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Sorry if I am being stupid. But I came to this lemma in the Appendix, written by Bourgain, Kontorovich and Magee, of the paper by Magee, Oh and Winter [MOW16]. It starts out like this:

Lemma 42 Let $\pi$ be a unitary $G$-representation on a Hilbert space $H$, and assume that the operator $A$ acts on $H$ via $$A\varphi=\sum_{j\in J}\pi(h_j)\varphi,$$ for some $h_j\in G$ and indexing set $J$. Assume that $A$ has the "spectral gap" property: there is some $C_0>0$ so that $$\langle A\varphi,\varphi\rangle\leq(1-C_0)|J|\|\varphi\|^2.$$ For sme positive coefficients $\kappa_j>0$, let $\widetilde {A}$ act on $H$ as $$\widetilde{A}\varphi=\sum_{j\in J}\kappa_j\pi(h_j)\varphi,$$ and assume that the $L^\infty$ norm of the coefficients is controlled by the $L^1$ norm, in the sense that for some $K\geq1$, $$\text{max}\kappa_j\leq K\bar{\kappa},$$ where $$\bar{\kappa}:=\frac{1}{|J|}\sum_{j}\kappa_j$$ is the coefficient average. Then $\widetilde{A}$ has the following "spectral gap": $$\langle\widetilde{A}\varphi,\varphi\rangle\leq\bar{\kappa}(1-C_0+\sqrt{K-1})|J|\|\varphi\|^2.$$

Here, the group $G$ is $\mathrm{SL}_2(\mathbb{Z}/q\mathbb{Z})$ with an arbitrary integer $q$.

In their paper, they said it was an exercise in Cauchy-Schwarz. However, I was wondering how they did it to make the final result consist of two terms, $(1-C_0)$ and $\sqrt{K-1}$, in the bracket.

I attempted a few times applying the Cauchy-Schwarz directly to the inner product, only to find terms with $K$ instead of $\sqrt{K-1}$.

This is a duplicate of the question asked in Mathstackexchange but without any further progress.

Any comments are appreciated.

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    $\begingroup$ Look at $\widetilde{A} - \kappa A$ and then use Cauchy-Schwarz. $\endgroup$
    – Lucia
    Nov 27, 2016 at 1:46
  • $\begingroup$ Looks like this is more combinatorial than it appears. Suppose there are m k's larger than average and n k's smaller than average. Then |J|>=m+n and K-1 >= n/m, so the coefficient in the inequality comes out as √(n/m).(m+n). By AM-GM (a special case of C-S!) This is at least √(m/n).2√(mn)=2n, which is what you get from the trivial bound. $\endgroup$
    – Fan Zheng
    Nov 27, 2016 at 5:14
  • $\begingroup$ @FanZheng: I see no reason for $K-1\ge n/m$. If half the $\kappa$s are very slightly more than average and half are very slightly less, then $K$ can be pretty close to $1$. $\endgroup$
    – Lucia
    Nov 27, 2016 at 18:16
  • $\begingroup$ Sorry but I missed a factor $\bar\kappa-\min\kappa_i$, but fortunately it appears on both sides of the inequality, so it still works. But I guess you may have a different argument in mind because it's really a stretch to call the above argument C-S. $\endgroup$
    – Fan Zheng
    Nov 27, 2016 at 19:34

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We can decompose $\langle\widetilde{A}\varphi,\varphi\rangle$ as follows.

\begin{split} \langle\widetilde{A}\varphi,\varphi\rangle&=\langle(\widetilde{A}-\bar{\kappa}A)\varphi,\varphi\rangle+\langle\bar{\kappa}A\varphi,\varphi \rangle\\ &=\langle\sum_{j\in\mathcal{J}} (k_j-\bar{\kappa})\pi(h_j)\varphi,\varphi\rangle +\bar{\kappa}\langle A\varphi,\varphi \rangle\\ &\overset{\text{(By C-S)}}\le\bigg\langle \left(\sum_{j\in\mathcal{J}}(k_j-\bar{\kappa})^2\right)^{1/2}\left(\sum_{j\in\mathcal{J}}(\pi(h_j)\varphi)^2\right)^{1/2},\varphi\bigg\rangle+\bar{\kappa}(1-C_0)|\mathcal{J}|\|\varphi\|^2. \end{split} The rest will be pure computation.

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