CauchySchwarzBunyakovsky inequality

JaneFairfax · 2009-02-05 11:11:19

Today I learned from Introduction to Metric and Topological Spaces by W.A. Sutherland a version of the CauchySchwarz inequality involving integrals. It goes by the funny name of CauchySchwarzBunyakovsky inequality.

First, lets recall the CauchySchwarz inequality. It states

for al real numbers

.

Proof:

The inequality is obviously true if

. Hence we may assume that at least one is not 0.

Then

Treating the LHS as a quadratic in

, we see that its discriminant cannot be positive.

This is the proof given in Introduction to Metric and Topological Spaces.

The integral version is as follows.

The proof is similar to the CauchySchwarz case, only this time you start with

Last edited by JaneFairfax (2009-02-06 03:32:01)

Daniel123 · 2009-02-05 11:53:01

That's really nice

TheDude · 2009-02-06 00:56:02

JaneFairfax wrote:

Treating the LHS as a quadratic in
, we see that its discrimant cannot be positive.

I don't follow this part. Why can't the discriminant be positive?

JaneFairfax · 2009-02-06 01:04:20

Because then the quadratic equation LHS = 0 would have two distinct real roots and the LHS would be negative between these roots.

Last edited by JaneFairfax (2009-02-06 01:36:09)

Daniel123 · 2009-02-06 01:09:12

TheDude wrote:

JaneFairfax wrote:
Treating the LHS as a quadratic in
, we see that its discrimant cannot be positive.
I don't follow this part. Why can't the discriminant be positive?

The fact that the quadratic is always greater than or equal to 0 means that it must have at most one real root.

EDIT: In other words, what Jane said. I clicked 'quote' and left the room for too long

Last edited by Daniel123 (2009-02-06 01:10:35)

TheDude · 2009-02-06 04:14:03

I'm an idiot. Thanks for the explanation.

Ricky · 2009-02-06 07:01:18

It goes by the funny name of CauchySchwarzBunyakovsky inequality.

That is rather odd, normally I've heard it referred to as the Cauchy-Schwarz special case of Holder's inequality, where Holder's inequality is:

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Where 1/p + 1/q = 1. Of course, Cauchy-Schwarz is the special case with p = q = 2.

George,Y · 2009-02-07 09:26:59

That's nice. I remember we have a similar proof in Calculus for multi-variable function's Taylor expansion. It adds in lamda as well. Such method is called "adding parameter", which shares the same delicacy as adding a line to solve geometry problems.

George,Y · 2010-09-10 16:14:34

A good one

tharun reddy · 2010-09-27 01:06:44

...hi jane .........but you assumed that "lambda" was real but by setting discriminant to less than zero ur making lambda value imaginary:)

tharun reddy · 2010-09-27 01:08:05

..iam really in need of a solution to this problem please reply soon

George,Y · 2010-10-04 22:24:11

Jane has already stopped discussing serious topics.

bobbym · 2010-10-04 22:40:37

Hi George,Y;

George you are a good man and I like you.

Jane has done a lot for the forum.
She has solved more than her share of tough problems here.
I know you mean you miss her input, so do I.

ChiaraM27 · 2010-11-16 12:56:45

Basically, the roots cannot be real. Consider the quadratic equation as a parabola. If the equation has real roots then it crosses the x axis twice and has negative values between them, but we know that our quadratic function cannot be negative, so the roots have to be imaginary and so the discriminant has to be less than zero.

JaneFairfax · 2010-11-26 00:46:50

tharun reddy wrote:

...hi jane .........but you assumed that "lambda" was real but by setting discriminant to less than zero ur making lambda value imaginary:)

Last edited by JaneFairfax (2010-12-21 07:37:00)

Math Is Fun Forum

#1 2009-02-05 11:11:19