Talk:Min-max theorem

[Untitled][edit]

What happens if a positive eigenvalue occurs with multiplicity 2? Also, the statement as it stands precludes the possibility that there is no minimal strictly positive eigenvalue. I'm changing it accordingly... L upin 15:49, 30 July 2005 (UTC)[reply]

compact operators[edit]

For infinitely dimensional spaces, there might well be only a finite number of positive eigenvalues or no eigenvalues at all, contrary to what is stated in the demonstration in the section devoted to compact operators. In this case, there is of course an infinite number of negative eigenvalues. The demonstration should be modified accordingly. For example when the number of positive eigenvalues is a finite n and $S_{k}$ has $k>n$ , then it seems to me that $\max _{S_{k}}\min _{x\in S_{k},\|x\|=1}(Ax,x)=0.$ Similar considerations should apply also for negative eigenvalues. Luca.Argenti (talk) 09:23, 29 May 2008 (UTC)[reply]

min max principle for singular values[edit]

I have found many other sources, which change the place of min and max in the equations, so It seems incorrect here.14.139.38.11 (talk) 05:10, 17 February 2014 (UTC)[reply]

However, we hould have a section on this. I will se if I can find good references. Kjetil B Halvorsen 14:47, 25 April 2014 (UTC) — Preceding unsigned comment added by Kjetil1001 (talk • contribs)

compact operators, question concerning the proof[edit]

In the article it is written: "But A is compact, therefore the function f(x) = (Ax, x) is weakly continuous.". I don't see how this follows. I can easily prove that f(x) is weakly sequentially continuous by using that a compact operator maps weakly convergent sequences into norm convergent sequences. However, it is well known that this is not true for nets. As far as I know, however, there is no guarantee that a functional attains its minimum if it is only sequentially (lower semi-)continuous. Maybe there is some theorem behind here that I don't know. If so, could someone please give a reference? I think it should be cited in the article then. — Preceding unsigned comment added by 138.246.2.201 (talk) 00:34, 30 January 2017 (UTC)[reply]

In fact, various claims in that proof are simply wrong.

First of all, it is not true that any bounded set in $H$ is weakly compact. If $H$ is infinite-dimensional, then the weak closure of $\{x\in H:\|x\|=1\}$ is all of $\{x\in H:\|x\|\leq 1\}$ (c.f. Conway, A Course in Functional Analysis, Exercise V.1.10, or this question on Mathematics Stack Exchange). As a concrete example, note that any orthonormal sequence converges weakly to zero; see Weak convergence (Hilbert space).
Secondly, to answer the question above, it is not true that the function f(x) = (Ax, x) is weakly continuous. If $H$ is infinite-dimensional, $A$ is injective, and we take the net from this answer on Mathematics Stack Exchange, then the image net is not eventually bounded (perpetually unbounded?), whereas the original net converges to zero.
To see that the minimum $\min _{x\in S_{k},\|x\|=1}(Ax,x)$ is attained, you may use that $S_{k}$ is finite-dimensional (so its unit sphere is compact in the norm topology).
Finally, it is simply not true that the maximum $\max _{x\in S_{k-1}^{\perp },\|x\|=1}(Ax,x)$ is always attained. If $H$ is infinite-dimensional and $A$ is negative definite on $S_{k-1}^{\perp }$ (by which I mean negative and injective; we still have $0\in \sigma (A)$ because $A$ is compact), then the supremum is 0 but we have $(Ax,x)<0$ for all $x\in S_{k-1}^{\perp }$ . This is not a big problem: the mere existence of a subspace $S_{k-1}$ with this property is enough to conclude that there are fewer than $k$ positive eigenvalues to begin with, so $\lambda _{k}$ doesn't even exist in this case. Nevertheless, it shows that the current proof is way too optimistic, and the compactness argument is beyond repair.

I suggest that the proof be revised or removed in favour of a reference to the literature. I am not sure how to deal with this; I delegate this to someone who is a little more experienced on Wikipedia and/or a little more "in" on the discussion regarding mathematical proofs on Wikipedia (see Wikipedia:WikiProject_Mathematics/Proofs).

Min-max vs Inf-sup[edit]

it seems to me the proof only argues for Inf-sup rather than Min-max (OK, the Min is quite obvious). I'd say the fact that the min is achieved is through the specific subspace span(u1,…uk) and the sup is achieved because R_A is continuous and we can restrict it to a compact (the closed unit ball). Doesn't that need to be mentionned? Jeanlesquare (talk) 08:44, 10 November 2018 (UTC)[reply]