       Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)

• To: mathgroup at smc.vnet.net
• Subject: [mg67399] Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)
• From: dh <dh at metrohm.ch>
• Date: Thu, 22 Jun 2006 06:21:10 -0400 (EDT)
• References: <e7589k\$l5d\$1@smc.vnet.net> <e7846p\$fmp\$1@smc.vnet.net> <e7aoeo\$8t5\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Please not that there is a typo in the foregoing msg.: Plus + should

dh wrote:
> Hi Ronnen,
> after you changed the question to (square in the denominator):
> r = (b-a) Integral[ f(x)^2 dx, {x, a, b} ] /
>                 Integral[ f(x) dx, {x, a, b} ]^2
> your guess is certanly right. The proof is done using Cauch-Schwarz
> inequality. If S is a (real) scalar product, then
> S[f1,f2]^2 <= S[f1,f1]+S[f2,f2]
> Now consider the scalar product: Integrate[f1[x] f2[x],{x,a,b}]:
>
> Integrate[1 f[x],{x,a,b}]^2 = S[1,f]^2 <= S[1,1] + S[f,f] = (b-a) +
> Integrate[f[x]^2,{x,a,b}]
>
> Daniel
>

```

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