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11:23
2
A: Operator bounded then adjoint is bounded

Theo BenditThe adjoint has a more general meaning in the context of Banach Spaces, which I think is actually more intuitive than the Hilbert Space case. Suppose $X$ and $Y$ are Banach spaces and $T : X \to Y$ is bounded and linear. Then, define $$T^* : Y^* \to X^* : f \mapsto f \circ T.$$ Then, because $T$...

This is a very beautiful proof, thank you! May I ask you in which context have you encontre such arguments?
@Surb Well, the adjoint is a standard tool in functional analysis, and the Riesz representation theorem is utterly fundamental to the study of Hilbert Spaces in particular, and is what makes the adjoint come into its own. Would that I could say that I did something special here, but the above arguments are standard in these fields. :-)
@Surb Functional analysis can be very beautiful though. I particularly enjoy the geometric flavour of functional analysis, where geometric properties of the space and its unit ball tell you deep properties about the space itself.
Thanks for your answer, sorry I wasn't precise enough. I'm particularly interested by the Banach space interpretation of the Adjoint and in particular your first estimate. I know it (quite) well for finite dimensional vector space and the proofs I know, hold for any norm but uses an inner product.
@Surb I don't tend to use the adjoint much in my work, so I can't really tell you where specifically to go further with it. I work more with nonlinear operators.
:) me too, I was about to say: "Functional analysis can be very beautiful though. I particularly enjoy the geometric flavour " I agree and actually my favorite is "nonlinear functional analysis"
11:23
What do you do in particular?
Well, I'm working on nonlinear Perron-Frobenius theory.
which are nonlinear eigenvalue problems
and you?
I'm tackling the Chebyshev conjecture: it's the conjecture that, if a subset of a Hilbert space admits unique closest points, then the set must be convex.
so for every y in V \ S there exists a unique z in S such that ||y-z|| is minimal, right?
Yes.
Or equivalently, just $y \in V$.
sure
11:28
It's a beautiful problem; so simple to explain, but so hard to prove or disprove.
Looks interesting :)
... can the problem be reformulated in terms of computing norm of linear operators?
How do eigenvalues for nonlinear functions work? I thought going into infinite dimensions and losing linearity would make eigenvalues fairly meaningless.
well it is just f(x) = lambda x
and typically you'd assume f to be homogeneous
of some degree
Ah OK, that makes some sense.
to keep eigenvectors as rays and possibly change the eigenvalue on the ray if f is not homogeneous of degree 1
the connection with your answer is the following:
Consider the p norm ||.||_p on R^n and A \in R^{m x n}, then you wan to compute:
max_x ||Ax||_p / ||x||_q
If you write the critical point equation of this optimization problem you obtain something of the form
(A^T(Ax)^(p-1))^(1/(q-1) = lambda x
where the exponents are take componentwise
(this is a nonlinear eigenvalue problem)
11:34
OK, cool!
I'm guessing there's plenty more applications to optimisation in there too.
indeed
In response to your earlier question, you can't really reformulate the Chebyshev conjecture in terms of just computing a norm of a linear operator. In a sense, computing the norm is akin to computing the distance function, which is an optimal scalar. The metric projection is more akin to the set of points with optimal norm.
but, this is the point. In fact, the matrix norm above is in general NP-hard to compute.
However, with Perron-Frobenius theory you can find conditions on the norm and the operator under which the global maximizer is unique up to sign.
"The metric projection is more akin to the set of points with optimal norm." This is what I had in mind
Something along the lines
There exists U^*= U^*(S) in V^* such that if ||L|| has unique maximizer for all L in U^*, then the conjecture is verified for S
(I'm going for a quick cigaret, I'll be back in 3min)
OK
My work specifically has been very focused on using Mobius-transform-like nonlinear operators to fluidly change between projection problems, furthest point problems, and functional maximisation problems (which is more general than finding a norm), which yields some nice results.
Actually, that's one thing I've been wondering for a bit now: if a bounded set has the property that every non-zero functional achieves its maximum uniquely, then is the set convex? If it's true, I can prove the Chebyshev conjecture.
6
Q: Non-trivial faces of the closed convex hull of a non-convex closed set with connected complement

Theo BenditI'm trying to prove or disprove a problem, but I'm struggling to make headway. Any help is appreciated. Suppose $X$ is a Hilbert Space, and $C \subseteq X$ is closed, bounded, non-convex, and $X \setminus C$ is connected. Does the boundary of $\operatorname{conv} C$ necessarily contain a line...

Interesting, I'll think about it. Could you recommend me a couple of papers on the problem?
11:48
For a gentle introduction, I can recommend math.auckland.ac.nz/~moors/chebyshevsurveyJAMS.pdf by my supervisor and his old masters student. For some really neat ideas (most of which are also included in the above survey) I can definitely recommend ams.org/journals/tran/1969-144-00/S0002-9947-1969-0253023-7
Thank you very much
Oh, also this jstor.org/stable/2046912 has a really good introduction to one of the more popular tools.
12:30
"if a bounded set has the property that every non-zero functional achieves its maximum uniquely, then is the set convex?"
I strongly believe that, in finite dimension, for each non-convex set, there exists a functional with 2 maxima.
is it known?
13:25
In finite dimensions (or indeed with any compact convex set) you can use Bishop-Phelps theorem to find a support point of cl(conv C) \ C. Such a support point cannot be an extreme point, because otherwise it'd belong to C, using Milman's Theorem. So, the support point must be part of a line segment.
In infinite dimensions, cl(conv C) may not be compact when $C$ is bounded, so Milman's Theorem may not be used with the strong topology. It also can't be used with the weak topology unless we assume $C$ is weakly closed. I have a feeling that the dentability of the space could make up the difference though; being an extreme point is not enough to guarantee membership in C, but being strongly exposed is!

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