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Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuous differentiable function, and suppose $\int_0^\infty f(r)dr<\infty$ and that $f(0)<\infty$. Is it true that $\lim_{r\rightarrow\infty}f(r)=0$?

My physics class seems to assume this. I see why, if the limit exists, it must be zero, but I don't see why the limit has to exist. Perhaps one line of logic would be that, since we assume the derivative exists, we can assume $\int_0^\infty f(r)\frac{df(r)}{dr}dr$ exists, and by integration by parts this is equal to $|_0^\infty f(r)^2-\int_0^\infty f(r)\frac{df(r)}{dr}dr$, so moving everything else to one side the limit is finite, and thus zero. But by using integration by parts aren't I already assuming the limit exists?

For context, this is an introductory quantum mechanics course. They state the only properties a wavefunction $\psi$ must have is that $\int_{-\infty}^\infty |\psi(x)|^2dx<\infty$ and $\psi$ is continuous and have continuous derivative. To prove the momentum operator is hermitian, we need $|_{-\infty}^\infty|\psi|^2=0$.

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No. Consider $$ f(x) = \sum_{n\geq 1}n\exp\left[-n^6(x-n)^2\right]. $$

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No, let consider for example

$$f(r)=\sin (r^2)$$

and refer to

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