jump to navigation

Cauchy-Schwarz via tensor product October 1, 2008

Posted by Scott Carnahan in Uncategorized.
trackback

I’m a teaching assistant for the freshman “multivariable calculus with extra pain” class this term, and we went over Cauchy’s inequality for \mathbb{R}^n a couple weeks ago. One of the homework problems from that week asked for a proof of Bunyakovsky’s generalization: that for functions f and g that are continuous on [0,1],
(\int_0^1 f(x)g(x) dx)^2 \leq (\int_0^1 f(x)^2 dx)(\int_0^1 g(x)^2 dx).
Most of the solutions involved translating the finite dimensional Pythagorean style proof in the text to the function setting. Some of the students used a clever trick with the discriminant of a quadratic polynomial, analogous to (and possibly derived from) the one in this wikipedia article. There was one solution that stood out to me, partly because it used material which we haven’t covered yet, but mostly because I hadn’t seen it before. My question for the audience is: does this proof appear elsewhere?

The engine behind the proof is Fubini’s theorem: (\int_0^1 f(x)g(x) dx)^2 = \int_0^1 \int_0^1 f(x)g(x)f(y)g(y) dx dy and (\int_0^1 f(x)^2 dx)(\int_0^1 g(x)^2 dx) = \int_0^1 \int_0^1 f(x)^2g(y)^2 dx dy. The inequality then follows from a symmetrization
(\int_0^1 f(x)^2 dx)(\int_0^1 g(x)^2 dx) = \int_0^1 \int_0^1 f(y)^2g(x)^2 dx dy
together with the observation that
\int_0^1 \int_0^1 (f(x)g(y) - f(y)g(x))^2 dx dy \geq 0.

Here’s a linear algebra translation: Define (a \otimes b, c \otimes d) := (a,c)(b,d). Then
(f \otimes g - g \otimes f, f \otimes g - g \otimes f) \geq 0 implies
(f \otimes g,f \otimes g) + (g \otimes f, g\otimes f) \geq (f \otimes g, g \otimes f) + (g \otimes f, f \otimes g),
which implies 2(f,f)(g,g) \geq 2(f,g)(g,f).

I actually prefer the Fubini version, because the last integral is somehow more manifestly non-negative than the inner product. The idea that tensor products preserve positive semidefiniteness seems less obvious than the non-negativity of squares.

I should compare this to other proofs online: the last integral is essentially the same as the sum of squares in Timothy Gowers’s first proof. Also the use of tensor products may remind you of Terry Tao’s post about tensor products that includes a proof of Cauchy-Schwarz, but this proof doesn’t actually use the tensor product to improve an existing estimate, and the proof there employs some really neat observations about symmetries rather than using tensor products.

You might be interested to know that quantum optics people seem to routinely violate Cauchy-Schwarz.

About these ads

Comments

1. Steve - October 1, 2008

Those quantum optics “violations” refer to an inequality that takes the same form as Cauchy-Schwartz, but using coherences of light, rather than vectors in a vector space. The inequality is obeyed in the classical theory of light, but quantum effects enable a violation.

2. Vishal Lama - October 1, 2008

I have a suspicion, based on a vague recollection, that the proof based on Fubini’s theorem has perhaps appeared in the book Putnam and Beyond by Titu Andreescu and Razvan Gelca. I don’t have a copy right now, and so, I can’t confirm immediately if it’s true.

3. Lior Silberman - October 2, 2008

This proof seems to be the continuous version of Lagrange’s well-known identity
\left( \sum_i a_i^2 \right) \left( \sum_j b_j^2 \right) - \frac{1}{2}\sum_{i,j} \left( a_i b_j - a_j b_i \right)^2.

4. Lior Silberman - October 2, 2008

Oops — here’s the correct version:

\left( \sum_i a_i^2 \right) \left( \sum_j b_j^2 \right) - \left( \sum_i a_i b_i \right)^2 = \frac{1}{2}\sum_{i,j} \left( a_i b_j - a_j b_i \right)^2.

5. Lior Silberman - October 2, 2008

[could someone explain how to get LaTeX to parse?]

[Wordpress doesn't seem to like line breaks - ed.]

6. Vishal Lama - October 2, 2008

Lior,

If you visit Terry Tao’s blog and scroll down to the end of the (front) page, you will find instructions for embedding LaTeX in comments.

7. Emmanuel Kowalski - October 2, 2008

Coincidentally, I’m teaching Functional Analysis this semester; to provve Cauchy-Schwarz in a general Hilbert space, I used the discriminant method (it’s a useful trick to know, and the students were supposed to have seen this earlier anyway, so I didn’t feel too bad with not motivating it better). Then I looked in Bourbaki, feeling confident he would argue in the same way, but no: the idea is the same (the norm of x+ty is non-negative for all t), but a specific value of t is chosen where this clearly implies the desired inequality. (I don’t have the book handy, but I guess this value of t must be the point where the quadratic expression is minimal).

8. Scott Carnahan - October 2, 2008

Gowers calls the Bourbaki proof “slightly less motivated” in his Cauchy-Schwarz page (linked above). He points out both that one has chosen t to minimize the distance from x+ty to the origin, and that this minimization argument shouldn’t be necessary, since it should be clear that the line described by x+ty hits the origin if and only if x and y are collinear.

9. Emmanuel Kowalski - October 4, 2008

I don’t really think of Bourbaki as a place to find a well-motivated argument usually (this was not their objective). Rather, I think it contains what the Bourbaki group (as it was at the time of writing a particular part…) thought as being the “right” underlying explanation for a particular mathematical fact. Somehow, I had the feeling they would have thought more natural to look at the norm of x+ty as a global function of t.

10. Scott Carnahan - October 5, 2008

Thanks for pointing out the name, Lior. It seemed rather unlikely that no one had thought of this before, but it doesn’t seem to have had much exposure.


Sorry comments are closed for this entry

Follow

Get every new post delivered to your Inbox.

Join 646 other followers

%d bloggers like this: