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This is probably a very obvious, but I am slightly confused.

Suppose $f(\xi)$ is such that when $f^2(\xi)$ is integrated from $\xi=-\infty $ to $\xi=+\infty$ equals to $1$ and correspondingly for $g(\xi)$.

Now $h(x,y)={1\over \sqrt 2}(f(x)g(y)+f(y)g(x))$ squared and integrated over $x,y$ would be $${1\over 2}\int\int f^2(x)g^2(y)+f^2(y)g^2(x)+2f(x)f(y)g(x)g(y)\,\,\,dxdy$$ and this is supposed to equal $1$ also. However, suppose $f=g$ then $h(x,y)=\sqrt 2f(x)f(y)$ giving an extra factor of $\sqrt 2$??

Thanks.

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You need a condition connecting $f$ and $g$ for this to hold. What I think you want is orthogonality $\int_{-\infty}^\infty f(\xi)g(\xi)\, d\xi = 0$. Exactly then the $2f(x)f(y)g(x)g(y)$ term above integrates to $0$, leaving you with $\frac 12(1+1)$. So orthogonality is necessary! –  martini Nov 7 '12 at 14:44
    
Thank you, @martini , but what about the factor $\sqrt 2$? –  Henry Nov 7 '12 at 14:52
    
If $f = g$ then $f$ and $g$ are orthogonal exactly iff $f = 0$. –  martini Nov 7 '12 at 14:57
    
Ah, of course, thank you, @martini –  Henry Nov 7 '12 at 15:20

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