Quick notes on the "Enhanced Cauchy Schwarz inequality and some of its statistical applications

I briefly read this nice paper by Sergio Scarlatti.

In this, the author proves an intermediate term between the classical bounds of the CS inequality.

Classical CS inequality

First things first, for all \(x,y\) in some Hilbert space \(H\) with inner product \((.,.)\) we have:

\[|(x,y)|\leq \|x\|\|y\|.\]

The author shares a nice proof of CS that I’m not sure if I’ve seen before (but looks neat). We define the matrix \(C=C(x,y)\) with \(c_{ij}=\frac{1}{\sqrt{2}}(x_iy_j-x_jy_i),\ i,j=1,\ldots,n\). Then, its second norm is


If we substitute the definition of \(c_{ij}\) to the above and carry out the algebra, we have \(\|C\|^2_2=\|x\|^2\|y\|^2-(x,y)^2,\) which proves CS as the norm is non-negative.

Enhanced CS

Now, suppose \(V\subseteq H\) is some closed subspace of \(H\). Then if \(P\) is the orthogonal projection onto \(V\) (i.e., \(PH=V\)), the author defines:

\[D(x,y|P):=\|Px\|\cdot \|Py\|+\|P^{\perp}x\|\cdot \|P^{\perp}y\|,\]

for all \(x,y\in H\) and where \(P^{\perp}\) is the projection on the orthogonal complement of \(V\). Then, \(|(x,y)|\leq D(x,y|P)\leq \|x\|\|y\|.\)

I like inequalities with free terms! We can pick \(P\) depending on the problem and the bounds would adapt correspondingly. If \(P\) is a trivial projection to \(H\) or its complement, we recover usual CS.

Proof: It’s a short argument.

\[|(x,y)|=|(Px,Py)+(P^{\perp}x, P^{\perp}y)|\leq |(Px,Py)|+|(P^{\perp}x, P^{\perp}y)|\leq \|Px\|\|Py\|+\|P^{\perp}x\|\|P^{\perp}y\|,\]

where we used bilinearity of inner product, triangle inequality, and CS inequality for each term. Now, if \(a=\|Px\|, b=\|Py\|, c=\|P^{\perp}x\|, d=\|P^{\perp}y\|\), the author shows that

\[ab+cd=\sqrt{(ab+cd)^2}\leq \sqrt{(ab+cd)^2+(ac-bd)^2}=\sqrt{(a^2+c^2)(d^2+b^2)},\]

which is important because with the definitions of \(a,b,c,d\) above, \(a^2+c^2=\|x\|^2\) and similarly for \(y\).

The author shows this with an appeal to algebra (as shown above), but you will notice a simpler way; this bound is just CS but applied to the vectors \(u=(a,c)\) and \(v=(b,d)\). \(\square\)

Going beyond \(P\)

What if we have more than one subspace? Suppose \(V, U\subseteq H\) with corresponding projections \(P,Q\), then \(x=Px+P^{\perp}x\) and similarly for \(y\), which gives

\[\begin{align} |(x,y)|&\leq |(Px,Qy)|+|(Px, Q^{\perp}y)|+|(P^{\perp}x,Qy)|+|(P^{\perp}x, Q^{\perp}y)|\\ &\leq (\|Px\|+\|P^{\perp}x\|)(\|Qy\|+\|Q^{\perp}y\|), \end{align}\]

Again, by bilinearity, triangle ineq., and CS.

The last bound is not as good; setting \(P=Q\) there does not recover the previous results (note \(\|x\|\leq \|Px\|+\|P^{\perp}x\|\)). That’s because some terms would be cancelled from the first bound but are not cancelled from the second bound. For example, \((Px, P^{\perp}y)=0\) regardless of \(x,y\), and so on.


  1. Scarlatti, S., 2024. Enhanced Cauchy Schwarz inequality and some of its statistical applications. arXiv preprint arXiv:2403.13964.