I have heard references to "hard" vs. "soft" analysis. What is the difference? It seems to do with generality versus nittygritty estimates, but I haven't gotten any responses more clear than that.

7$\begingroup$ <a href="terrytao.wordpress.com/2007/05/23/… post</a> at Tao's blog gives a handful of examples illustrating the two concepts, and also provides an indepth account of one instance. $\endgroup$– Ed DeanNov 20, 2010 at 2:07

$\begingroup$ Oops, thought I could use tags in the comment; that looks dreadful, sorry. $\endgroup$– Ed DeanNov 20, 2010 at 2:08

6$\begingroup$ Ed: because of the formatting error, your link is not working. Below is the link that Ed wanted to use: terrytao.wordpress.com/2007/05/23/… $\endgroup$– Willie WongNov 20, 2010 at 2:17

3$\begingroup$ I don't think there is a neat separation. Maybe we should introduce a Mohs scale of analysis hardness. $\endgroup$– Pietro MajerAug 8, 2020 at 14:50

$\begingroup$ See also: math.stackexchange.com/q/141364/442 for the same question. $\endgroup$– Gerald EdgarAug 8, 2020 at 17:25
4 Answers
Disclaimer: I'm no expertthis is really a question for the analysts and historians of mathematics.
As far as I know,the terminology came into existence in the early 20th century distinguish the classical "calculus" type analysis (hard analysis) from the new point set topology/functional analysis approach (soft analysis). A hard analytic argument uses a direct calculation or construction of an exact estimate bounds of specific function or function types to prove a statement. A soft analytic arguement uses the general topological or geometric properties of a space in which a function or function class is defined to prove a result indirectly without a precisely calculated "bound".
For example, the fact that the Cantor set has measure zero is a "hard" analytic arguement; it uses an epsilondelta arguement to show the limit of the sequence of "slices" of the lengths of it's component intervals on the real line converges to 0.
An example of a "soft" analytic arguement: (IVT) Let $f$ be a continuous function defined on a connected subset of the real line i.e. an interval with a well defined least upper bound and greatest lower bound. Then the function is defined at every point inbetween the lub and the glb. A soft proof would be as follows: Since an interval $I$ of $\Bbb R$ is a connected subset of $\Bbb R$ and $f$ is continuous, then $f(I)$ is also connected. Therefore, for every $x \in I$, $f(x)$ is in $f(I)$. Notice this proof does not involve a direct computation of bounds that proves $f(x)$ is in the image set of $f$ (although it certainly COULD be proven that way).
Anyway, that's how Gerald Itzkowitz taught it to me and I learned a long time ago to trust him on these matters........

2$\begingroup$ I think that your answer about where the line is drawn is good, but I would like to point out that Sobolev spaces are in the realm of functional analysis but the various embedding theorems are part of hard analysis. I think that is right. I personally draw the line by seeing what i find appealing and what i don't. Also, I think you have to get a bit deeper into the field before you start to see the stark differences between the two, in that the examples that best exemplify it are further down the rabbit hole. I think you are right about the epsilon delta stuff though. $\endgroup$ Nov 20, 2010 at 3:15

1$\begingroup$ Well,my knowledge of functional analysis and operator theory is still rather cursory,Seanhence the disclaimer at the beginning.I hope to rectify this by independent study after getting my Master's this January. $\endgroup$ Nov 20, 2010 at 4:25

$\begingroup$ I was not able to quickly find a reference, but I remember that also Gaetano Fichera described the concept in the same way in several historical papers on the mathematics (in Italy) at the beginning of the 20th century. In particular, in the school of Mauro Picone, the “hard analysis” approach was made concrete for example in the emphasis on the explicit calculation of the constants involved in a priori estimates for PDEs, due to their importance for the construction of solutions. $\endgroup$ Aug 8, 2020 at 18:12
A historical note.
Hermann Weyl mentioned in his talk "Felix Kleins Stellung in der mathematischen Gegenwart" that the dichotomy of "hard vs. soft analysis" had been suggested by Hardy. According to Hardy, there is the function theory of the "hard, sharp, narrow" kind (due to Bohr, Landau or Littlewood) as opposed to the "soft, large, vague" kind (due to Birkhoff or Koebe).
Edit. Apparently, Hardy's musings are contained in his paper "Prolegomena To a Chapter on Inequalities" (unfortunately, I don't have access to it at the moment).
Edit 2. Indeed, here's the quotation from Hardy's paper.
A thorough mastery of elementary inequalities is today one of the first necessary qualifications for research in the theory of functions; at any rate, in function theory of the "hard, sharp, narrow" kind as opposed to the "soft, large, vague" kind (I do not use any of these adjectives as words either of praise or blame), the functiontheory of Bohr, Landau, or Littlewood, as opposed to the functiontheory of Birkhoff or Koebe. It is essential to anyone working in this field to be master both of the main results and of the tricks of the trade.

8$\begingroup$ Wouldn't it be nice if some charitable soul would translate that into a less barbarous language? $\endgroup$ Nov 20, 2010 at 2:43

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3$\begingroup$ I confess that by default, I interpret words in their modern meaning if one exists, in this particular case e.g. as in thefreedictionary.com/barbarous. I am glad this is cleared up now and I shall certainly refrain from flagging. $\endgroup$– Alex B.Nov 20, 2010 at 3:41

2$\begingroup$ To be honest, I am quite puzzled by the fact that usage of the word barbarous in its meaning in a language I don't understand (originally, in a language which is not Greek, later, not in Greek nor in Latin, and so on...) could rise such opposition as evinced by the upvotes on Alex's initial commentwhich can only be understood as downvotes on mine. I don't expect anyone to catch a jokingly reference to Herodotus, but what do people really find so downvotable? $\endgroup$ Nov 21, 2010 at 2:20

6$\begingroup$ Mariano, the most likely reason people find your comment downvotable is that barbarous is essentially a synonym for barbaric in English, i.e. savage, violent, cruel, brutal, uncivilized, and uncultured. So, the obvious interpretation of your statement is something to the effect of "wouldn't it be nice if someone translated that ugly German text into a language actually spoken by civilized people?" I assume this was not your intention, of course! It is also somewhat common for people to say that German is an ugly language, probably a sentiment that has carried over from the 1940s. $\endgroup$ Nov 30, 2010 at 20:03
Consider the problem of shuffling a deck of cards using some shuffling technique. One may wonder, "If I use this or that shuffling technique, will I shuffle the deck?" The problem can be transferred to a question of whether a particular finite Markov process converges to the uniform distribution or not. Omitting some details, a classical theorem says that, yes, the process will converge (to the uniform distribution) as long as your technique is reasonable. Not only that, the convergence will eventually be exponential. This seems like a useful theorem, but it is actually rather deficient. The problem is that your practical side may wonder how many shuffles are required to get the deck reasonably randomized, and this theorem doesn't help. So you might say that the original analysis was soft because the result does not help in solving this quantitative problem. It tells you that shuffling is a good idea, but it doesn't give you any clue whether a given technique could shuffle the deck in your lifetime or not. A hard analysis would tell you, for example, "If one defines reasonably randomized by measure blahblah, then $2\log_2(52)$ riffle shuffles are sufficient to randomize the deck."

$\begingroup$ A probablistic arguement is a good idea,Peterbut I'm not sure if as it stands it'll fly. Any of the probablists or analysts in the room want to chime in on this? $\endgroup$ Nov 20, 2010 at 18:25

$\begingroup$ I'm not exactly sure what you mean that this is a probabilistic argument. One can phrase it in a probabilistic way, but really it boils down to a hard linear algebra problem since the Markov process has finitely many states (the measures here are just vectors/functions). The statement that convergence happens is just the PerronFrobenius Theorem. It is rather convenient to use probabilistic methods in the hard analysis of shuffling, though. I think this example rather clearly demonstrates the struggle between generality and quantitativeness. By the way, I myself happen to be an analyst... $\endgroup$ Nov 20, 2010 at 23:44
Cantor sets, then. I would expand the ternary Cantor set by a factor of three, note that this makes two disjoint copies, and conclude the measure was zero that way. A "soft" argument indeed. That does make the point that "facts" need not belong to hard or soft varieties; arguments may do. Other Cantor sets don't have the exact structure to carry out the enlargement. So you need a "hard" argument to include them, at least on the face of it.


4$\begingroup$ @Andrew: if you wanted to ask what Charles meant, anything specific, then do so? "Huh?" is not a sensible comment nor question. $\endgroup$ Nov 21, 2010 at 4:02

4$\begingroup$ To quote Andrew L: the fact that the Cantor set has measure zero is a "hard" analytic argument. That is something I contest in two ways. Firstly, the fact is not either "hard" or "soft". Second, the argument I sketch by enlarging the ternary Cantor set is what I'd call "soft" since it is based on selfsimilarity. As I mentioned, there are other Cantor sets and selfsimilarity doesn't work directly for those; and at this point an "epsilondelta" or at least "let epsilon > 0" argument may be the obvious way to proceed. $\endgroup$ Nov 21, 2010 at 8:41

$\begingroup$ A hard argument for this would give more: For any integer k calculate integers m, n with km<n such that the Cantor set is covered by fewer than m intervals all of length 1/n. $\endgroup$ May 7, 2018 at 11:01