f(x)$ such that $f(x)e^{-px}$ wouldn't tend to $0$ as $x$ tends to infinity?

This question is motivated when studying Laplace Transform when I encountered the following result

Suppose $f$ and $f'$ both have Laplace Transform on some half plane $\Re(p)>p_0$, provided that $f(x)e^{-px}\to 0$ as $x\to \infty$ for $p$ such that $\Re(p)>p_0$. Then we have $\hat {f'}(p)=p\hat {f}(p)-f(0)$.

Where the hat notation is meant to be the Laplace Transform of function.

Now, I am just quite curious if there exists a function $f$ such that the condition of "provided that $f(x)e^{-px}\to 0$ as $x\to \infty$ for $p$ such that $\Re(p)>p_0$" would fail. Surely we can just choose $p_0$ to be very large in its real part unless we can construct something that can grow more rapidly than exponentials?

Many thanks in advance!

UnsinkableSam
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3 Answers3

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How about $f(x) = \exp(x^2)$.

EDIT: Note that $$ \dfrac{d}{dx} \left(f(x) e^{-px}\right) = f'(x) e^{-px} - p f(x) e^{-px} $$

so $$\int_0^b f'(x) e^{-px}\; dx - p \int_0^b f(x) e^{-px}\; dx = f(b) e^{-pb} - f(0)$$

If $f(x)$ and $f'(x)$ both have Laplace transforms at $p$, the left side must go to a finite limit as $b \to \infty$, so the right side must also, and in particular $f(x) e^{-px}$ must be bounded, and for $\text{Re}(c) > \text{Re}(p)$ we must have $f(x) e^{-cx} \to 0$ as $x \to \infty$. So it is not possible for both $f$ and $f'$ to have a Laplace transform for $\text{Re}(p) > p_0$ without $|f(x) e^{-px}| \to 0$ as $x \to \infty$ for $\text{Re}(p) > p_0$.

Robert Israel
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    That is a good idea! However though, I dont think this one has a Laplace transform since the integral would just diverge? Is it possible that both $f$ and $f'$ have a LT but it does not have the aforementioned property? – UnsinkableSam Jun 29 '20 at 20:33
  • $f(x)=e^{x^2}$ for $x$ an integer, and 0 otherwise. – NeitherNor Jun 29 '20 at 20:47
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$f(x)=\Gamma(x)$ works as well.

K.defaoite
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A way to go is to find a function that is guaranteed to grow faster than $-px$ descends, regardless of $p$. For instance, $x^2$, $x^3$, ..., $\mathrm{e}^x$, $x \ln x$, and so on. So each of the following is a potential $f(x)$: \begin{align*} &\mathrm{e}^{x^2} \text{,} \\ &\mathrm{e}^{x^3} \text{,} \\ &\vdots \text{,} \\ &\mathrm{e}^{\mathrm{e}^x} \text{, and } \\ &\mathrm{e}^{x \ln x} = x^x \text{.} \end{align*}

And of course, once one sees these, ideas for even more rapidly growing functions should spring to mind.

Eric Towers
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