1
$\begingroup$

I am trying to understand a proof of a simple lemma from this paper (lemma 3) which I paste here (refer to the 1st inequality, $H$ is a Hilbert space):

The first step in the proof is to use assumption no. 2 and Cauchy-Schwarz so that $\langle x_i,Te_k \rangle \leq c\|x_i\|$, then there is no longer dependence on $T$ so the $sup$ can be removed. But now, the proof seems to use the following step: $$\sum_i\gamma_{ik}\|x_i\|\leq\|\sum_i\gamma_{ik}x_i\|$$ which I fail to understand - it looks like the triangle inequality but the sign is reversed. (The next steps of the proofs are understandable).

What am I missing?

EDIT the last inequality I wrote is obviously wrong, so the proof is not using it, but something else which I don't understand

$\endgroup$
6
  • $\begingroup$ What does orthogaussian mean? $\endgroup$
    – Roberto Rastapopoulos
    Mar 12, 2018 at 13:52
  • $\begingroup$ That's a great question, I couldn't find anywhere a definition, but I'm 99% sure it's a series of i.i.d standard normal variables. $\endgroup$
    – Itamar Katz
    Mar 12, 2018 at 13:55
  • $\begingroup$ I feel like the inclusion of terms from probability such as expected value and Gaussian are a mistake. The proof by convexity and induction is much easier to understand. Also - I have never heard the term "orthogaussian". $\endgroup$
    – Oria Gruber
    Mar 13, 2018 at 10:18
  • $\begingroup$ You manually typed Schwarz's name in two ways, both are wrong, when you have the correct name in both the image you included and the tag that you're using. $\endgroup$
    – Asaf Karagila ♦
    Mar 13, 2018 at 10:28
  • $\begingroup$ @AsafKaragila thanks, fixed. $\endgroup$
    – Itamar Katz
    Mar 13, 2018 at 10:38

1 Answer 1

Reset to default
3
$\begingroup$

First write $\sum_{k=1}^{K} \sum_{i=1}^{n} \gamma_{ik} \langle x_i,Te_k \rangle$ as $\sum_{k=1}^{K} \langle \sum_{i=1}^{n} \gamma_{ik} x_i,Te_k \rangle$ before applying Cauchy - Schwartz.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.