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I want to show the following inequality is true: $|\int_{a}^{b}fg|^2\leq \int_{a}^{b}f^2\int_{a}^{b}g^2$.

My first thought was to use the tagged partition definition of a Riemann integral combined with the Schwarz inequality.

So let $P=\{a=x_0<x_1<...<x_n=b\}$ and $S_p=\{t_1,...,t_n\}$ where $t_i \in [x_{i-1}, x_i]$.

By the Schwarz inequality we have:

$$\left|\sum\limits_{i=1}^{n}f(t_i)g(t_i)\Delta x_i \right|^2\leq\sum\limits_{i=1}^{n}f(t_i)^2\Delta x_i\sum\limits_{i=1}^{n}g(t_i)^2\Delta x_i.$$

Essentially I was going to follow the proof for the Cauchy-Schwarz inequality but by making proper substitutions.

I have two questions: is this the best way of proving this, and is this even a correct start.

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Need absolute values around the integrals on the right hand side. And the absolute value in your integral of a product is unbalanced. –  vonbrand Feb 8 '13 at 2:11
    
In the statement of the problem there are no absolute values on the right side. –  abet Feb 8 '13 at 2:12
    
google "cauchy schwarz inequality" or en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality –  user61426 Feb 8 '13 at 19:49

2 Answers 2

up vote 4 down vote accepted

You seem to know the "sum" version of the Cauchy-Schwarz inequality. The standard proof of the "integral" version is exactly the same as that for the "sum" version: just change sums to integrals. It's really a theorem about inner products.

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This is the $p=q=2$ case of Holder's inequality, which has a very simple proof.

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1  
No need to generalize: it is the Cauchy-Schwarz inequality (integral version). –  Robert Israel Feb 8 '13 at 2:24

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