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.