# Proof of Khintchine's inequality on $\mathbb{C}$?

I'm trying to understand this proof of Khintchine's inequality for the complex case ($$a_n\in\mathbb{C}$$). The author claims that the complex case follows from the real case by taking absolute values where appropriate, but I don't see where to place those absolut values, even after filling in the details of the proof:

Proof (for $$a_n\in\mathbb{R}$$): Let $$\epsilon_1,\dots,\epsilon_n$$ be some i.i.d. random variables with $$\mathbb P(\epsilon_k=\pm1)=\frac12$$ and let $$a_1,\dots,a_n\in\mathbb{R}$$ (w.l.o.g. $$\sum_{k=1}^na_k^2=1$$). For every $$\lambda>0$$ we have \begin{align*} \mathbb E(\exp(\lambda\epsilon_ka_k)) &=\sum_{j=\pm1}\exp(j\lambda a_k)\cdot\overbrace{\mathbb P(\epsilon_k=j)}^{=1/2} =\cosh(\lambda a_k) =\sum_{\ell=0}^\infty\frac{(\lambda a_k)^{2\ell}}{(2\ell)!} \\ &\leq\sum_{\ell=0}^\infty\frac{(\lambda^2a_k^2)^\ell}{2^\ell\cdot\ell!} =\sum_{\ell=0}^\infty\frac{(\lambda^2a_k^2\,/\,2)^\ell}{\ell!} =\exp(\lambda^2a_k^2\,/\,2), \end{align*} which gives us $$\mathbb E\bigg(\exp\bigg(\lambda\sum_{k=1}^n\epsilon_ka_k\bigg)\bigg) =\prod_{k=1}^n\mathbb E(\exp(\lambda\epsilon_ka_k)) \leq\prod_{k=1}^n\exp(\lambda^2a_k^2\,/\,2) =\exp\bigg(\frac{\lambda^2}2\sum_{k=1}^na_k^2\bigg) =\exp(\lambda^2\,/\,2),$$ and thus \begin{align*} \mathbb P\bigg(\bigg|\sum_{k=1}^n\epsilon_ka_k\bigg|\geq\lambda\bigg) &=2\mathbb P\bigg(\sum_{k=1}^n\epsilon_ka_k\geq\lambda\bigg) =2\mathbb P\bigg(\exp\bigg(\lambda\sum_{k=1}^n\epsilon_ka_k\bigg)\geq\exp(\lambda^2)\bigg) \\ &\overset{(*)}\leq\frac2{\exp(\lambda^2)}\mathbb E\bigg(\exp\bigg(\lambda\sum_{k=1}^n\epsilon_ka_k\bigg)\bigg) \leq\frac2{\exp(\lambda^2)}\exp(\lambda^2\,/\,2) =\exp(-\lambda^2\,/\,2), \end{align*} where $$(*)$$ uses the fact that for every $$\alpha>0$$ and every real or complex (integrable) random variable $$X$$ we have $$\mathbb P(|X|\geq\alpha) =\mathbb E\Big(1\cdot1_{\big\{\frac{|X|}\alpha\geq1\big\}}\Big) \leq\mathbb E\Big(\frac{|X|}\alpha\cdot1_{\big\{\frac{|X|}\alpha\geq1\big\}}\Big) \leq\mathbb E\Big(\frac{|X|}\alpha\Big) =\frac1\alpha\mathbb E|X|.$$ [... The rest of the proof is irrelevant for my question ...] $$\blacksquare$$

Now, my question is where do I have to put absolut values in the inequalities above to prove the complex case? In order for the first block to make sense in $$\mathbb{C}$$ we need to replace $$(\lambda a_k)^{2\ell}$$ with $$|\lambda a_k|^{2\ell}$$. We have $$|\mathbb E(\exp(\lambda\epsilon_ka_k))| =|\cosh(\lambda a_k)| \leq\sum_{\ell=0}^\infty\frac{|\lambda a_k|^{2\ell}}{(2\ell)!} \leq\cdots\leq\exp(\lambda^2|a_k|^2\,/\,2),$$ which implies $$\bigg|\mathbb E\bigg(\exp\bigg(\lambda\sum_{k=1}^n\epsilon_ka_k\bigg)\bigg)\bigg| =\prod_{k=1}^n|\mathbb E(\exp(\lambda\epsilon_ka_k))| \leq\cdots\leq\exp\bigg(\frac{\lambda^2}2\sum_{k=1}^n|a_k|^2\bigg) =\exp(\lambda^2\,/\,2).$$ But this last result leads me nowhere. I don't see how to translate the third block into complex numbers such that I can make use of the last result.

• I don’t get the linked proof. It looks like it has its inequality the wrong way around in Jensen’s inequality, which would be $\mathbb{E}\exp(X) \geq \exp(\mathbb{E}X)$, not $\leq$. – WimC Jun 10 at 11:17
• Well, maybe you understand my more-fleshed-out part of the proof, which doesn't require the knowledge of the article (I just linked to it as a source for the claim that the complex case should be as easy as the real case) – Cubi73 Jun 10 at 11:29
• I see it now, thanks. – WimC Jun 10 at 13:29