If $\operatorname{supp}(f)\subset |0,a]$ then $\mathcal L\{f\}(s)$ can't go to zero faster than $e^{-sa}$

Question

Let $a>0, \operatorname{supp}(f)=[0,a],\, f\in L^1(0,a)$ and dénote $\mathcal L\{f\}$ the Laplace transform of $f$.

Question: I want to show, only with real analysis and without additional conditions on $f$, the fact:

If $\displaystyle \lim_{s\to +\infty} e^{sa} \mathcal L\{f\}(s)=0$ then $f=0$ a.e.

This shows that if $\operatorname{supp}(f)\subset |0,a]$ then $\mathcal L\{f\}(s)$ can't go to zero faster than $e^{-sa}$.

We can prove it using complex analysis: let $g(x)=f(a-x) $ then $e^{sa}\mathcal L\{f\}(s)=e^{sa}\int_0^a f(t)e^{ts}dt=e^{sa}\int_0^a f(a-x)e^{-(a-x)s} dx=\int_0^a g(x)e^{sx}dx$. But by assymption $\displaystyle \lim_{s\to +\infty}\int_0^a g(x)e^{sx} dx=0$ then $g=0$ a.e. (see If $\lim_{t\to +\infty} \int_{0}^{\pi} f(x)e^{xt} \, dx=0$ then $f=0$?) then $f=0$ a.e.

We can prove it by additional conditions on $f$:

1. If $f(0^+)$ exists and is not zero and if the initial value theorem of Laplace transform is valid then $\lim_{s\to +\infty} s\mathcal L\{f\}(s)=f(0^+)$ and $\mathcal L\{f\}(s)\sim \frac{f(0^+)}s$ and then $\lim_{s\to +\infty} e^{as} \mathcal L\{f\}(s)\ne 0$ except if $f=0$ a.e.

2. If $\exists n\in \mathbb N$, such that $f^{(k)}(0^+)$ exists forall $k\leq n$, $f^{(n)}(0^+)\ne 0$, $\forall k<n,\, f^{(k)}(0^+)= 0$, then with some regularity of $f$, $\mathcal L\{f\}(s)\sim n!\frac{f^{(n)}(0^+)}{s^n}$ and $\lim_{s\to +\infty} e^{as} \mathcal L\{f\}(s)\ne 0$ except if $f=0$ a.e.

3. If $f$ is smooth near $0^+$ and $\forall n\in\mathbb N,\, f^{(n)}(0^+)=0$ then $f=0$ near $0^+$, there exist $\alpha\in\mathopen]0,a\mathclose[$ such that $f=0$ on $[0,\alpha]$ and $f$ not identically 0 near $\alpha^+$ (if $\alpha=a$ nothing to prove); put $h(x)=f(x)$ on $[\alpha, a]$ then $e^{sa}\mathcal L\{f\}(s)=e^{sa}e^{-s\alpha} \mathcal L\{h\}(s)=e^{s(a-\alpha)} \mathcal L\{h\}(s)$, with case 2 (since $h$ not identically 0 on $[\alpha, a]$), there exists $n\in\mathbb N$ such that $\mathcal L\{h\}(s)\sim n!\frac{h^{(n)}(\alpha^+)}{s^n}$ and thus $\lim_{s\to +\infty} e^{as} \mathcal L\{f\}(s)\ne 0$ except if $f=0$ a.e.