Does a function in $L^2(\mathbb{R_+},\mathbb{C}$ necessarily be $0$ at $\pm \infty$?

Question

if I have $f \in L^2(\mathbb{R_+};\mathbb{C})$ and want to study something like this

\begin{equation} \int_0^{\infty}f'(x)\overline{f(x)}\,dx \end{equation}

Then I integrate by parts getting \begin{equation} |f(x)|^2 |_0^{\infty}-\int_0^\infty f(x)\overline{f'(x)} \,dx \end{equation}

can I say that '$f(\infty)=0'$ if I know that $f \in L^2(\mathbb{R}_+;\mathbb{C})$?

EDIT:

I'm sorry I meant that my $f \in H^1(\mathbb{R}_+;\mathbb{C})$ so it has weak derivative

Answer 1

If $ f \in L^2 $, then it is not clear that $ f $ admits limit as $ x \to \infty $.

Answer 2

No, in general you cannot. Also, what does $f'$ mean for an $L^2(\mathbb{R}_+,\mathbb{C})$? Most functions in this space not differentiable.

If however, you have that $f$ has an almost everywhere derivative $f'$ and the fundamental theorem of calculus holds, i.e. $f$ is locally absolutely continuous, and this derivative is also $L^2$, then you can argue as you have that $|f(x)|^2$ has a limit, in which case yes it must be $0$.

Details: The argument is that since $f$ is weakly differentiable, and this weak derivative is also $L^2$, you have the existence of the following limit, $$ \langle f,f'\rangle=\lim_{M\to \infty}\int_0^M f'(x)\overline{f(x)}\mathrm dx $$ Now you integrate by parts and see that $$ \lim_{M\to \infty}\left(|f(x)|^2 \vert_0^M-\int_0^M f(x)\overline{f'(x)}\mathrm dx\right) $$ must also exist, being equal to the first. But the second term has an existent limit, as again an inner product of $f$ and $f'$, so the first limit exists (why?). If it does, it must be zero (suppose it isn't, and $||f||_{L^2}$ is finite....).