A question on use of square integrable functions I'm approaching this from a physicist's perspective, so apologies for any inaccuracies (and lack of rigour).
As far as I understand it, a square-integrable function $f(x)$ satisfies the condition $$\int_{-\infty}^{\infty}\big\vert f(x)\big\vert^{2}\;dx\;<\;\infty$$ Often in physics textbooks, especially in quantum mechanics (QM), the author states that this implies that $$\lim_{x\rightarrow\pm\infty}f(x)=0$$ However, I have come to understand that this statement is not necessarily true. This being the case, how does one argue, for example, that $$\mathscr{F}[f'](x)=\int_{-\infty}^{\infty}f'(x)e^{-ikx}\;dx=(ik)\mathscr{F}[f](x)$$ as doesn't this rely on the fact that $f(x)\rightarrow 0$ as $x\rightarrow\pm\infty$, such that the boundary term (introduced through integration-by-parts) $$f(x)e^{-ikx}\big\vert_{-\infty}^{\infty}$$ equals zero?
[The fact that $\lim_{x\rightarrow\pm\infty}f(x)=0$ (such that the boundary term vanishes) is also used in QM to argue that the derivative operator is anti-Hermitian]
 A: If $f$ and $f'$ are in $L^{2}$, then $ff'$ is absolutely integrable because
$$
         2|ff'| \le |f|^{2}+|f'|^{2}.
$$
Therefore, in this case, the following has finite limits as $x\rightarrow\pm\infty$:
$$
            f^{2}(x)-f^{2}(0)=2\int_{0}^{x}f(t)f'(t)dt.
$$
In this case, the limits $\lim_{x\rightarrow\pm\infty}f^{2}(x)=L_{\pm}$ must be $0$ because, otherwise, $f$ would remain bounded away from $0$ for large positive or negative $x$, which would contradict $f \in L^{2}$.
With that in mind, if you assume $f,f'\in L^{2}$, then the following converges in $L^{2}$:
\begin{align}
 \widehat{f'}(s)
    & = \lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f'(x)e^{-ixs}dx \\
    & = \lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\left[\left.f(x)e^{-isx}\right|_{-R}^{R}+is\int_{-R}^{R}f(x)e^{-isx}dx\right] \\
    & = \lim_{R\rightarrow\infty}\frac{is}{\sqrt{2\pi}}\int_{-R}^{R}f(x)e^{-isx}dx \\
    & = is\widehat{f}(s).
\end{align}
So assuming $f,f'\in L^{2}$ is enough to justify the above, assuming that $f$ is locally the integral of $f'$, which is to say that $f$ is locally absolutely continuous, which is equivalent to $f'$ being the weak derivative of $f$.
A: I am not familiar with your context, but I may be able to help out understand the definition and implications on $\lim_{x \rightarrow \pm \infty} f(x)$.
A function is square integrable on $(-\infty, \infty)$, hence the name, if 
$$
\int_{-\infty}^\infty f(x)^{\color{red}{2}} \mathrm{d}x \lt \infty
$$
(… and the above makes actually sense, ie. $f$ is measurable. However this is true for many, many functions, for example for all continuous functions. I do not think that this is the point of your question, so I will not go into the details here.)
In general
$$
\int_{-\infty}^\infty f(x)^2 \mathrm{d}x \lt \infty
\Rightarrow \lim_{x \rightarrow \pm \infty} f(x) = 0
$$
is wrong – even worse: $f$ does not even need to be bounded. However there are two nice results:
Result 1:
If $f$ is uniformly continuous, one has indeed
$$
\int_{-\infty}^\infty |f(x)| \mathrm{d}x \lt \infty
\Rightarrow \lim_{x \rightarrow \pm \infty} f(x) = 0
$$
One way to show uniform continuity of $f$, is to show that $f'$ is bounded. (Note however, that there are uniformly continuous functions with unbounded $f'$.)
In your case that means of course $f^2$ needs to be uniformly continuous.
Result 2:
If you are unable to use result 1, there is also a weaker but more general result: There exists a sequence $(x_n)_{n \in \mathbb{N}}$ with $\lim_{n \rightarrow \infty} x_n = \infty$ and
$$
\int_{-\infty}^\infty |f(x)| \mathrm{d}x \lt \infty
\Rightarrow \lim_{n \rightarrow \infty} f(x_n) = 0
$$
Of course you can apply this to $f^2$ as well.
Also you get a similar result for $x_n \rightarrow -\infty$.
