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If a function $f: \mathbb{R}\rightarrow \mathbb{C}$ is continuous on $[0, \infty)$, is it true that $f(- \log x)$ is continuous on $(0, 1]$?

If yes, then how would I prove it?

Thanks and best regards.

babakks
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A.A
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    Yes that is true, in general the composition of continuous functions is continuous. – Joel Cohen Sep 21 '16 at 09:07
  • Dear Cohen, its true that the composition of function is continuous, but in this particular case, as the lower limit value of interval (0, 1] is 0 and - log x would be evaluated on it. i .e., log 0 which is indeterminate. So how would I manipulate it. – A.A Sep 21 '16 at 09:16
  • I actually want to prove it using the epsilon-delta definition of continuity. – A.A Sep 21 '16 at 09:21

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The previous answers are correct: although $\log 0$ is indeterminate, we do not need it to prove continuity. For continuity of $f(- \log x)$ it suffices to prove continuity at every value in the interval $(0, 1]$.

Continuity atna value is a local condition, which means that to prove continuity at a value we only have to consider a very small interval around it. For example, we can show that $f(- \log x)$ is continuous at every value $\epsilon$ in the interval $(0,1]$ by only considering the interval $(1/2 \epsilon, 1]$. Perhaps this helps you realize that $x=0$ plays no role in this story?

Gnuk
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You even don't have to consider continuity in particular values. Consider the following functions:

  • $\log: (0, 1] \to (-∞, 0]$,
  • $(-): (-∞, 0] \to [0, ∞)$,
  • $f: [0, ∞) \to ℂ$.

All of them are continuous, and your function $x \mapsto f(-\log x)$ defined on $(0, 1]$ is their composition, and hence continuous.

user87690
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