Cauchy-Schwartz Inequality for Step Functions

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Homework Help Overview

The problem involves proving the Cauchy-Schwartz inequality for step functions defined on a real interval. Participants are tasked with demonstrating that the square of the integral of the product of two step functions is less than or equal to the product of their individual integrals squared.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the hint involving a quadratic function and its properties, including non-negativity and differentiation. There are attempts to find a turning point and to express the integral in terms of a variable.

Discussion Status

Some participants have expressed confusion regarding the initial steps and the integration variable. Others have provided guidance on expanding the quadratic function and minimizing it, although there is still uncertainty about how to derive the inequality from the established expressions.

Contextual Notes

There is a mention of integrating with respect to the variable associated with the step functions, which may not have been clearly defined initially. Participants are navigating through the implications of the hint and the structure of the problem.

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Homework Statement



Let

\phi,\psi : [a,b] \rightarrow \Re

be step functions.

Prove that

(\int \phi\psi)^{2} \leq (\int\phi^{2})(\int\psi^{2}) .

Hint: Consider the quadratic function of a real variable t defined by

Q(t)=\int(t\phi+\psi)^2 .

The Attempt at a Solution



I really don't know where to start with this, and the hint only confuses me more! :p

Any help appreciated, thanks!
 
Last edited:
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Q(t)>=0, since it's the integral of a nonnegative function (a square). Expand Q(t) out and differentiate with respect to t. Solve Q'(t)=0 for t and put that value of t back into the expression Q(t)>=0 and see what you get.
 
Yeah I get a similar thing. So we get a turning point of Q at some value t=-psi/phi, and when you put this back into Q you get

\int0 = constant.

Am I being really dumb cos I can't seem to get anything like the inequality from this :((((

Cheers.
 
I meant integrate first. I.e.
<br /> t^2 \int \phi^{2} + 2t \int\phi \psi + \int\psi^{2} \geq 0.<br />

Now minimize that. The minimum occurs at a value of t that is a ratio of two integrals.
 
Ah yeah I got it :-) Thanks!
 
Deano10 said:
Sorry to be a pain, but I am still a little confused!

Just to check, what exactly are we integrating with respect to?

Whatever variable phi and psi are functions of. Call it x. So write psi(x) and phi(x).
 

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