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Let $f(x)=x^\alpha$ where $\alpha\in(0,1)$

I want to show to that $f(x)$ is continuous on the interval $I=[0,+\infty)$

Let $x_{0}\in[0,+\infty)$

$|f(x)-f(x_{0}|=|x^\alpha-{x_{0}}^\alpha|$

For $\alpha\in\mathbb{Q}\cap(0,1)$, we can write $\alpha=\frac{a}{b}$ where $a,b\in\mathbb{N}$ and $a<b$

I know how to prove that $x^{\frac{1}{b}}$ is continuous on $I$ and that $x^a$ is continous on $I$. So by composition, so is $x^{\frac{a}{b}}$.

However, what do I do in the case where $\alpha\in\mathbb{Q^\complement}\cap(0,1)$?

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    $\begingroup$ How do you define $x^\alpha$ for $\alpha\notin\Bbb Q$? $\endgroup$ Commented May 12, 2018 at 22:20
  • $\begingroup$ @Jean-ClaudeArbaut, I didn't understand what you mean by define. $\endgroup$
    – Omrane
    Commented May 12, 2018 at 23:40
  • $\begingroup$ @Jean-Claude Arbaut how about this as a sketch - first define $x^\alpha$ as $e^{\alpha\ln x}$ to address your question re general $\alpha$. Then prove continuity of $e^x$ (can pbbly find a link to a good proof somewhere on MSE) whence we have continuity of $\ln x$ via inverse function theorem then invoke composition of continuous functions - that work? (if so pbbly too heavy-duty) $\endgroup$
    – Mehness
    Commented May 12, 2018 at 23:53
  • $\begingroup$ @MathVandal That's a possibility. There is also a definition of $a^b$ using Dedekind cuts, if I remember well. That would lead to a different proof. But the OP must first understand that the existence of $x^\alpha$ is not obvious, and that this operation must be given a meaning in one way or another. It's also true of $x^{1/n}$ (for positive integer $n$), but there are easy proofs of existence of this. $\endgroup$ Commented May 13, 2018 at 0:07
  • $\begingroup$ @Jean-ClaudeArbaut thanks will look into the Dedekind cut definition, dimly remember sth abt this but very dimly! Agree re non-triviality of general $x^\alpha$ $\endgroup$
    – Mehness
    Commented May 13, 2018 at 0:29

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