cauchy schwarz inequality states that: (case of real numbers)

$$ \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right) $$

and we get an equality when {ak} and {bk} are linearly dependant.

my question is:

what should be the conditions for {ak} and {bk} so we get:

$$ \left( \sum_{k=1}^n a_k b_k \right)^2 \ll \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right) $$



  • $\begingroup$ Hint: the smallest value of $\displaystyle \left( \sum_{k=1}^n a_k b_k \right)^2$ is $0$ and it occurs when the standard inner product of the two (non-zero) vectors is $0$. If either vector is allowed to be the zero vector, then you will have equality (instead of $\ll$) in your desired result. $\endgroup$ – Dilip Sarwate Jul 10 '14 at 22:28
  • $\begingroup$ What do you mean by $\ll$? $\endgroup$ – egreg Jul 10 '14 at 22:38
  • $\begingroup$ the left term "very small", compared to the right term $\endgroup$ – Dan Ohev Zion Jul 11 '14 at 15:47
  • $\begingroup$ Dilip Sarwate, thank you, but in my application non of the vectors is the zero vector. also, the standard inner product (left term) is not zero in my case. I just need the left term to be much smaller than the right term (but not zero) $\endgroup$ – Dan Ohev Zion Jul 11 '14 at 15:50

Define the following two vectors \begin{align} \textbf{a}&=(a_1,a_2,...,a_n),\\ \textbf{b}&=(b_1,b_2,...,b_n) \end{align} the dot product is defined such that $|\textbf{a}\cdot \textbf{b}|^2 = |\textbf{a}|^2|\textbf{b}|^2\cos^2\theta$ where $\theta$ is the angle between the two vectors $\textbf{a}$ and $\textbf{b}$. So the desired condition holds if $\cos^2\theta\ll 1$, i.e. if $\textbf{a}$ and $\textbf{b}$ are close to being at right-angles.

  • $\begingroup$ OK, but note that the elements of the vectors a and b are real numbers. how do you define the angle between these vectors? $\endgroup$ – Dan Ohev Zion Jul 11 '14 at 15:45
  • $\begingroup$ @DanOhevZion I have updated my answer - does that clarify things? $\endgroup$ – lemon Jul 11 '14 at 17:26
  • $\begingroup$ yes, thank you. can you give a set of conditions for vectors a and b, such that the angle between them satisifies $\cos^2\theta\ll 1$ ? $\endgroup$ – Dan Ohev Zion Jul 13 '14 at 20:09
  • $\begingroup$ Any condition that gives $\cos^2\theta\ll 1$ would just be a reformulation of the inequality that you presented. For example, one condition would be $|\textbf{a}\times\textbf{b}|\approx|\textbf{a}||\textbf{b}|$ where $\textbf{a}\times\textbf{b}$ is the vector product. What exactly is it you're trying to achieve? $\endgroup$ – lemon Jul 13 '14 at 21:47

If you form a 2x2 symmetrical matrix of the three sums as a covariance matrix

$M= \begin{bmatrix} \sum{a_k^2} & \sum{a_k b_k} \\ \sum{a_k b_k} & \sum{b_k^2} \end{bmatrix} $

... the Cauchy-Schwarz inequality is equivalent to stating that $M$ is positive semi-definite, i.e. eigenvalues of $M$ are >=0.

If the value of $(\sum{a_k b_k})^2$ is no more than a negligible fraction of $\sum{a_k^2} \sum{b_k^2}$, it means:

  • the eigenvalues of the matrix are nearly equal; or
  • the eigenvectors are nearly aligned to the axes.

So you are looking for the case where either of these conditions is true, which means the case where the $a_k$ are statistically independent of the $b_k$, i.e. there's little or no correlation between $a_k$ and $b_k$ (over the set of $k$ ).

The opposite extreme - where the Cauchy-Schwarz holds to the equality - occurs when you have linear dependence: $\alpha a_k = \beta b_k$ for some $\alpha, \beta$ and all $k$. In this case one of the eigenvalues is zero, and the corresponding eigenvector is in the direction $(\alpha, -\beta)$.

  • $\begingroup$ Thank you! This gives me a new perspective for my problem $\endgroup$ – Dan Ohev Zion Nov 9 '14 at 8:06

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