Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\sqrt 2$??


share|cite|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.