I was working on Fourier transform when I see this expression :
Can someone explain :
-- What means the bar under the function f ? Does a complex conjugate make sens here ?
-- What can be the uses of this expression ? Is it usefull for something about Fourier transform ?
Thank !
On
Yes, it means complex conjugate.
If you put $g(x)=f(x)$ then you obtain the nice result that if a function is in $L^{2}\bigcap L^{1}$ then its Fourier transform is in $L^{2}$.
Moreover, the theoerem implies that the Fourier transform is a unitary operator in $L^{2}$ which implies it is an isometry.
The Plancherel formula shows that the inner product between two functions can be taken in either the time domain - or the frequency domain. This is an amazing fact - and can be exploited to develop inversion formulas for integral equations (such as the Wavelet Transform) and fast cross correlations. When $f(t) = g(t)$ you get the energy in $f(t)$ - Energy is always a positive quantity, although the function (or vector) may be complex (this is why we use the complex conjugate: for a complex number $z=R\exp(i\theta)$, we compute its magnitude via $\bar{z}z = R^2 \ge 0$). Similarly, $\int_{-\infty} ^\infty \bar{f(t)}f(t)dt = \|f(t)\|^2 \ge 0$ gives us the energy (square of the norm) in $f(t)$. This is also true for dot products of vectors in $\mathbb{C^n}$. Note that in many books the conjugate is denoted with an asterisk: $<f,g> = \int_{-\infty}^\infty f(t)^* g(t)dt = \int_{-\infty}^\infty \hat{f(\xi)}^* \hat{g(\xi)} d\xi\ge 0$.