I was wondering with triangle inequality is valid for p-norm like this.

$x\in\Bbb{R}^n$, for all $p\ge1$,$$\Vert{x}\Vert_p=\left(\sum_{i=1}^{n} \vert{x_i}\vert^p\right)^{1/p}.$$

And I found a good repo for this Why is every $p$-norm convex?.

Months ago I started to learn measure theory and I notice the Minkowski inequality is stated over a different space (measure space) and in different format of the norm(at least is not the same format with that in $\Bbb{R}^n$), here it is:

Consider we have a measure space $(\Omega,A,\mu)$, and we define r-th norm as $$\Vert{x}\Vert_r= \{E|x|\}^{1/r}$$ this E only makes sense when we are talking about some measurable function X over this measure space. So my question is why we can use it to prove something not over this space(just in Rn), since the ways to define them are different IMO.

This might seem stupid but really confuse my here. Thank you!


1 Answer 1


$\mathbb{R}^n$ is the space of measurable functions on the measure space $(\Omega, \mathcal{F})$ where $\Omega = \{1,\dots,n\}$ and $\mathcal{F}$ is the power set of $\{1,\dots,n\}$. To see this, think of a vector $x$ as the function $x:\{1,\dots,n\} \to \mathbb{R}$ defined by $x(n) = x_n$.

Then the p-norm on $\mathbb{R}^n$ is the p-th integral norm arising from the counting measure (defined by setting $\mu(\{k\}) = 1$ for every $k$) since if $f: \Omega \to \mathbb{R}$ then $\int_\Omega f d\mu = \sum_{k=1}^n f(k)$. So $$\|x\|_{L^p(\Omega,\mathcal{F},\mu)} = \bigg(\int_\Omega |x|^p d\mu \bigg)^\frac1p = \bigg(\sum_{k=1}^n |x_n|^p \bigg)^\frac1p = \|x\|_p$$

  • $\begingroup$ Important notational pedantry: the operator $E$ is equivalent to the operator $\int$, which is where the connection comes from. $\endgroup$ Commented May 27, 2018 at 22:23
  • $\begingroup$ @DavidKraemer What is $E$ if not just the integral? $\endgroup$ Commented May 27, 2018 at 22:27
  • $\begingroup$ I just wanted to mention explicitly that they were the same to head off any confusion about why the question uses $E$ and the answer uses $\int$. $\endgroup$ Commented May 27, 2018 at 22:30
  • 1
    $\begingroup$ @DavidKraemer You just mean they are different notation for precisely the same thing? If so I misunderstood you, sorry! I probably should have been more aware of this potential source of confusion when writing the answer. $\endgroup$ Commented May 27, 2018 at 22:32
  • $\begingroup$ @DavidKraemer yeah I do know that; Is this the formal equivalent form for the Rn in measure theory language if there is any? Because I see any n element can serve as sample space in this specific example. $\endgroup$
    – Li haonan
    Commented May 28, 2018 at 0:19

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