Let ${\bf x}=(x_1,...,x_n)$, the p-norm of x is $(|x_1|^p+...+|x_n|^p)^{1/p}$. If one of the components of x is 0, there will be exponential of the form $0^p$. If p is an irrational, $x^p$ is only definable for x>0 (see e.g. page 181 of baby Rudin), that is, $0^p$ is not defined. So how do we handle this problem when working with p-norm? Assume that p does not take irrational value, or prescribe that $0^p$ means 0 when it occurs? Thanks!

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    $\begingroup$ $0^p$ is taken to mean 0 in this context. There is no problem in defining the nonnegative p^th power of nonnegative reals for p>0 including $0^p=0$, and thus no ambiguity in defining $x\mapsto |x|^p$ on $\mathbb{R}$. $\endgroup$ – Jonas Meyer Jun 4 '10 at 1:29
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    $\begingroup$ Zero to any positive power is zero. Voting to close. $\endgroup$ – Steve Huntsman Jun 4 '10 at 1:32
  • $\begingroup$ @ Jonas Meyer. where is a definition for $0^p$, p irrational? can you refer me to a textbook, or a webpage? $\endgroup$ – zzzhhh Jun 4 '10 at 1:36
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    $\begingroup$ In the FAQ: "MathOverflow's primary goal is for users to ask and answer research level math questions, the sorts of questions you come across when you're writing or reading articles or graduate level books." The FAQ also lists some sites that are more appropriate for other sorts of mathematics questions. $\endgroup$ – Steve Huntsman Jun 4 '10 at 2:45
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    $\begingroup$ If 0<p, then 0^p = exp(p*ln(0)) = exp(p*(-∞)) = exp(-∞) = 0. I would vote to close if I knew how. $\endgroup$ – user5810 Jun 4 '10 at 3:37

The question is basically jumping to conclusions about page 181 of baby Rudin. (May he rest in peace.) He says there, "We now define $x^\alpha$ for any real $\alpha$ and any $x > 0$. The continuity and monotonicity of $E$ and $L$ show that this definition leads to the same result as the previously suggested one."

Well, Rudin does not say what might or might not go wrong when $x = 0$. The answer is that when $\alpha > 0$, then you can equally well ask for $f(x) = x^\alpha$ to be continuous and monotonic, and then it has a continuous extension to $x = 0$. I have not seen a reasonable alternative definition of $0^\alpha$ for any question in real analysis. This is so even though when $x < 0$ and $\alpha$ is either irrational or has an even denominator, then you can only reasonably say that $x^\alpha$ is either undefined or complex and multivalued. The only reason that Rudin would exclude $x = 0$ is that his formula $e^{\alpha \log x}$ doesn't work, and because $0^\alpha = \infty$ (unsigned, not $+\infty$) when $\alpha < 0$.

A less trivial version of the same issue occurs when you look at the entropy of the vector $\vec{x}$ when $||\vec{x}||_1 = 1$, i.e. $$H(\vec{x}) = \sum_k -x_k (\log x_k).$$ This is related to the derivative of the $p$-norm as $p \to 1$. In this case when $x_k = 0$, you still define the term in the sum as $0$, by continuous extension of $f(x) = x(\log x)$.

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    $\begingroup$ I also think that the other commenters aren't wrong to call it a naive question. $\endgroup$ – Greg Kuperberg Jun 4 '10 at 3:45

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