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I have read the Wikipedia page on the topic, and the answers on this post focusing solely on $L^1$. The integral definitions are very clear, and the fact that $L^2$ has a natural norm induced by the scalar product, which is important in Fourier tranforms, and other bits of knowledge around these definitions, still leave me wanting for a couple of plots of functions, one for each space, as well as a short definition in English, even if only approximate.

To some degree, and leaving aside measure theory, which would likely have pre-empted this post to begin with, the absolute value of the functions may seem to serve a similar purpose to squaring. Yet, these are completely separate spaces.

If it wasn't clear enough from the tone of my answer, or my self-deprecating profile, I have no formal training in mathematics, so I'm looking for answers along the vein of the notable Terry Tao's explanation here:

I’ll start today with my article on “Function spaces“. Just as the analysis of numerical quantities relies heavily on the concept of magnitude or absolute value to measure the size of such quantities, or the extent to which two such quantities are close to each other, the analysis of functions relies on the concept of a norm to measure various “sizes” of such functions, as well as the extent to which two functions resemble to each other. But while numbers mainly have just one notion of magnitude (not counting the $p$-adic valuations, which are of importance in number theory), functions have a wide variety of such magnitudes, such as “height” ($L^\infty$ or $C^0$ norm), “mass” ($L^1$ norm), “mean square” or “energy” ($L^2$ or $H^1$ norms), “slope” (Lipschitz or $C^1$ norms), and so forth. In modern mathematics, we use the framework of function spaces to understand the properties of functions and their magnitudes; they provide a precise and rigorous way to formalise such “fuzzy” notions as a function being tall, thin, flat, smooth, oscillating, etc.

Evidently my question here is a bit more concrete, but not far off from what Professor Tao is addressing in the most natural, uncondescending and didactic way to capture as broad a segment of his blog's readers as possible. Clearly he is not talking at the audience, seeking acclaim from the initiated, but rather communicating effectively.

The first $1/2$ of the answer I am looking for would be this motivation for $L^2$:

Roughly speaking, $L^2$ space is the only functional space among $L^p$ spaces which is a Hilbert space, i.e. it has an inner product (and also complete)! One can imagine this spaces as a generalization of $\mathbb R^n$ to infinite dimensional cases. So many trends like finding minimum/maximum of function from $\mathbb R^n$ to $\mathbb R$ can be generalized to these spaces in a similar way...

But I am looking for more motivation behind the comment, "$L^1$ and $L^2$ are not "completely separate", in that $L^1 \cap L^2$ is a "very large" set."

Is $\{L^1\} > \{L^2\}$? How much bigger (or smaller)? What does it mean that the former measures the mean, while them latter, the energy? Etc.

There is this post in the Physics SE, also delving into $L^2,$ but again, not making an intuitive comparison with $L^1.$

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    I don't see what is self-depreciating about "We have a hotel in Paris, and we'll fly to see Italy on Friday, and fly back home on Saturday", and can read nothing from the tone of your question (unless you are referring to an answer somewhere?) I get a very similar feel from the quote of Tao and from the answer of Gyu Eun Lee. The main difference being explicit functions are mentioned, and calculations are alluded to. Can you try to specify what answer you are instead looking for? I'll also note that $L^1$ and $L^2$ are not "completely separate", in that $L^1\cap L^2$ is a "very large" set – Calvin Khor Apr 30 '21 at 01:29
  • @CalvinKhor You can't see Italy in a day; and you can't get more than a passing motivation of advanced mathematics without a life of dedication and effort. But there are tourists, and they still like to say they loved seeing Italy, even if it was for one day; and there are those that are just curious about mathematics, acknowledging that we are utterly unqualified. As for the substance of your comment... – Antoni Parellada Apr 30 '21 at 01:36
  • Fair enough, such between-the-lines if not explicitly said would be certainly far beyond my comprehension :) mirroring your current question in some ways i suppose – Calvin Khor Apr 30 '21 at 01:38
  • @CalvinKhor I am trying to find an example (hopefully not on some technical measure theoretical issue) of an $L^1$ function (integrable over a set wrt some measure), which turns out not to be in $L^2.$ – Antoni Parellada Apr 30 '21 at 01:39
  • This is in fact provided by the answer of G.E.L. , though perhaps implicitly. Consider $|x|^{-\alpha}$ for some $\alpha>0$. Squaring a function makes big bigger, and small smaller. So if it is $L^1$ for $|x|>1$, then its square which is smaller, would certainly be integrable. But for $|x|<1$, it turns out you can square and make it too big. Specifically you can consider the function $$f:\mathbb R\to\mathbb R, \quad f(x) := \begin{cases}1/\sqrt{|x|} & |x|<1 \ 0 & |x|\ge 1\end{cases}$$ You can do the same in $\mathbb R^n$ by replacing the exponent $-1/2$ with $-n/2$ – Calvin Khor Apr 30 '21 at 01:42
  • Absolutely not! $5^2$ is much bigger than $5$. $0.001^2$ is much smaller than $0.001$. The behavior flips around $1$; $x>1$ implies $x^2>x>1$, and $0<x<1$ implies $0<x^2<x<1$. – Calvin Khor Apr 30 '21 at 01:54
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    @CalvinKhor This is exactly what I needed to get a bit of the motivation behind all this. Thank you! – Antoni Parellada Apr 30 '21 at 01:54
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    for $|x|<1$, $\frac1{|x|^{\alpha}} > 1$, so $\left(\frac1{|x|^\alpha }\right)^2 > \frac1{|x|^{\alpha}}$. – Calvin Khor Apr 30 '21 at 01:59
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    @CalvinKhor Right! No confusion. I think it was just the wording in this case: In "Squaring a function makes big bigger, and small smaller" you were referring to below $1.$ Thanks again. – Antoni Parellada Apr 30 '21 at 02:02
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    @Mathtourist: Whether one uses $L_1$ or $L_2$ also depends on specific applications. For example, in predictive modeling, regression models for examples, $L_2$ is used more that $L_1$ since $L_2$ works better with means, while $L_1$ dorks better with medians. In some applications in optics, $L_1$ estimates works better than $L_2$ estimates. – Mittens May 01 '21 at 00:59

3 Answers3

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On $\mathbb{R}^d$, all quantitative properties of $L^p$ spaces can be summarized by the functions $f(x) = |x|^{-\alpha}$. An $L^p$ function cannot decay (on average) at infinity at the rate $|x|^{-d/p}$ or slower, and cannot have any singularities which blow up at the rate $|x|^{-d/p}$ or faster.

This can be encapsulated by the use of weak $L^p$ spaces, which are essentially $L^p$ spaces supplemented by functions which are barely not in $L^p$. The function $f(x) = |x|^{-\alpha}$ is in weak $L^p$ if and only if $\alpha = \frac{d}{p}$. This means that, in some sharp sense, $|x|^{-d/p}$ is barely on the edge of belonging to $L^p$. This can be thought of as a quantitative characterization of $L^p$: it consists of functions which behave better than $|x|^{-d/p}$ for the purposes of integration (i.e. on average).

On $L^p$ spaces over other measure spaces it can be a little harder to make this precise, but the basic idea is the same: $L^p$ functions have a threshold blowup rate and spatial distribution rate that they cannot exceed.

Gyu Eun Lee
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One way to look at the difference is through the Fourier transform (https://en.wikipedia.org/wiki/Fourier_transform). The Fourier transform can be viewed as taking a function whose variable is time and returning a function whose variable is frequency. That is, it moves us from time variables to frequency variables. The Fourier transform of an $L^2$ function is an $L^2$ function (see the Plancherel theorem), so $L^2$ functions have the same behavior in time as they do in frequency. On the other hand, the Fourier transform of an $L^1$ function must be continuous and decay to zero at infinity (see the Riemann-Lebesgue Lemma), so the behavior in frequency is very different than in time.

cmk
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  • That is a cool take. I understand that the Fourier transform maps $L^1$ to $C^0$ (continuous functions), and $L^2\to L^2$? – Antoni Parellada Apr 30 '21 at 03:07
  • @Mathtourist That's correct, but the even more striking result is that the Fourier transform maps $L^1$ to $C_0$ (continuous functions that go to 0 at $\pm\infty$). So, the behavior in frequency is very different for $L^1$ functions. – cmk Apr 30 '21 at 13:52
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Although this comment/answer is not directly about intuition, it may be worthwhile to say that there are many useful spaces of functions, with various properties. The "why" of the properties is sometimes that we can prove the theorems, and sometimes tha we find ourselves compelled to consider these spaces.

Comparisons between various spaces of functions are sometimes easy, sometimes difficult. Sometimes profound.

In most cases of interest to me, the "definitions" of spaces of functions are really characterizations, motivated by some intention. And the "comparisons" are the basic theorems I need to get started on some project. (And they may be the main point, but it's hard to know in advance.)

paul garrett
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