The tauberian final value Theorem

I take it you're talking about the result that says $\lim_{t\to\infty}f(t)=\lim_{s\to0}sF(s)$. Yes, we need to assume that the limit of $f$ exists.

Regarding the word "tauberian" in the title: I really think this is an abelian theorem; the corresponding tauberian theorem, which is not true without an extra "tauberian" hypothesis, would assume that $\lim sF(s)$ exists and deduce the existence of $\lim f(t)$. (For an example showing that the bare converse is false, let $f(t)=\sin(t)$.) Edit: Seems to me that the extra condition $tf'(t)\to0$ is enough. See Aha below...

Don't know a reference, but the FVT is trivial (meaning that although it's new to me I saw a proof immediately, just by starting with $\epsilon>0$.) We also need to assume that $f$ is continuous (measurable would do), so that the Laplace transform $$F(s)=\int_0^\infty f(t)e^{-st}\,dt$$exists for $s>0$. (If we assumed only that $f$ is meaasurable we'd also need to assume that $f$ is bounded.)

FVT. Suppose $f:[0,\infty)\to\Bbb C$ is continuous and $\lim_{t\to\infty}f(t)=L\in\Bbb C$. Then $F(s)$ exists for $s>0$ and $\lim_{s\to0^+}sF(s)=L$.

Proof: Note first that the hypothesis implies that $f$ is bounded, hence $F(s)$ exists for every $s>0$. Say $|f(t)|\le c$ for all $t\ge0$.

Since $s\int_0^\infty e^{-st}\,dt=1$ we have $$sF(s)-L=s\int_0^\infty (f(t)-L)e^{-st}\,dt$$for every $s>0$; hence $$|sF(s)-L|\le s\int_0^\infty |f(t)-L|e^{-st}\,dt$$Let $\epsilon>0$. Choose $A$ so $$|f(t)-L|<\epsilon\quad(t>A).$$Now $$|sF(s)-L|\le s\int_0^A |f(t)-L|e^{-st}\,dt +s\int_A^\infty|f(t)-L|e^{-st}\,dt=I+II.$$

Since $|f(t)-L|\le 2c$ and $e^{-ts}<1$ we have $$I\le 2cAs.$$Similarly $$II\le\epsilon s\int_0^\infty e^{-ts}\,dt=\epsilon,$$so $I+II<2\epsilon$ if $s>0$ is small enough. QED.

Bonus: "Abelian" versus "Tauberian":

I'm saying FVT is an "abelian" theorem because as we see above it follows from the fact that $sF(s)$ is just a weighted average of the values of $f(t)$. On the other hand a "tauberian" converse would require an extra "tauberian condition", because $f(t)$ is not just an average of the values of $sF(s)$.

This goes back to the original abelian/tauberian pair, Abel's Theorem and Tauber's Theorem. Suppose $$f(z)=\sum a_n z^n$$is holomorphic in the unit disk.

Abel's Theorem If $\sum a_n=s$ then $\lim_{z\to1^-}f(z)=s$.

Tauber's Theorem If $na_n\to0$ and $\lim_{z\to 1^-}f(z)=s$ then $\sum a_n=s$.

Abel's theorem is trivial once you do the summation by parts to show that if $0<z<1$ then $f(z)$ is an average of the partial sums $s_n=\sum_0^n a_j$. Tauber's theorem is not trivial in the same way.

(Hint regarding why FVT seemed obvious to me: The argument is really the same as the part of the proof of Abel's theorem after the summation by parts. Look up the proof of Abel's theorem; when you really "understand" that proof a lot of other things will be obvious to you as well.)

Aha: It seemms to me that $\lim tf'(t)=0$ is enough for the converse. Of course we also have to assume $f$ is bounded to ensure that $F(s)$ exists for $s>0$:

Tauberian Converse FVT. Suppose $f:[0,\infty)\to\Bbb C$ is bounded, differentiable, and satisfies $\lim_{t\to\infty}tf'(t)=0$. If $\lim_{s\to0}sF(s)=L$ then $\lim_{t\to\infty} f(t)=L$.

First a technicality:

Lemma. Suppose $tf'(t)\to0$. Then for $t>0$ fixed we have $\lim_{\lambda\to\infty}(f(\lambda)-f(\lambda t))=0.$

Proof. Let $\alpha=\min(1,t)$. Then MVT shows that there exists $\xi\ge\alpha\lambda$ with $$f(\lambda)-f(\lambda t)=\lambda(1-t)f'(\xi).$$If $\lambda$ is large enough then $|f'(\xi)|\le\frac\epsilon\xi\le\frac\epsilon{\alpha\lambda}$.

Proof of the theorem: Say $|f(t)|\le c$ for all $t\ge0$. As before, if $s>0$ then$$\begin{aligned}f\left(\frac1s\right)-sF(s) &=s\int_0^\infty\left(f\left(\frac1s\right)-f(t)\right)e^{-st}\,dt \\&=\int_0^\infty\left(f\left(\frac1s\right)-f\left(\frac ts\right)\right)e^{-t}\,dt.\end{aligned}$$So the Lemma implies that $$ f\left(\frac1s\right)-sF(s)\to0\quad(s\to0),$$by dominated convergence.

Hmm: That was obviously inspired by the condition $na_n\to0$ in Tauber's theorem. Considering Littlewood's improvement, one might conjecture that assuming $tf'(t)$ bounded is enough, but that that makes it much harder.

Yet one more edit: Yes, it turns out that $tf'(t)$ bounded is enough. Things are getting seriously tauberian; we can't expect to prove this by the sort of trivial arguments we use above, instead we pull out a big gun: Wiener's Tauberian Theorem.

Note this is much more than long enough already - I'm going to assume the reader knows WTT. In fact the results below contain the results above, but it seems worthwhile keeping the stuff above anyway, since it's so much more elementary.

Mentioning WTT raises the question "Huh? WTT talks about convolutions; I don't see any convolutions here..." There are in fact convolution operators everywhere above, just not convolutions on $\Bbb R$. To begin:

Let $G=((0,\infty),\times)$ be the multiplicative group of positive reals. Note that $dt/t$ is a Haar measure on $G$. The notation $f*g$ below will refer to convolution on $G$: $$f*g(s)=\int_0^\infty f(t)g(s/t)\,\frac{dt}t.$$

POV Integral operators on $(0,\infty)$ are often convolution operators on $G$.

And so it is with the Laplace transform:

Lemma. If $s>0$ then $sF(s)=f*K\left(\frac1s\right)$, where $K(t)=\frac1te^{-1/t}$.

Proof: Just work out the definition of the convolution.

So now we wonder about the zeroes of the Fourier transform $\hat K$. The dual group of $G$ is $$\hat G=\{\chi_\alpha:\alpha\in\Bbb R\},$$where $$\chi_\alpha(t)=t^{i\alpha}.$$And so $$\begin{aligned}\hat K(\chi_\alpha)&=\int_0^\infty t^{-i\alpha}\frac1t e^{-1/t}\,\frac{dt}t \\&=\int_0^\infty t^{1+i\alpha}e^{-t}\,\frac{dt}t \\&=\Gamma(1+i\alpha)\ne0.\end{aligned}$$

So WTT implies the following. (Note that WTT holds on any LCA group; if that's bothersome note that $G$ is isomorphic to $\Bbb R$, so in fact WTT for $\Bbb R$ is enough.)

WTT-FVT. Given $f\in L^\infty(G)$, we have $\lim_{s\to0^+}sF(s)=0$ if and only if $\lim_{s\to\infty}f*g(s)=0$ for every $g\in L^1(G)$.

And that's the whole story regarding what $sF(s)\to0$ really says about $f$. Now for $\lambda>1$ define $$f_\lambda(s)=\frac1{\log\lambda}\int_s^{\lambda s}f(t)\,\frac{dt}t.$$

Exercise there exists $m_\lambda\in L^1(G)$ such that $f_\lambda=f*m_\lambda$.

Exercise If $f$ and $tf'(t)$ are bounded on $(0,\infty)$ then $f_\lambda\to f$ uniformly as $\lambda\to1$.

Combining this and that we see

Improved Tauberian Converse FVT. If $f$ and $tf'(t)$ are bounded on $(0,\infty)$ and $sF(s)\to L$ then $\lim_{t\to\infty} f(t)=L$.

(Of course WLOG $L=1$. Now WTT-FVT and the first exercise show that $f_\lambda(t)\to0$ as $t\to\infty$, so the second exercise shhows that $f(t)\to0$.)

In fact assuming just that $f$ is bounded and uniformly continuous on $G$ is enough. Note of course that "uniformly continuous on $G$" is not the same as "uniformly continuous on $(0,\infty)$"; a function is uniformly continuous on $G$ if for every $\epsilon>0$ there exists $\lambda>1$ so that $1/\lambda<s/t<\lambda$ implies $|f(s)-f(t)|<\epsilon$. The proof above applies more or less without change; part of what's above is simply verifying that if $tf'(t)$ is bounded then $f$ is in fact uniformly continuous on $G$.

For a result with yet weaker hypotheses look up Pitt's version of WTT and translate that condition from $\Bbb R$ to $G$.