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This is an excerpt from a textbook on quantum mechanics. $u(x)$ is a square-integrable complex-valued function of a real variable.

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The assumption that $u(x)\neq 0$ is probably there, too. It seems to imply that these two statements are equivalent:

  • $u(x)$ and $u'(x)$ are continuous at $x=a$.
  • $\frac{u'(x)}{u(x)}$ is continuous at $x=a$.

It is easy to prove that the former implies the latter, but what about the other way around? Is it true? If so, how to prove it? If not, could there be an assumption that I've missed, or have I misinterpreted the text?

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    I'm somewhat curious why the author(s) of the book consider this _useful_.2011-06-26

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We should assume an existence of u'(a) to talk about \frac{u'(x)}{u(x)} in the neighborhood of $a$, which leads itself to the continuity of $u(x)$ at $a$. If f(x) = u'/u is continuous, then it's true for u'(x) = f(x)u(x) since it is a multiplication of two continuous functions.

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    Thanks, your first sentence makes it clear.2011-06-26