In trying to understand a result of D. Rider (Trans. AMS, 1973) I've got stuck on a lemma that he uses. At one point he makes a step without comment or explanation, but I can't see why it works.

Here is a paraphrase of what I think he is claiming. Let $d_1,\dots, d_n$ be complex numbers of modulus $1$; let $b_1,\dots, b_n$ be complex numbers of modulus $\leq 1$. (In the intended application the numbers $b_1,\dots, b_n$ are actually the diagonal entries of a unitary $n\times n$ matrix, but I strongly suspect that is not used in the argument.) Then Rider seems to assert, without further comment, that $$\left\vert\sum_{j=1}^n (d_jb_j-1)\right\vert^{1/2} \leq \left\vert\sum_{j=1}^n (d_j-1)\right\vert^{1/2} + \left\vert\sum_{j=1}^n (b_j-1)\right\vert^{1/2}$$

Probably I am just being dense, and have failed to spot a routine estimate that does the job. If anyone could get me started on the right track that would be most helpful, as this is starting to get very irritating, and is not even the main part of Rider's argument.

Update Firstly, thanks to everyone who answered, but in particular to Will Jagy for various off-MSE exchanges, and to George Lowther for his elegant argument (the key part that I had failed to think of, was the use of $\rm Re$ and its linearity rather than $\vert\cdot\vert$ and its sub-additivity, allowing one to boost up the observation made by Gerry Myerson).

In case it's of interest, here are some more details on how the question I asked corresponds to Lemma 5.1 in Rider's paper. The paper is dealing with central trigonometric polynomials on compact groups, which means: linear combinations of traces of irreducible representations. Paraphrased lightly, and with a fairly trivial reduction step thrown in, the aforementioned lemma says the following:

Let $\phi: G \to U(n)$ be an (irreducible) unitary representation of a (compact) group $G$. If $$\vert n^{-1}\operatorname{\rm Tr}\phi(g_i)-z_i\vert \leq \delta_i \qquad(i=1,\dots, p)$$ where $\vert z_i\vert=1$ for all $i$, then $$\left\vert n^{-1}\operatorname{Tr}\phi(g_1\dotsb g_n) - \prod_{i=1}^p z_i \right\vert \leq \left(\sum_{i=1}^p \delta_i^{1/2}\right)^2$$

The proof given in the paper goes as follows: reduce to the case $p=2$ (if this can be done then induction will do the rest); then observe that WLOG $\phi(g_1)$ is a diagonal unitary matrix, w.r.t. an appropriate basis. If we let $d_1,\dots, d_n$ be the diagonal entries of $\phi(g_1)$ and $b_1,\dots, b_n$ be those of $\phi(g_2)$, then the trace of $\phi(g_1g_2)$ is just $\sum_{i=1}^n d_ib_i$, and so we have to prove

Claim. If $|z_1|=|z_2|=1$, $\left\vert\sum_{i=1}^n (d_i - z_1)\right\vert\leq n\delta_1$, and $\left\vert\sum_{i=1}^n (b_i - z_2)\right\vert\leq n\delta_2$, then $$\left\vert \sum_{i=1}^n (d_ib_i - z_1z_2) \right\vert \leq n(\delta_1^{1/2}+\delta_2^{1/2})^2.$$

This is where Rider is content to say "done"; and it is hopefully clear that this is equivalent to my original question.

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Note also: ams.org/notices/200809/tx080901112p.pdf – Will Jagy Apr 22 '12 at 2:53

Yes, the result holds. First, for any complex number $z$ with $\vert z\vert\le1$, we have $$2\Re(1-z)=1-\vert z\vert^2+\vert z-1\vert^2\ge\vert z-1\vert^2.$$ It follows that if $Z$ is a complex random variable with $\vert Z\vert\le1$ then, $$\mathbb{E}\left[\left\vert Z-1\right\vert^2\right]\le2\Re\left(\mathbb{E}\left[1-Z\right]\right)\le2\left\vert\mathbb{E}\left[Z-1\right]\right\vert.$$ Therefore, if $X,Y$ are complex random variables with $\vert X\vert\le1$ and $\vert Y\vert\le1$ then, using Cauchy–Schwarz, \begin{align} \left\vert\mathbb{E}\left[(X-1)(Y-1)\right]\right\vert^2&\le\mathbb{E}\left[\left\vert X-1\right\vert^2\right]\mathbb{E}\left[\left\vert Y-1\right\vert^2\right]\cr &\le4\left\vert\mathbb{E}\left[X-1\right]\mathbb{E}\left[Y-1\right]\right\vert \end{align} So, using the identity $XY-1=(X-1)+(Y-1)+(X-1)(Y-1)$ (as in Gerry Myerson's answer), \begin{align} \left\vert\mathbb{E}\left[XY-1\right]\right\vert&\le\left\vert\mathbb{E}\left[X-1\right]\right\vert+\left\vert\mathbb{E}\left[Y-1\right]\right\vert+\left\vert\mathbb{E}\left[(X-1)(Y-1)\right]\right\vert\cr &\le\left\vert\mathbb{E}\left[X-1\right]\right\vert+\left\vert\mathbb{E}\left[Y-1\right]\right\vert+2\left\vert\mathbb{E}\left[X-1\right]\mathbb{E}\left[Y-1\right]\right\vert^{1/2}\cr &=\left(\left\vert\mathbb{E}[X-1]\right\vert^{1/2}+\left\vert\mathbb{E}[Y-1]\right\vert^{1/2}\right)^2. \end{align} Take the square root of each side to obtain $$\left\vert\mathbb{E}\left[XY-1\right]\right\vert^{1/2}\le\left\vert\mathbb{E}[X-1]\right\vert^{1/2}+\left\vert\mathbb{E}[Y-1]\right\vert^{1/2},$$ which is equivalent to the inequality in the question. Note that it is only necessary to have $\vert d_i\vert\le1$ and $\vert b_i\vert\le1$ (we do not need $\vert d_i\vert=1$).

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Let's see, is $\; \; n \; \mathbb E (\mbox{thing}) \; = \; \sum \; \mbox{thing}\; ?$ – Will Jagy Apr 23 '12 at 20:35
Yes, that's what I mean when I say that it is equivalent to the inequality in the question. Dividing both sides of the original inequality by $\sqrt{n}$ converts it to an inequality for means, which seems to be a natural way to look at it it -- so I did that. – George Lowther Apr 23 '12 at 20:40
...I could put summations back in, in place of the expectations if that makes people happier. – George Lowther Apr 23 '12 at 20:44
Also, have you looked at Lemma 5.1, pages 465-466 in the Rider article? This is just one case that Yemon massaged into this shape, by taking all $z_i = 1$ and a single normalized character. – Will Jagy Apr 23 '12 at 20:47
George, no, if this is how you thought of it, that is information right there. I think I will write out some alternate formulations for myself, as, put simply, I never got near proving anything. Also, Hardy, Littlewood, Polya "Inequalities" write in terms of means when possible. – Will Jagy Apr 23 '12 at 20:52

Well, I can do the case $n=1$.

$$|db-1|=|(d-1)+(b-1)+(d-1)(b-1)|\le|d-1|+|b-1|+|(d-1)(b-1)|$$ and $$|(d-1)(b-1)|\le2|(d-1)(b-1)|^{1/2}$$ taken together and taking the square root on both sides gives $$|db-1|^{1/2}\le|d-1|^{1/2}+|b-1|^{1/2}$$

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Bill Johnson asked about truth for other exponents, and I finally realize that the exponent $1/2$ really is correct and not arbitrary. $$\left| \sum_{j=1}^n (d_jb_j-1)\right|^\alpha \leq \left| \sum_{j=1}^n (d_j-1)\right|^\alpha + \left| \sum_{j=1}^n (b_j-1)\right|^\alpha$$ is false for some data with $n=2$ as soon as $\alpha > 1/2.$ For some $0 \leq \theta \leq \pi/2,$ take $$d_1 = b_1 = e^{i \theta}, \; \; d_2 = b_2 = \bar{d_1} = \bar{b_1} = e^{-i \theta}.$$ Then the inequality with exponent $\alpha$ is false for $$0 < \theta < \arccos \left( \frac{2^{1/\alpha} - 2}{2} \right).$$ For $\alpha = 1,$ which is my previous "answer," I used $\theta = \pi/3$ as optimal in one sense, but I could have used any $0 \leq \theta \leq \pi/2.$

$\theta = \pi/3$ gives falsehood as long as $\alpha > \frac{\log 2}{\log 3} \approx 0.6309...$

As $\alpha$ decreases to $1/2,$ the window for $\theta$ becomes narrower, until finally we get to $\alpha = 1/2$ and $0 < \theta < \arccos 1 = 0,$ no such thing.

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What if $\alpha\lt1/2$? – Gerry Myerson Apr 22 '12 at 4:14
@Gerry Continuous $f(0) = 0.$ For $x > 0,$ $f$ is $C^2,$ with $f > 0, f' > 0, f'' < 0.$ Then, for $x,y > 0,$ we have $f(x+y) < f(x) + f(y).$ PROOF:For $x > 0, \;0 < \lambda < 1, \; f((1 - \lambda) \cdot 0 + \lambda \cdot x) > (1 - \lambda) f(0) + \lambda f(x),$ so $\lambda f(x) < f(\lambda x).$ For $x,y > 0,$ take $\lambda = \frac{x}{x+y},$ with $\frac{x}{x+y} f(x+y) < f(x).$ Next $\frac{y}{x+y} f(x+y) < f(y).$ Together $f(x+y) < f(x) + f(y).$ Taking $f(x) = x^\beta$ with $0 < \beta < 1,$ and $A,B,C \geq 0$ with $A \leq B + C,$ we get $A^\beta \leq (B + C)^\beta \leq B^\beta + C^\beta.$ – Will Jagy Apr 22 '12 at 20:48

I am beginning a sequence of experiments. My first, incorrect, answer, talked about what happened if the $1/2$ exponents are erased. I have done a bunch of programming, and have a $$\mbox{KOMPUTER KONJECTURE:}$$ $$\left| \sum_{j=1}^n (d_jb_j-1)\right| \leq \left| \sum_{j=1}^n (d_j-1)\right| + \left| \sum_{j=1}^n (b_j-1)\right| + \frac{n}{2}$$ For even $n,$ equality can occur. Take $n=2k,$ then $$d_1 = b_1 = d_2 = b_2 = \cdots = d_k = b_k = e^{i \pi / 3}$$ and $$d_{k+1} = b_{k+1} = d_{k+2} = b_{k+2} = \cdots = d_{2k} = b_{2k} = e^{-i \pi / 3}$$

For odd $n=2k-1,$ equality does not occur. The most extreme case, with some angle $0 \leq \theta < \pi / 3,$ occurs at $$d_1 = b_1 = d_2 = b_2 = \cdots = d_k = b_k = e^{i \theta}$$ and $$d_{k+1} = b_{k+1} = d_{k+2} = b_{k+2} = \cdots = d_{2k-1} = b_{2k-1} = e^{i \theta} e^{4 i \pi / 3}.$$ For $n=1$ it does not matter, I proved the thing is true, no $1/2$ needed, and we may take $\theta = 0.$ For $n=3, \theta \approx 46.53^\circ \approx 0.2585 \pi,$ instead of $1.5$ the excess is about $1.34886.$ For $n=5, \theta \approx 52.01^\circ \approx 0.28895 \pi,$ instead of $2.5$ the excess is about $2.4097595.$ For $n=7, \theta \approx 54.31^\circ \approx 0.30173 \pi,$ instead of $3.5$ the excess is about $3.435627.$ For $n=9, \theta \approx 55.58^\circ \approx 0.30878 \pi,$ instead of $4.5$ the excess is about $4.449959.$

What was striking was all the $d_j = b_j,$ then the consistent presence of the $4 \pi / 3$ in both the odd and even cases.

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I think the 'trick' here is to introduce a term in the left hand side that will help to get the right hand side. Specifically if you use the fact that the left hand side is less than what you get when you bring in the absolute value signs inside the summation, then add and subtract $d_j$ to the term $|d_jb_j-1|$... you're looking at the sum of $|d_jb_j-d_j+d_j-1|\le|d_j||b_j-1|+|d_j-1|$ and then I think you're getting nearer to the final thing you want (I didn't distribute the sqrt nicely, but I'm new)!

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Tried this kind of approach, but I think it's too blunt. – user16299 Apr 21 '12 at 8:28

Using the concept of means and of residuals (=deviations from the mean) I come up to a part of a proof and which converts the problem into one of scalar variables (n=1) like in Gerry Myerson's answer; the last step is still missing.

We use $\small \bar d, \bar b$ for the means, $\small \hat d_j, \hat b_j$ for the deviations and $\small \bar c_{jj}$ for the mean of the crossproduct of the deviations. Also we rewrite the original equation $$\small \left\vert \sum_{j=1}^n (d_jb_j-1)\right\vert^{1/2} \leq \left\vert\sum_{j=1}^n (d_j-1)\right\vert^{1/2} + \left\vert\sum_{j=1}^n (b_j-1)\right\vert^{1/2}$$ in squared terms $$\small \left\vert \sum_{j=1}^n (d_jb_j-1)\right\vert \leq \left\vert\sum_{j=1}^n (d_j-1)\right\vert + \left\vert\sum_{j=1}^n (b_j-1)\right\vert +2 \sqrt{\left\vert\sum_{j=1}^n (d_j-1)\right\vert \left\vert\sum_{j=1}^n (b_j-1)\right\vert}$$ After the means and residuals are worked out and cancelled we get it in this form $$\small \left\vert (\bar c_{jj} +(\bar b -1)(\bar d-1) +(\bar b -1)+(\bar d-1) \right\vert \leq \left\vert \bar d -1\right\vert + \left\vert \bar b-1\right\vert + 2\sqrt{|\bar d-1||\bar b-1|}$$ From the construction of the means (and the initial conditions on the $\small d_j$ and $\small b_j$) we have that the moduli of the means as well as that of the residuals are always $\small \le 1$.

It is crucial to note,

1. that if the modulus of the mean has its maximum 1 (because all j'th values must be the same), then we cannot have residuals ($\small \bar b=1 \text{ or } \bar d=1 \to \bar c_{jj}=0$ )

2. and if the residuals take their maximum (their moduli are all 1) then both means must be zero. ($\small \bar |c_{jj}|=1 \to \bar b=0 \text{ and } \bar d=0$ )

We can now begin with the case distinctions based on that extremes and see, that the equation holds in that cases.

May be this suffices also for values inside the range for the extremes, I don't see it at the moment. But if this does not yet suffice, we can at least see, that when we choose a value for the means $\small \bar b,\bar d$, there is always an upper bound for the possible value of $\small \bar c_{jj}$ and increasing $\small |\bar b| , |\bar d|$ decreases $\small \vert\bar c_{jj}\vert$. However to settle the question completely this functional relation must be made explicite.

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Thank you for the attention, but I am afraid that I can't see how this "partial solution" advances towards a full solution. Of course one can show that the inequality holds in special cases; if I did not already know this I would not have asked. – user16299 Apr 22 '12 at 21:02
@Yemon :Sure; I'm still working on it. What we see is that we have a similar equation to that of Gerry Myerson (where n=1 and Gerry has provided a solution). Unfortunately, for n>1 his formula is no more sufficient. Here we have n=1 but for exactness one "correction-factor" (the covariance $\small \bar c_{jj}$ which decreases the lhs, if the correlation is not perfect. Thus I think the focus for the finding of the final proof should be on that functional relation between the mean $\small \bar b \bar d$ and that covariance-term. I'm still trying to make this idea an argument... – Gottfried Helms Apr 22 '12 at 22:12