Cauchy-Schwarz via tensor product October 1, 2008Posted by Scott Carnahan in Uncategorized.
I’m a teaching assistant for the freshman “multivariable calculus with extra pain” class this term, and we went over Cauchy’s inequality for 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],
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: and . The inequality then follows from a symmetrization
together with the observation that
Here’s a linear algebra translation: Define . Then
which implies .
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.