3
$\begingroup$

Here $P:H \to \mathcal{K}$ is a projection operator from a Hilbert space onto a closed convex subset.

enter image description here

I don't follow the hypothesis of the proof by contradiction argument for the uniform convergence (all else is fine). Would someone tell me how exactly the contradiction hypothesis is formed?

I don't really understand why the last quantity is greater than or equal to $\epsilon$ for all $n$. Isn't the point the uniformity of the convergence of $h$ -- for a contradiction argument, isn't $o(h_n)/\lVert h_n\rVert$ supposed to still go to zero but at a rate that depends on the sequence $h_n$?

  • 0
    You need to explain the notation. What is $P(x)$ and what is equation (2.5) – 2017-03-15
  • 0
    @NathanaelSkrepek I think both are not relevant since my question is on the set up of the contradiction argument. Just assume (4.15) is true and the only thing I don't follow is how he sets up the contradiction argument for the uniformity of the convergence. – 2017-03-15
  • 0
    I still think it is necessary to know what (2.5) is because if I assume (4.15) I already have that it converges uniformly as far as I see. – 2017-03-15
  • 0
    What is $S_K(x )$ in 4.15? – 2017-03-15
  • 0
    This is part of the paper: Zarantonello, Eduardo H. Projections on convex sets in Hilbert space and spectral theory. I. Projections on convex sets. Contributions to nonlinear functional analysis (Proc. Sympos., Math. Res. Center, Univ. Wisconsin, Madison, Wis., 1971), pp. 237–341. Academic Press, New York, 1971. – 2017-03-16

1 Answers 1

1

Let's try to formalize Lemma 4.5 a bit, so the hypothesis for the contradiction argument can be deduced more easily.

The lemma states that

  • for all locally compact cones of increments $S_K(x)$ and
  • for all null sequences h_n

we have that $o(h)/\|h\|$ goes to zero as well. This is what is meant by uniformity of the limit, i.e. no matter which cone and which null sequence you choose, you will always end up with the desired convergence behaviour of $P_K$.

Now, suppose for the sake of contradiction that the convergence is not uniform in the above sense. Then,

  • there exists a locally compact cone of increments $S_K(x)$ and
  • there exists a null sequence h_n

such that $o(h)/\|h\|$ doesn't go to zero, i.e. there exists an $\varepsilon > 0$ such that $\|P_K(x + h_n) - x - h_n\| / \|h_n\| \geq \varepsilon$ for all $n \in \mathbb{N}$.

  • 0
    where is the contradiction? – 2017-03-15
  • 0
    @NathanaelSkrepek: as usual, the contradiction is at the end of the proof... Note that OP didn't post the full proof. See https://books.google.de/books?id=3LziBQAAQBAJ&pg=PA299&lpg=PA299&dq=over+any+locally+compact+cone+of+increments&source=bl&ots=-6kt-gee80&sig=8AyRvaEE5PBiujoPldXIBrlRD5A&hl=de&sa=X&ved=0ahUKEwjAjdj8_djSAhVGhSwKHRY_DRAQ6AEIHDAA#v=snippet&q=cone&f=false – 2017-03-16
  • 0
    @NathanaelSkrepek: Note that also OP is asking how the hypothesis is "formed" and this should be what is shown in my post. – 2017-03-16