Let $H$ be a Hilbert space and $a,b,x$ be vectors in $H$.

I'm looking for a sharp upper bound on $|\langle a,x \rangle \langle b,x \rangle - \langle x,x \rangle \langle a,b\rangle|$.

By the triangle inequality and Cauchy Schwarz this is clearly bounded by $2\|a\|\|b\|\|x\|^2$.

I think this is quite brutal, is there a bound with a constant less than $2$ ?

I tried rewriting the quantity as $\langle \langle a,x \rangle b - \langle a,b \rangle x, x \rangle$ or $\langle \langle a,x \rangle x - \langle x,x \rangle a, b \rangle$ but I can't get an improved upper bound this way.

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    $\begingroup$ A better bound is provided by the square root of the product between |a|^2|x|^2-<a,x>^2 and the same expression with a replaced by b $\endgroup$ Commented Aug 13, 2022 at 10:39
  • $\begingroup$ In R^3 that is just |a^x||b^x| with ^ being the cross product. Sorry for the bad formatting, I am writing from my phone. $\endgroup$ Commented Aug 13, 2022 at 10:43
  • $\begingroup$ @JackD'Aurizio How do you obtain this upper bound ? A further bound would then be $\|a\|\|b\|\|x\|^2$, thus reducing the constant from $2$ to $1$. $\endgroup$ Commented Aug 13, 2022 at 10:44
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    $\begingroup$ This is exercise 4.12 from The Cauchy-Schwarz Master Class by Steele. There is an excellent solution there using Gram-Schmidt. The bound provided there is sharper than the ones in the answers below. $\endgroup$
    – V.S.e.H.
    Commented Aug 13, 2022 at 14:15
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    $\begingroup$ @V.S.e.H. There is no shaper bound than $\lvert \langle a,x \rangle \langle b,x \rangle - \langle x,x \rangle \langle a,b\rangle \lvert \le \lVert a \rVert \lVert b \rVert \lVert x \rVert^2 $ as equality is attained when $b=-a$ and $x$ orthogonal to $a$. $\endgroup$ Commented Aug 13, 2022 at 15:02

5 Answers 5


Let's prove that $\lvert \langle a,x \rangle \langle b,x \rangle - \langle x,x \rangle \langle a,b\rangle \lvert \le \lVert a \rVert \lVert b \rVert \lVert x \rVert^2$.

Without loss of generality, we can suppose that $\lVert a \rVert = \lVert b \rVert = \lVert x \rVert=1$. Now take an orthonormal basis $\{e_1, e_2, \dots\}$ of $H$ such that $a= \cos \theta e_1 + \sin \theta e_2$ and $b= \cos \theta e_1 - \sin \theta e_2$, where $0 \le \theta \le \frac{\pi}{2}$. We get

$$\begin{aligned} \langle a,x \rangle \langle b,x \rangle - \langle x,x \rangle \langle a,b\rangle &= (\cos \theta \langle e_1,x \rangle + \sin \theta \langle e_2,x \rangle)(\cos \theta \langle e_1,x \rangle - \sin \theta \langle e_2,x \rangle) - \cos 2\theta\\ &=\cos^2 \theta\langle e_1,x \rangle^2 - \sin^2 \theta \langle e_2,x \rangle^2 - \cos 2 \theta\\ &=(\langle e_1,x \rangle^2 + \langle e_2,x \rangle^2) \cos^2 \theta - \langle e_2,x \rangle^2 - \cos 2 \theta\\ &=\frac{\langle e_1,x \rangle^2 - \langle e_2,x \rangle^2}{2}-\left(1 - \frac{\langle e_1,x \rangle^2 + \langle e_2,x \rangle^2}{2}\right) \cos 2 \theta \end{aligned}$$

Even if it means swapping $a$ and $b$, we can suppose that $\lvert \langle e_1,x \rangle \rvert \ge \lvert \langle e_2,x \rangle \rvert$. The quantity above depends on $\theta$ and is positive and maximum for $\theta = \pi$, i.e. $b=-a$. In that case

$$\begin{aligned} \lvert \langle a,x \rangle \langle b,x \rangle - \langle x,x \rangle \langle a,b\rangle \lvert &\le 1 - \langle e_2,x \rangle^2 \le 1 \end{aligned}$$ and we get the inequality

$$\lvert \langle a,x \rangle \langle b,x \rangle - \langle x,x \rangle \langle a,b\rangle \lvert \le \lVert a \rVert \lVert b \rVert \lVert x \rVert^2$$ which is an equality if and only if $b= \pm a$ and $x$ is orthogonal to $a$.

  • $\begingroup$ Thank you for the thorough answer. Can you please add why it is possible to find a basis such that $a= \cos \theta e_1 + \sin \theta e_2$ and $b= \cos \theta e_1 - \sin \theta e_2$ ? $\endgroup$ Commented Aug 13, 2022 at 16:07
  • $\begingroup$ Just take $e_1 =\frac{a+b}{\lVert a+b \rVert}$, $e_2 =\frac{a-b}{\lVert a-b \rVert}$ and let $\cos 2 \theta = \langle a, b \rangle$. $\endgroup$ Commented Aug 13, 2022 at 16:13

The span of the three vectors $\,a,b,x\,$ generate a subspace of at most three dimensions. Thus, without loss of generality we can assume that they are in $\,\mathbb R^3.\,$ Use the Binet-Cauchy identity

$$(a \times b)\cdot(c \times d) = (a \cdot c)(b \cdot d)-(a \cdot d)(b \cdot c) \tag1 $$

specialized to

$$(a \times x)\cdot(b \times x)=(a \cdot b)(x \cdot x)-(a \cdot x)(x \cdot b) \tag2 $$

and using $\,(a \cdot x) = (x \cdot a)\,$ to get

$$ \lVert(a \times x)\rVert^2 = (a \times x)\cdot(a \times x)= (a \cdot a)(x \cdot x)-(a \cdot x)(x \cdot a) \le \lVert a\rVert^2 \lVert x\rVert^2 \tag3 $$

which implies

$$ \lVert(a \times x)\rVert \le \lVert a\rVert \lVert x\rVert \qquad \text{ and } \qquad \lVert(b \times x)\rVert \le \lVert b\rVert \lVert x\rVert. \tag4$$

Rearrange the terms of equation $(2)$ to get

$$ (a \cdot x)(x \cdot b)-(a \cdot b)(x \cdot x)=-(a \times x)\cdot(b \times x) \tag5. $$

Change notation to

$$ \langle a,x\rangle \langle x,b\rangle - \langle a,b\rangle \langle x,x\rangle = -(a \times x)\cdot(b \times x). \tag6 $$

Use the results of equation $(4)$ on the right side and $\, \langle x,b\rangle = \langle b,x\rangle \,$ to get

$$ |\langle a,x\rangle \langle b,x\rangle -\langle a,b\rangle \langle x,x\rangle| \le \lVert a \rVert \lVert b \rVert \lVert x \rVert^2 \tag7. $$


We may assume without loss of generality that $a,b,x\in\mathbb{R}^3$ have unit modulus and $$ x=(1,0,0)\qquad a=(\cos\theta,\sin\theta,0),\qquad b=(\cos\lambda\cos\phi,\sin\lambda\cos\phi,\sin\phi). $$

Here we have $\langle a,x\rangle=\cos\theta$, $\langle b,x\rangle=\cos\lambda\cos\phi$ and $\langle a,b\rangle = \cos\theta\cos\lambda\cos\phi+\sin\theta\sin\lambda\cos\phi$, so our purpose is to find $$ \max_{\theta,\lambda,\phi}\left|\cos\phi\right|\cdot\left|\cos\theta\cos\lambda-\cos(\theta-\lambda)\right|=\max_{\theta,\lambda}\left|\cos\theta\cos\lambda-\cos(\theta-\lambda)\right|. $$ The stationary points of the function $f(\theta,\lambda)=\cos\theta\cos\lambda-\cos(\theta-\lambda)$ occur at the points such that $$ -\sin\theta\cos\lambda+\sin(\theta-\lambda) = 0 = -\cos\theta\sin\lambda-\sin(\theta-\lambda)$$ in particular along the curves described by $\sin(\theta+\lambda)=0$. Since $$f(\theta,\pi n-\theta)=(-1)^n\left(\cos^2\theta-\cos(2\theta)\right)=(-1)^n\sin^2\theta$$ we have that $$ \left|\langle a,x\rangle \langle b,x\rangle - \langle a,b\rangle \langle x,x\rangle \right| \leq |a||b||x|^2.$$ The condition $|\cos\phi|=1$ is equivalent to the fact that $a,b,x$ lie on the same plane through the origin. If that is not the case the RHS can be further improved.

Alternative approach. This proves just a weaker inequality, but I want to outline it nevertheless since I promised in the comments. Assuming $a,b,x\in\mathbb{R}^n$ (even though it is sufficient to tackle the case $n=3$) we want to bound the absolute value of $$ \sum a_j x_j \sum b_k x_k - \sum x_j^2 \sum a_k b_k = \sum_{j\neq k} x_j b_k(a_j x_k-x_j a_k) $$ where by Lagrange's identity $$ \sum_{1\leq j < k\leq n}(a_j x_k-x_j a_k)^2 = \sum a_j^2 \sum x_k^2-\left(\sum a_j x_j\right)^2$$ such that $$\left|\sum_{j\neq k}x_j b_k(a_j x_k-x_j a_k)\right|\leq \sqrt{\sum_{j\neq k}x_j^2 b_k^2\sum_{j\neq k}(a_j x_k-x_j a_k)^2}$$ where the RHS equals $$ \sqrt{\left(\sum x_j^2\sum b_k^2-\sum x_j^2 b_j^2\right)\cdot 2\left(\sum x_j^2\sum a_k^2-\left(\sum a_j x_j\right)^2\right)} $$ such that $$ \left|\langle a,x\rangle \langle b,x\rangle - \langle a,b\rangle \langle x,x\rangle \right| \leq \color{blue}{\sqrt{2}} |a||b||x|^2.$$

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    $\begingroup$ The vector space generated by $a,b,x$ has at most dimension $3$: if $a,b,x$ live in a space $H$ with more than three dimensions, up to isometries we may assume that all the coordinates of $a,b,x$ from the fourth one to the last one are zero. I.e. we may assume that $a,b,x\in\mathbb{R}^3$. $\endgroup$ Commented Aug 13, 2022 at 15:20
  • $\begingroup$ Well, this is basically the proof which mathcounterexample gave above ... $\endgroup$
    – Salcio
    Commented Aug 13, 2022 at 17:06

$ \newcommand\form[1]{\langle#1\rangle} \renewcommand\Re{\operatorname{Re}} \renewcommand\Im{\operatorname{Im}} \newcommand\conj\bar \newcommand\Ext{{\bigwedge}} $

The inner product $\form{\cdot,\cdot}$ extends naturally to an inner product on the exterior algebra $\Ext H$ via $$ \form{v_1\wedge\cdots\wedge v_k,\: w_1\wedge\cdots\wedge w_l} = \delta_{kl}\det\Bigl(\form{v_i,v_j}\Bigr)_{i,j=1}^k, \tag{$*$} $$ i.e. take the determinant of the matrix with those entries. To see this, note that $\form{v,w}' := \form{v,\conj w}$ is a bilinear form; then is is well known that this extends naturally to a bilinear form $\form{\cdot,\cdot}'$ on $\Ext H$ via ($*$) but using $\form{\cdot,\cdot}'$, and we define $\form{X, Y} := \form{X, \conj Y}'$ for $X, Y \in \Ext H$. It is the easy to determine that ($*$) is satisfied.

Since $\form{\cdot,\cdot}$ is an inner product, it satisfies Cauchy-Schwarz. We see $$ |\form{x,x}\form{a,b} - \form{a,x}\form{b,x}| = |\form{x\wedge a, x\wedge b}| \leq \sqrt{\form{x\wedge a, x\wedge a}\form{x\wedge b, x\wedge b}}. $$ But $$ \form{x\wedge a, x\wedge a} = |x|^2|a|^2 - \form{x, a}\form{a, x} = |x|^2|a|^2 - |\form{x, a}|^2 \leq |x|^2|a|^2, $$ and the same for $\form{x\wedge b, x\wedge b}$. Hence $$ |\form{x,x}\form{a,b} - \form{a,x}\form{b,x}| \leq |x|^2|a||b|. $$

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    $\begingroup$ I see that this is essentially the same as Somos' answer; however, both he and others seem to be assuming the Hilbert space is real, where my answer does not. $\endgroup$ Commented Aug 13, 2022 at 18:06

Assume WLOG, $\left\langle a, a\right\rangle = 1$. Let $A = \frac12 \left(ab^T + ba^T\right)$, $t = \left\langle a, b\right\rangle$ and $c = b - ta$. Then $\left\langle c, a\right\rangle = 0$, $\left\langle b, b\right\rangle = t^2 + \left\langle c, c \right\rangle$ and $$A = taa^T + \frac{1}{2}\left(ac^T + ca^T\right).$$

Let's find the set of eigenvalues of $A$. Since $\text{rank}(A) \le 2$ ($\text{Im}(A) \subset \text{Vect}(a, c)$), it has at most two non-zero eigenvalues. Since $$Aa = ta + \frac{1}{2} c$$ and $$Ac = \frac{1}{2}\left\langle c, c \right\rangle a,$$ the non-zero eignevalues (if they exist) are solutions of the equation:

$$\mu^2 - t\mu - \frac{1}{4}\left\langle c, c\right\rangle$$

This proves that:

$$\lambda \left(A\right) = \left\{0, \frac{t \pm \sqrt{t^2 + \left\langle c, c\right\rangle}}{2}\right\} = \left\{0, \frac{\left\langle a, b\right\rangle \pm \left\|b\right\|}{2}\right\}.$$

Now back to the question:

$$\left|\left\langle a, x\right\rangle\left\langle b, x\right\rangle - \left\langle a, b\right\rangle\left\langle x, x\right\rangle \right| = \left|\left\langle x, \left(A - \left\langle a, b\right\rangle I\right)x\right\rangle\right| \le \left(\max\limits_{\mu \in \lambda\left(A - \left\langle a, b\right\rangle I\right)} |\mu|\right)\left\|x\right\|^2 $$

However, \begin{align} \max\limits_{\mu \in \lambda\left(A - \left\langle a, b\right\rangle I\right)} |\mu| &= \max\limits_{\mu \in \lambda\left(A\right)} |\mu - \left\langle a, b\right\rangle|\\ &= \max \left\{\left|\left\langle a, b\right\rangle\right|, \left| \frac{\left\langle a, b\right\rangle \pm \left\|b\right\|}{2}-\left\langle a, b\right\rangle \right|\right\}\\ &= \max \left\{\left|\left\langle a, b\right\rangle\right|, \left|\frac{\left\|b\right\| + \left\langle a, b\right\rangle}{2}\right|, \left|\frac{\left\|b\right\| - \left\langle a, b\right\rangle}{2}\right|\right\} \le \frac{1}{2}\left(\left\|b\right\| + \left|\left\langle a, b\right\rangle \right|\right). \end{align}

And you will find the better inequality:

$$\left|\left\langle a, x\right\rangle\left\langle b, x\right\rangle - \left\langle a, b\right\rangle\left\langle x, x\right\rangle \right| \le \frac{1}{2}\left(\left\|a\right\|\left\|b\right\| + \left|\left\langle a, b\right\rangle\right|\right)\left\|x\right\|^2$$


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