Starting with Cauchy

Summary of Chapter 1 of Steele’s The Cauchy-Schwarz Master Class
Published

April 1, 2023

Conventions

  • \(a_i\)’s and \(b_i\)’s will denote reals.
  • \(V\) will denote an inner product space over \(\mathbb R\), and \(\|\cdot\|\) will denote the induced norm.

Notes

We start with an inductive proof of the following version of Cauchy’s inequality:

Proposition 1 One has \[\begin{equation*} a_1 b_1 + \cdots + a_n b_n\le \sqrt{a_1^2 + \cdots + a_n^2}\sqrt{b_1^2 + \cdots + b_n^2}. \end{equation*}\]

Proof. For \(n = 1\), we have equality. Let’s check the \(n = 2\) case, which will also be used in the inductive step. We have the following chain of equivalent statements: \[\begin{align*} (a_1 b_1 + a_2 b_2)^2 & \le (a_1^2 + a_2^2) (b_1^2 + b_2^2)\\ 2a_1 b_1 a_2 b_2 & \le a_1^2 b_2^2 + a_2^2 b_1^2\\ 0 & \le (a_1 b_2 - a_2 b_1)^2 \end{align*}\]

We now move on to the \((n + 1)\)-th case: \[\begin{align*} a_1 b_1 + \cdots + a_{n + 1} b_{n + 1} & = a_1 b_1 + \cdots + a_n b_n + a_{n + 1} b_{n + 1}\\ & \le \sqrt{a_1^2 + \cdots + a_n^2}\sqrt{b_1^2 + \cdots + b_n^2} + a_{n + 1} b_{n + 1}\\ & \le \sqrt{a_1^2 + \cdots + a_{n + 1}^2}\sqrt{b_1^2 + \cdots + b_{n + 1}^2} \end{align*}\] The second inequality uses the induction hypothesis, and the third uses the \(n = 2\) case.

We thus get:

Corollary 1 For \(a_1, a_2, \ldots, b_1, b_2, \ldots\), if \(\sum_i a_i^2\) and \(\sum_i b_i^2\) are finite, then so is \(\sum_i |a_i b_i|\).

However, we give an alternate proof not relying on Proposition 1:

Proof. We are motivated to investigate the smallness of \(|ab|\) when \(a^2\) and \(b^2\) are small. We’ll be done if we get hold of a constant \(C\) such that \[\begin{equation*} |ab|\le C(a^2 + b^2)\tag{1}\label{EQN: ab and a^2 b^2} \end{equation*}\] for then, that will imply a stronger result: \[\begin{equation*} \sum\nolimits_i |a_i b_i|\le C\left(\sum\nolimits_i a_i^2 + \sum\nolimits_i b_i^2\right) \end{equation*}\] Indeed, \(\ref{EQN: ab and a^2 b^2}\) holds for \(C = 1/2\) since \((|a| - |b|)^2\ge 0\).

The following stronger result also oozes out of the above proof:

Proposition 2 If \(\sum_i a_i^2\) and \(\sum_i b_i^2\) are finite, then \(\sum_i a_i b_i\) is absolutely convergent with \[\begin{equation*} \sum_i a_i b_i\le \frac{1}{2}\sum_i a_i^2 + \frac{1}{2}\sum_i b_i^2. \end{equation*}\]

Note that Proposition 2 can also be derived from Proposition 1 since \(\sqrt{x}\sqrt y\le (x^2 + y^2)/2\) for \(x, y\ge 0\). Conversely, we may derive Proposition 1 (even for the infinite case) from Proposition 2 as well by normalizing. Just make the following replacements: \(a_j\to a_j\big/\sqrt{\sum_i a_i^2}\) and \(b_j\to b_j\big/\sqrt{\sum_i b_i^2}\).

“Normalization gives us a systematic way to pass from an additive to a multiplicative inequality.”

The above discussion motivates a new way to prove the Cauchy-Schwarz inequality:

Theorem 1 (Cauchy-Schwarz) For \(u, v\in V\), one has \[\begin{equation*} \langle u, v\rangle\le \|u\|\|v\| \end{equation*}\] with equality holding \(\iff\) \(u\), \(v\) are linearly dependent.

Proof. It’s clear if any one of \(u\) or \(v\) is \(0\). Thus, we may without loss of generality assume that neither is \(0\). Note that the starting point for the proof of Corollary 1 (which was also the proof of Proposition 2) was that \((|a| - |b|)^2\ge 0\). Along the same lines, we observe that \[\begin{alignat*}{2} && 0 & \le \langle u - v, u - v\rangle &\\ &\stackrel{\text{\scriptsize w}}{\implies}& 0 & \le \|u\|^2 - 2\langle u, v\rangle + \|v\|^2 \\ &\stackrel{\text{\scriptsize w}}{\implies}& \langle u, v\rangle & \le (\|u\|^2 + \|v\|^2) / 2 \end{alignat*}\]

which is the generalization of Proposition 2. Replacing \(u\to u/\|u\|\) and \(v\to v/\|v\|\) (“normalizing”), we get \[\begin{equation*} \langle u, v\rangle\le\|u\|\|v\|. \end{equation*}\] Now, we show the fact about equality. “\(\Leftarrow\)” is evident. For the converse, let \(\langle u, v\rangle = \|u\|\|v\|\) which implies \(\langle\hat u, \hat v\rangle = 1\) where \(\hat u:= u/\|u\|\) and \(\hat v := v/\|v\|\) are the “normalized” variables. Then it follows that \(\|\hat u - \hat v\|^2 = 0\) so that \(v = \frac{\|v\|}{\|u\|} u\).

The popular quick proof involving the quadratic polynomial is due to Schwarz (see Exercise 11).

Exercises

Exercise 1 (The \(1\)-trick and the splitting trick) For any reals \(a_1, \ldots, a_n\), one has:

  1. \(\sum_i a_i\le \sqrt n \, \sqrt{\sum_i a_i^2}\)
  2. \(\sum_i a_i\le \sqrt{\sum_i |a_i|^{2/3}}\, \sqrt{\sum_i |a_i|^{4/3}}\)

Just note that \(a_i = 1\, a_i\) and that \(a_i\le |a_i| = |a_i|^{1/3}\, |a_i|^{2/3}\). Now apply Cauchy-Schwarz.

Exercise 2 (Products of averages and averages of products) Let \(a_i, b_i, p_i\ge 0\) for \(i = 1, \ldots, n\) with \(\sum_i p_i = 1\) and each \(a_i b_i\ge 1\). Then we also have \[\begin{equation*} \left(\sum_{i = 1}^n p_i a_i\right) \left(\sum_{i = 1}^n p_i b_i\right)\ge 1 \end{equation*}\]

\(\text{LHS}\ge \bigl(\sum_i p_i\sqrt{a_i b_i}\bigr)^2\ge \bigl(\sum_i p_i\bigr)^2 = 1\).

Exercise 3 (Why not three or more?) For reals \(a_i, b_i, c_i\)’s for \(i = 1, \ldots, n\), one has: \[\begin{equation*} \sum_i a_i b_i c_i\le \sqrt{\sum_i a_i^2}\, \sqrt{\sum_i b_i^2} \, \sqrt{\sum_i c_i^2}. \end{equation*}\]

Note

Putting each \(c_i = 1\) in the second yields \(\sum_i a_i b_i\le \sqrt n\, \sqrt{\sum_i a_i^2}\, \sqrt{\sum_i b_i^2}\), which is not as strong as Cauchy.

We have \[\begin{align*} \biggl(\sum_i a_i b_i c_i\biggr)^2 & \le \biggl( \sum_i a_i^2 \biggr)^2 \biggl( \sum_i b_i^2 c_i^2 \biggr)^2\\ & \le \biggl( \sum_i a_i^2 \biggr)^2 \biggl( \sum_i b_i^4 \biggr) \biggl( \sum_i c_i^4 \biggr)\\ & \le \biggl( \sum_i a_i^2 \biggr)^2 \biggl( \sum_i b_i^2 \biggr)^2 \biggl( \sum_i c_i^2 \biggr)^2 \end{align*}\] where the last inequality follows because \(\sum_i \alpha_i^2\le (\sum_i \alpha_i)^2\) for any \(\alpha_1, \ldots, \alpha_n\ge 0\).

Exercise 4 (Some help from symmetry) For \(x, y, z > 0\), one has: \[\begin{gather*} \left( \frac{x + y}{x + y + z} \right)^{1/2} + \left( \frac{y + z}{x + y + z} \right)^{1/2} + \left( \frac{z + x}{x + y + z} \right)^{1/2}\le \sqrt 6\\ x + y + z\le 2\left( \frac{x^2}{y + z} + \frac{y^2}{z + x} + \frac{z^2}{x + y} \right) \end{gather*}\]

For the first, just note that the sum of the summands squared on the LHS is equal to \(2\), and use the \(1\)-trick. For the second, use Cauchy on \(x + y + z\) by writing \(x = (x/\sqrt{y + z}) \sqrt{y + z}\) and so on.

Exercise 5 (A crystallographic inequality with a message) Let \(f\colon \mathbb{R\to R}\) be such that \(f(x)^2\le af(bx) + c\) where \(a, b, c\in\mathbb R\) (for instance, \(\cos(x)^2 = \cos(2x)/2 - 1/2\)). Let \(p_1 + \ldots + p_n = 1\) for \(p_i\ge 0\) and define \(g(x) := \sum_i p_i f(\alpha_i x_i)\) for \(\alpha_i\in\mathbb R\). Then \[\begin{equation*} g(x)^2\le a g(bx) + c. \end{equation*}\]

Deploy Cauchy on \(g(x) = \sum_i \sqrt{p_i}(\sqrt{p_i} f(\alpha_i x))\).

Exercise 6 (A sum of inversion preserving summands) For \(p_1, \ldots, p_n > 0\) with \(p_1 + \cdots + p_n = 1\), one has \[\begin{equation*} \sum_{i = 1}^n \left( p_i + \frac{1}{p_i} \right)^2\ge n^3 + 2n + \frac{1}{n} \end{equation*}\] with the equality holding \(\iff\) each \(p_i = 1/n\).

This is equivalent to showing that \(n\sum_i (p_i + 1/p_i)^2\ge (n^2 + 1)^2\). Indeed, \[\begin{align*} n\sum_i (p_i + 1/p_i)^2 & \ge \biggl(\sum_i (p_i +1/p_i)\biggr)^2\\ & = \biggl(1 + \sum_i 1/p_i\biggr)^2\\ & \ge (1 + n^2)^2 \end{align*}\] where the last inequality follows from Lemma 1 below. For equality, “\(\Leftarrow\)” is clear. For the converse, note that all the inequalities must be equalities so that \(p_i + 1/p_i\) is independent of \(i\) which implies that \(p_i\)’s are all equal.

Lemma 1 Let \(0 < p_1, \ldots, p_n \le 1\) with \(\sum_i p_i = 1\). Then for any \(k\ge 0\), one has \[\begin{equation*} \sum_{i = 1}^n \frac{1}{p_i^k}\ge n^{k + 1} \end{equation*}\] with the equality holding \(\iff\) each \(p_i = 1/n\).

Proof. We induct strongly. Assume for all \(k'\) less than \(k\). We have two cases:

  1. \(k\) is even: Then \(n\sum_i 1/p_i^{k}\) \(\ge\) \(\left(\sum_i 1/p_i^{\frac{k}{2}}\right)^2\) \(\ge\) \(\bigl(n^{\frac{k}{2} + 1}\bigr)^2 = n^{k + 2}\) (without loss of generality, assume that \(k/2 < k\)) so that \(\sum_i 1/p_i^k \ge n^{k + 1}\).
  2. \(k\) is odd: Then \(\sum_i 1/p_i^k\) \(=\) \(\left(\sum_i p_i\right) \left( \sum_i 1/p_i^k \right)\) \(\ge\) \(\left( \sum_i p_i^{\frac{1}{2}}/p_i^{\frac{k}{2}} \right)^2\) \(=\) \(\left( \sum_i 1/p_i^{\frac{k-1}{2}} \right)^2\) \(\ge\) \(\bigl(n^{\frac{k+1}{2}}\bigr)^2\) \(=\) \(n^{k + 1}\).

Finally, note that if the equality holds, then all the inequalities above become equalities, so we may again deploy induction.

Exercise 7 (Flexibility of form) Let \(A\in\mathbb R^{n\times n}\). Then \[\begin{equation*} \langle x, y\rangle := x^t A\, y \end{equation*}\] defines an inner product on \(\mathbb R^n\) \(\iff\) \(A\) is symmetric and positive definite. Further, for \(n = 2\), the symmetric matrix \(\begin{bsmallmatrix} a & b\\ b & d \end{bsmallmatrix}\) is positive definite \(\iff\) \(a, d > 0\) and \(b^2 < ad\).

Easy. We check the positive definiteness of \(A := \begin{bsmallmatrix} a & b\\ b & d \end{bsmallmatrix}\). Note that \(x^t A\, x = a x_1^2 + 2b x_1 x_2 + d x_2^2\). If \(x = 0\), then it’s clearly zero. Thus, , let \(x_1\ne 0\) so that \(x^t A\, x = x_1^2\, p(x_2/x_1)\) where \(p\) is the quadratic polynomial \(p(t) := a + 2bt + dt^2\). Thus, \(x^t A\, x > 0\) for all \(x\) with \(x_1\ne 0\) \(\iff\) \(p\) is always positive \(\stackrel{\text{\scriptsize w}}{\iff}\) \(b^2 < ad\) and \(d > 0\). Similarly, one can analyze when \(x_2\ne 0\).

Exercise 8 (Doing the sums) For real \(a_i\)’s and \(-1 < x < 1\), one has: \[\begin{align*} \sum_{i = 0}^n x^i a_i & \le \frac{1}{\sqrt{1 - x^2}} \left(\sum_{i = 0}^n a_i^2\right)^{1/2}\\ \sum_{i = 1}^n \frac{a_i}{i} & \le \sqrt 2 \left(\sum_{i = 0}^n a_i^2\right)^{1/2}\\ \sum_{i = 1}^n \frac{a_i}{n + i} & \le \sqrt{\ln 2} \left(\sum_{i = 0}^n a_i^2\right)^{1/2}\\ \sum_{i = 0}^n \binom{n}{i} a_i & \le \sqrt{\binom{2n}{n}}\left(\sum_{i = 0}^n a_i^2\right)^{1/2} \end{align*}\]

These follow from the following:

  1. \(\sum_{i = 0}^n x^{2i} = (1 - x^{2(n + 1)})/(1 - x^2)\) \({}\le 1/(1 - x^2)\) (since \(|x| < 1\)).
  2. \(\sum_{i = 1}^n 1/i^2 \le 1 + \sum_{i = 2}^n \frac{1}{(i - 1)i}\) \({}= 2 - 1/n\le 2\).
  3. \(\sum_{i = 1}^n 1/(n + i)\le \int_0^n 1/(n + x)\) \({}= \sqrt 2\).
  4. \(\sum_{i = 0}^n \binom{n}{i}^2 = \sum_{i = 0}^n \binom{n}{i}\binom{n}{n - i}\) \({}= \binom{2n}{n}\) where the last equality follows by noting that choosing \(n\) objects out of \(2n\) is equivalent to first dividing them into two groups of \(n\) each and then choosing from among those groups.

Exercise 9 (Beating the obvious bounds) Let \(a_1, \ldots, a_n\) be reals. Then \[\begin{equation*} \left(\sum_{i = 1}^n a_i\right)^2 + \left( \sum_{i = 1}^n (-1)^i a_i \right)^2\le (n + 2)\sum_{i = 1}^n a_i^2. \end{equation*}\]

\[\begin{align*} \text{LHS} & = \sum_{i, j = 1}^n \left(a_i a_j + (-1)^{i + j} a_i a_j\right)\\ & = 2\ \sum_{\mathclap{i, j : i + j\text{ even}}} a_i a_j\\ & = 2\sum_i a_i\ \sum_{\mathclap{j : i + j\text{ even}}} a_j\\ & \le 2\sum_i a_i \sqrt{\left\lceil \frac{n}{2} \right\rceil} \sqrt{\ \ \sum_{\mathclap{j : i + j\text{ even}}} a_j^2}\\ & = 2 \sqrt{\left\lceil \frac{n}{2} \right\rceil} \left( \sum_{\text{even $i$}} a_i \sqrt{\sum_{\text{even $j$}} a_j^2} + \sum_{\text{odd $i$}} a_i \sqrt{\sum_{\text{odd $j$}} a_j^2}\right)\\ & \le 2 \left\lceil \frac{n}{2} \right\rceil \left( \sum_{\text{even $j$}} a_j^2 + \sum_{\text{odd $j$}} a_j^2\right)\\ & \le (n + 2)\sum_{i = 1}^n a_i^2 \end{align*}\]

Exercise 10 (Schur’s lemma—the \(R\) and \(C\) bounds) Let \(A\in\mathbb R^{m\times n}\). Then for any \(x\in\mathbb R^m\) and \(y\in \mathbb R^n\), one has \[\begin{equation*} \left| x^t A\, y \right|\le \sqrt{RC} \left( \sum_{i = 1}^m x_i^2 \right)^{1/2} \left( \sum_{i = 1}^n y_i^2 \right)^{1/2} \end{equation*}\] where \(R := \max_i \sum_j |A_{i, j}|\) and \(C := \max_j \sum_i |A_{i, j}|\).

\[\begin{align*} |x^t A\, y| & \le \sum_{i, j = 1}^n |x_i A_{i, j} y_j|\\ & \le \left(\sum_{i, j} x_i^2 |A_{i, j}|\right)^{1/2} \left( \sum_{i, j} |A_{i, j}| y_j^2 \right)^{1/2}\\ & = \left( \sum_i x_i^2 \sum_j |A_{i, j}| \right)^{1/2} \left( \sum_{j} y_j^2 \sum_i |A_{i, j}| \right)^{1/2}\\ & \le \text{RHS} \end{align*}\]

Exercise 11 (Schwarz’s argument in an inner product space) Prove Theorem 1 by considering the quadratic polynomial \(p(t) := \|u + tv\|^2\) on \(\mathbb R\).

Note that \(p\) is indeed quadratic, for \(\|u + tv\|^2 = \|u\|^2 + 2t\langle u, v\rangle + t^2\|v\|^2\). Since this is always nonnegative, its determinant must be nonpositive which yields \(|\langle u, v\rangle|\le\|u\|\|v\|\). Further, is the equality holds, then the discriminant of \(p\) vanishes so that it has (precisely) one root ensuring that \(u + tv = 0\) for some real \(t\).

Exercise 12 (Example of a self-generalization) If \(V_i\)’s are inner product spaces with inner products \(\langle\cdot, \cdot\rangle_i\) and \(w_i > 0\) for \(i = 1, \ldots, n\), then the following defines an inner product on \(\prod_i V_i\): \[\begin{equation*} \langle u, v\rangle := \sum_{i = 1}^n w_i \langle u_i, v_i\rangle_i. \end{equation*}\]

Easy.

Exercise 13 (Application of Cauchy’s inequality to an array) If \(A\in\mathbb R^{m\times n}\), then one has \[\begin{equation*} m\sum_{i = 1}^m r_i^2 + n\sum_{j = 1}^n c_j^2 \le \Bigl(\sum_{i, j} a_{i, j}\Bigr)^2 + mn\sum_{i, j} a_{i, j}^2 \end{equation*}\] where \(r_i := \sum_{j = 1}^n a_{i, j}\) and \(c_j := \sum_{i = 1}^m a_{i, j}\). Aldo, the equality holds \(\iff\) there exist \(\alpha_1, \ldots, \alpha_m, \beta_1, \ldots, \beta_n\in\mathbb R\) such that \(a_{i, j} = \alpha_i + \beta_j\).

Note that for \(X\in \mathbb R^{m\times n}\), one has \[\begin{equation*} \Bigl(\sum_{i, j} x_{i, j}\Bigr)^2\le mn\sum_{i, j} x_{i, j}^2. \end{equation*}\] Putting \(x_{i, j} = a_{i, j} - r_i/n - c_j/m\), we have: \[\begin{align*} \text{LHS} & = \Bigl(\sum_{i, j} a_{i, j} - \sum_i r_i - \sum_j c_j \Bigr)^2\\ & = \Bigl( \sum_{i, j} a_{i, j}^2 \Bigr)\\[2ex] \frac{\text{RHS}}{mn} & = \sum_{i, j}\left( a_{i, j}^2 + \frac{r_i^2}{n^2} + \frac{c_j^2}{m^2} - 2\frac{a_{i, j} r_i}{n} - 2\frac{a_{i, j} c_j}{m} + 2\frac{r_i c_j}{mn} \right)\\ & = \begin{aligned}[t] \sum_{i, j} a_{i, j}^2 + \frac{1}{n}\sum_i r_i^2 + \frac{1}{m}\sum_j c_j^2 - \frac{2}{n}\sum_i r_i^2 - \frac{2}{m}\sum_j c_j^2 \\[-1ex] {} + \frac{2}{mn}\Bigl(\sum_i r_i\Bigr)\Bigl(\sum_j c_j\Bigr) \end{aligned}\\ & = \sum_{i, j} a_{i, j}^2 - \frac{1}{n}\sum_i r_i^2 - \frac{1}{m}\sum_j c_j^2 + \frac{2}{mn}\Bigl( \sum_{i, j} a_{i, j} \Bigr)^2 \end{align*}\] This immediately yields the desired inequality.

Finally, the equality holds \(\iff\) \(x_{i, j}\) is independent of \(i\), \(j\) \(\stackrel{\text{\scriptsize w}}{\iff}\) \(a_{i, j} - r_i/n - c_j/m\) is independent of \(i\), \(j\) \(\stackrel{\text{\scriptsize w}}{\iff}\) \(\alpha_{i, j} = \alpha_i + \beta_j\) for some \(\alpha_1, \ldots, \alpha_m, \beta_1, \ldots, \beta_n\in\mathbb R\).

Exercise 14 (A Cauchy triple and Loomis-Whitney)  

  1. Let \(A, B, C\in\mathbb R^{n\times n}\). Then one has \[\begin{equation*} \left(\sum_{i, j, k = 1}^n a_{i, j}\, b_{j, k}\, c_{k, i}\right)^2 \le \left(\sum_{i, j = 1}^n a_{i, j}^2 \right) \left(\sum_{j, k = 1}^n b_{j, k}^2 \right) \left(\sum_{k, i = 1}^n c_{k, i}^2 \right). \end{equation*}\]
  2. Let \(S\subseteq\mathbb Z^3\) be finite and let \(S_x\), \(S_y\), \(S_z\) be projections on the \(yz\)-, \(zx\)-, \(xy\)-planes respectively. Then these are also finite with \[\begin{equation*} |S|\le |S_x|^{1/2} |S_y|^{1/2} |S_z|^{1/2}. \end{equation*}\]
  1. Solution to the first part: \[\begin{align*} \text{LHS} & = \sum_{i, k}\Bigl(\sum_j a_{i, j}\, b_{j, k}\Bigr) c_{k, i}\\ & \le \sum_{i, k} \biggl(\Bigl( \sum_j a_{i, j}^2 \Bigr)^{1/2} \Bigl( \sum_j b_{j, k}^2 \Bigr)^{1/2} c_{k, i}\biggr)\\ & = \sum_i \biggl(\Bigl( \sum_j a_{i, j}^2 \Bigr)^{1/2} \sum_k \Bigl( \sum_j b_{j, k}^2 \Bigr)^{1/2} c_{k, i}\biggr)\\ & \le \sum_i \biggl(\Bigl( \sum_j a_{i, j}^2 \Bigr)^{1/2} \Bigl( \sum_{j, k} b_{j, k}^2 \Bigr)^{1/2} \Bigl(\sum_k c_{k, i}^2\Bigr)^{1/2}\biggr)\\ & = \Bigl( \sum_{j, k} b_{j, k}^2 \Bigr)^{1/2} \sum_i \biggl(\Bigl( \sum_j a_{i, j}^2 \Bigr)^{1/2} \Bigl(\sum_k c_{k, i}^2\Bigr)^{1/2}\biggr)\\ & \le \Bigl( \sum_{j, k} b_{j, k}^2 \Bigr)^{1/2} \Bigl( \sum_{i, j} a_{i, j}^2 \Bigr)^{1/2} \Bigl( \sum_{k, i} c_{k, i}^2 \Bigr)^{1/2}\\ & = \text{RHS} \end{align*}\]
  2. Without loss of generality, let \(S\subseteq [1, n]^3\). Define \(A, B, C\in \{0, 1\}^{n\times n}\) by \(a_{i, j}\) to be \(1\) if \((i, j, k)\in S\) for some \(k\) and \(0\) otherwise, and similarly define \(b_{j, k}\) and \(c_{k, i}\). Thus, \(|S_z| = \sum_{i, j} a_{i, j}^2\) and similarly for \(|S_y|\) and \(|S_x|\). Define \(f\colon [1, n]^3\to\{0, 1\}\) by \((i, j, k)\mapsto a_{i, j}\, b_{j, k}\, c_{k, i}\). Then \(S\subseteq f^{-1}(\{1\})\). Thus, \[\begin{align*} |S| & \le \bigl|f^{-1}(\{1\})\bigr|\\ & = \sum_{\alpha\in [0, n]^3} f(\alpha)\\ & = \sum_{i, j, k = 1}^n a_{i, j}\, b_{j, k}\, c_{k, i}\\ & \le \Bigl( \sum_{i, j} a_{i, j}^2 \Bigr)^{1/2} \Bigl( \sum_{j, k} b_{j, k}^2 \Bigr)^{1/2} \Bigl( \sum_{k, i} c_{k, i}^2 \Bigr)^{1/2}\\ & = |S_z|^{1/2} |S_y|^{1/2} |S_x|^{1/2}. \end{align*}\]
Note

Note that the equality holds for taking \(S\) to be a cube aligned along the axes.

Exercise 15 (An application to statistical theory) Let \(D\) be finite a finite set and \(\Theta\subseteq\mathbb R\) be an interval. Let \(p\colon D\times\Theta\to[0, +\infty)\) be such that for each \(k\in D\), we have that \(\sum_{k\in D} p(k, \theta) = 1\) and that \(p(k, \cdot)\colon \Theta\to\mathbb R\) is differentiable. Let \(g\colon D\to \mathbb R\) such that \(\sum_{k\in D} g(k) p(k, \theta) = \theta\). Then \[\begin{equation*} \sum_{k\in D} (g(k) - \theta)^2 p(k, \theta) \ge \frac{1}{I(\theta)} \end{equation*}\] where \(I(\theta) := \sum_{k\in D}\left(\frac{p_\theta(k, \theta)}{p(k, \theta)}\right)^2 p(k, \theta)\).

\[\begin{align*} I(\theta) \sum_{k\in D} (g(k) - \theta)^2 p(k, \theta) & \ge \biggl(\sum_k p_\theta(k, \theta) (g(k) - \theta)\biggr)^2\\ & = \biggl(1 - \theta\sum_k p_\theta(k, \theta)\biggr)^2\\ & = 1 \end{align*}\] Here, the equalities follow by respectively differentiating \(\sum_k g(k) p(k, \theta) = \theta\) and \(\sum_k p(k, \theta) = 1\) with respect to \(\theta\).

Errata

  1. p. 11: “binomial quadratic formula” in the third paragraph.
  2. p. 17: \(\Theta\) must be given to be an appropriate set since differentiability is being talked of. For instance, \(\Theta\) may be taken to be an interval.
  3. p. 228: In the solution to Ex 7, \(\text{``}\langle x, y\rangle = 5x_1 y_1 + x_1 y_2 + x_2 y_1\) \({}+ \cancel{3y_2^2} {\color{YellowOrange}{3x_2 y_2}}\text{"}\) and \(\text{``}5z^2 + \cancel{3z}{\color{YellowOrange}{2z}} + 3 =0\text{"}\).