# Inequality proof using Cauchy-Schwarz and Jensen's inequality

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

• What does orthogaussian mean? – Roberto Rastapopoulos Mar 12 '18 at 13:52
• 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. – Itamar Katz Mar 12 '18 at 13:55
• 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". – Oria Gruber Mar 13 '18 at 10:18
• 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. – Asaf Karagila Mar 13 '18 at 10:28
• @AsafKaragila thanks, fixed. – Itamar Katz Mar 13 '18 at 10:38

## 1 Answer

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.