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Who first defined the norm in $L_p$ space as $$\left(\int{\lvert f(x) \rvert^p}\right)^{1/p}$$

Is there any reference for this? Is it just an simple extension from $L_2$?

$L_p$ space has some really nice properties. I think it relies on the definition of this norm, but this definition does not seem to come naturally or maybe I just can't see how it comes naturally. I understand this definition gives a norm. I just want to know the underlying thought of it.

Are those nice properties just by chance or does it actually have a relationship with this definition?

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  • $\begingroup$ What!? No mention of Cauchy Schwarz or Holder on this question? $\endgroup$ – abnry Mar 10 '16 at 23:18
  • $\begingroup$ My 2 cents (a related answer) $\endgroup$ – Giuseppe Negro Mar 10 '16 at 23:34
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In a certain sense, there are basically $4$ $L^p$ spaces with $p \geq 1$: $L^1,L^2,L^\infty$, and everything else. Specifically:

  • $L^2$ is the only "Hilbertable" $L^p$ space, which gives it a wealth of additional properties.
  • For "typical" measure spaces, all $L^p$ spaces except $L^\infty$ are separable.
  • For any measure space, $L^p$ is reflexive for all $1<p<\infty$. For "typical" measure spaces, $L^1$ and $L^\infty$ are both not reflexive.
  • $L^1$ loses some theorems held by $L^p$ spaces with $1<p<\infty$ because for "typical" measure spaces it is not isomorphic to the dual space of any Banach space. (It "would" be the dual of $L^\infty$ but it is usually too small.) This means that the theorem "the unit ball in a space which is the dual of something is weak-* compact" does not apply to $L^1$, and thus neither do many of its consequences.
  • $L^\infty$ also loses a lot of other theorems held by other $L^p$ spaces...and frankly I have some trouble giving a simple explanation for why. In PDE theory many of the relevant theorems fail because approximation by compactly supported smooth functions fails in $L^\infty$, but there is something more general that makes theorems in $L^\infty$ more elusive.

Here any time I say "typical" my statements include the Lebesgue measure, all measures which are mutually absolutely continuous with respect to it, and restrictions of such measures to positive measure subsets.

I know that one place where "unusual" values of $p$ (i.e. not $1,2$ or $\infty$) pop up is in Sobolev space theory, specifically with Sobolev embedding. Here a Sobolev space in sufficiently high dimension will embed into a Sobolev space with less regularity and more integrability. For example, in dimension $6$, $W^{1,2}$ embeds into $L^3$. In dimension $2$, $W^{1,2}$ embeds into $L^p$ for all $1 \leq p < \infty$.

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  • $\begingroup$ Small nitpick: if one allows $L^p$ spaces for $0<p<1$ (for whatever reason -- triangle inequalities are sometimes dispensable?), then there should be 5 types: $L^1,L^2,L^\infty$, pathological/non-metric ones, and everything else. $\endgroup$ – Clement C. Mar 8 '16 at 0:19
  • $\begingroup$ @ClementC. Valid point. Rather than actually inviting these into the discussion, I revised slightly to specifically exclude them from the discussion. $\endgroup$ – Ian Mar 8 '16 at 0:41
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The cases $p=1,2$ are quite natural: the former is a very reasonable definition of "how large a function is" and the latter arises from Hilbert space theory: a well developed and useful theory. If you want to generalize these for arbitrary $p$, you are forced to consider $\int |f(x)|^p$. This is not a norm for arbitrary $p$ because $\int |a f(x)|^p = |a|^p \int |f(x)|^p$, and there is exactly one way to remedy this: by introducing the power $\frac{1}{p}.$ Miraculously (or maybe not?) the $L^p$ norm we constructed is actually a norm.

The most curious case is $p=\infty$. At a glance, there is no reason to call the essential supremum norm $L^\infty$ as it has nothing to do with integrals. The following result shows that under certain conditions the infinity norm is the limit case of the $p$ norm:

If $f\in L^p \cap L^\infty$, $p<q<\infty$ then $f \in L^q$ and $\lim_{q \to \infty} \|f\|_q = \|f\|_\infty$ So, naming $L^{\infty}$ like that is justified.

That being said, $L^{\infty}$ is often the worst space to work in: see @Ian's answer for details.

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If p is close to 1, the p-norm gives more importance to the regions where the function is small and less importance to regions where it is large. If P is large, then the norm gives less importance to the regions where the function is small and more to those where it is large.

Thus varying to allows us to decide what we want to focus on.

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  • $\begingroup$ That's basically true on an infinite measure space, but there is a distinct monotonicity in the finite measure space context. $\endgroup$ – Ian Mar 8 '16 at 1:34
  • $\begingroup$ The norms for different ps give different weight to the different regions independently of finiteness of the measure of the space. $\endgroup$ – Mariano Suárez-Álvarez Mar 8 '16 at 1:37
  • $\begingroup$ In terms of the actual value of the $L^p$ norm, yes, but in terms of "which functions are in which spaces", no. $\endgroup$ – Ian Mar 8 '16 at 1:38
  • $\begingroup$ But that is irrelevant. A norm is a way of measuring functions: it allows you to say "this function is larger than that one" and the choice of P fixes a criterion to do that, based on different priorities. $\endgroup$ – Mariano Suárez-Álvarez Mar 8 '16 at 1:47
  • $\begingroup$ Eh, that depends on how you think about it I guess. "Pure" real analysts often don't care what a number is as long as they know it's finite and doesn't depend on any parameters. I guess in regards to your point, what I am saying is that if the measure space is finite, it is not that small regions are weighted that much less as it is that large regions are weighted that much more, when $p$ is large. To be sure, small regions are indeed weighted considerably less in $\| f \|_p^p$ when $p$ is large, but if you remove that exponent that effect essentially goes away. $\endgroup$ – Ian Mar 8 '16 at 1:50
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In a very rough and informal way, this is how I understand this definition (and I find it very natural)

Starting from finite dimension space, suppose a function $f$ has finite components, say $f_1, ...f_n$. You want to understand its certain behavior thru its components. One way of doing this is by averaging all its components in some way (here we choose $|f_i|^p$). You may choose "arithmetic average" $\displaystyle \frac {1}{p} \sum\limits_{i=0}^n |f_i|^p $, or "geometric average" $\displaystyle \big(\sum\limits_{i=0}^n |f_i|^p\big)^{1/p} $. The former is not so suitable because the usual average is not efficient to control the growth of exponential function, whereas the latter controls that much better, thus be a better candidate for a norm.

Now move to infinite dimension space. You would have "infinite" components in the norm, so you would end up with something like $||f||_p = \big(\sum\limits_{i=0}^\infty |f_i|^p\big)^{1/p}$. This makes you think of integral notation, because integral is essentially an infinite sum (think of how Riemann integral is constructed). Also you will have no more "components" but $f$ itself. Finally, you end up with something like the definition.

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  • $\begingroup$ Neither of those is a mean (there are $n$ terms, not $p$ terms). $\endgroup$ – Ian Mar 10 '16 at 23:22
  • $\begingroup$ Yes, they are not. But I would say they "look" like mean. $\endgroup$ – SiXUlm Mar 10 '16 at 23:24
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    $\begingroup$ Not really: the $1/p$ is just the right exponent for homogeneity, that's all. $\endgroup$ – Ian Mar 10 '16 at 23:25

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