Let ${\bf x}=(x_1,...,x_n)$, the pnorm 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 pnorm? Assume that p does not take irrational value, or prescribe that $0^p$ means 0 when it occurs? Thanks!

closed as off topic by Steve Huntsman, Andrew Stacey, Robin Chapman, gowers, Gerald Edgar Jun 4 '10 at 11:42
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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 $f(x) = x(\log x)$. 

