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Taking the $L^p$ space as an example, $L^\infty$ is the natural limit of the $L^p$ spaces in an informal sense.

For the p adic numbers, s there anyway that the usual norm on $\mathbb{R}$ is a natural limit of the p adic norms?

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    You shouldn't think about $p$-adic numbers like $L^p$ spaces. For the latter, $p$ does not even have to be a rational number. – 2017-02-21
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    I'm not, it was ust an example to show what I meant. – 2017-02-21

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The answer to your question is no. As you may know, the $p$-adic norms satisfy the strong triangle inequality, i.e. $|x+y|_p\leq\max\{|x|_p,|y|_p\}$ for all $x,y\in\mathbb{Q}$. Suppose that $(p_n)_n$ is a sequence of norms (or seminorms) defined on $\mathbb{R}$ such that for each $n\in\mathbb{N}$, $p_n$ satisfy the strong triangle inequality. If $p$ is a norm on $\mathbb{R}$ such that $\lim_{n\to \infty} p_n(x)=p(x)$ for all $x\in\mathbb{R}$, then it's clear that $p$ satisfies the strong triangle inequality. Since the usual norm on $\mathbb{R}$ does not satisfy the strong triangle inequality, we deduce that it cannot be the limit of norms that satisfy such inequality, in particular, it cannot be limit of $p$-adic norms (or extensions of these).

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    Is the usual norm on $\mathbb{R}$ then an outlier in the set of norms on it? It seems weird that all norms follow a pattern except for this one outlier that has nothing to do with the others. – 2017-02-21
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    Every norm can be classified as Archimedean (does not satisfy the strong triangle inequality) or non-Archimedean (does satisfy the strong triangle inequality). Ostrowski's Theorem states that every norm in $\mathbb{Q}$ is either a power of the usual norm or a power of a $p$-adic norm, so any norm on $\mathbb{R}$ is an extension of one of these. – 2017-02-21
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    Your question and your comment make it look as if you think the $p$-adic norm(s) are on $\Bbb R$. This is far from the case. These are norms on $\Bbb Q$, none of them extendable to $\Bbb R$. With the modification that you’re asking about various norms on $\Bbb Q$, then, one might indeed say that the archimedean norm is an outlier, but its existence is exactly what makes Arithmetic so interesting (and hard!). – 2017-02-21
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    @Lubin Actually any valuation on $\mathbb{Q}$ (in particular, any p-adic norm on $\mathbb{Q}$) can be extended to $\mathbb{C}$. For example, the valuation $|\cdot|_p$ can be extended to $\mathbb{C}_p$. Since $\mathbb{C}$ and $\mathbb{C}_p$ are isomorphic you can extend $|\cdot|_p$ to $\mathbb{C}$ in a natural way. Of course, this construction cannot be done explicitly since the isomorphism is not explicit. – 2017-02-21
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    Extend to $\Bbb C$, yes; extend to $\Bbb R$, no, no, a thousand times no. – 2017-02-21
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    If $|\cdot|$ is a valuation on $\mathbb{C}$ that extends $|\cdot|_p$, then the restriction of $|\cdot|$ to $\mathbb{R}$ is a valuation on $\mathbb{R}$ that extends $|\cdot|_p$. – 2017-02-21
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    @Lubin That's not what I think. It's just that in my course the usual norm is being labelled as the infinity norm, and I was wondering if that was deeper than just notation. I am well aware that the norms are only on $\mathbb{Q}$. – 2017-02-22
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    Why the archimedean absolute value is called the “infinite prime” is dealt with in the answers to this question – 2017-02-22
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    Your objection to my ill-considered remark was well-taken, @Chilote. I guess I was trying to force some continuity on the valuation-extension, but in the context of this discussion, that was not right. – 2017-03-01