# What is the 0 - standard?

On $\mathbb{R}^n$ and also $p\ge 1$ the $p$ - standard is specified as $$\|x\|_p=\left ( \sum _{j=1} ^n |x_j| ^p \right ) ^{1/p}$$ and also there is the $\infty$ - standard which is $\|x\|_\infty=\max _j |x_j|$. It is called the $\infty$ standard due to the fact that it is the restriction of $\|\cdot\|_p$ for $p\to \infty$.

Currently we can make use of the definition over for $p<1$ too and also specify a $p$ - "norm" for these $p$. The triangular inequality is not completely satisfied, yet I will certainly make use of the term "norm" however. For $p\to 0$ the restriction of $\|x\|_p$ is clearly $\infty$ if there go to the very least 2 nonzero access in $x$, yet if we make use of the adhering to changed definition $$\|x\|_p=\left ( \frac{1}{n} \sum _{j=1} ^n |x_j| ^p \right ) ^{1/p}$$ after that this need to have a restriction for $p\to 0$, which need to be called 0 - standard. What is this restriction?

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2019-12-02 02:51:42
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When $p$ is tiny, $$x^p = \exp(p \log x) \approx 1 + p\log x.$$ Therefore $$\frac{1}{n} \sum_{j=1}^n x_j^p \approx 1 + p\frac{1}{n} \sum_{j=1}^n \log x_j = 1 + p \log \sqrt[n]{\prod_{i=1}^n x_j}.$$ On the various other hand, we have $$(1+py)^{1/p} \longrightarrow \exp(y),$$ therefore we conveniently get that the standard comes close to the geometric mean, as Raskolnikov commented.