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The expression for Fourier transform is given as $$ F(\omega) = \int\limits_{-\infty}^{+\infty} f(t) \ e^{-j \omega t} \ dt \tag{1}$$

Now, let one function be $f(x)$ and other be $e^{j\omega t}$ then their convolution at $p(0)$ is given as

$$ p(0) = \int\limits_{-\infty}^{+\infty} f(t) \ e^{-j \omega t} \ dt \tag{2}$$ same as in $(1)$.

Also, if we split the exponential term into sinusoidal,the expression $(1)$ becomes

$$ F(\omega) = \int\limits_{-\infty}^{+\infty} (f(t) \cos{\omega t} -j f(t)\sin{\omega t})\ dt \tag{3}$$

Let $f(t)$ and $\cos{\omega t}$ be 2 functions. So their correlation at value $q(0)$ is given as

$$ q(0) = \int\limits_{-\infty}^{+\infty} (f(t) \cos{\omega t}) \ dt \tag{4}$$ same as in $(3)$. (we can do with f(t) and sine function also.)

So I want to know

  1. Whether Fourier transform formula be considered as Convolution or Correlation?

  2. Also, what is meaning of $p(0)$ and $q(0)$ terms ?

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Correlation and convolution are basically the same operations. You can express the cross-correlation of two functions $f(t)$ and $g(t)$ by a convolution:

$$R_{fg}(\tau)=f(\tau)\star g^*(-\tau)$$

where $\star$ denotes convolution, and $*$ denotes complex conjugation.

If you evaluate the cross-correlation at $\tau=0$ you get the inner product of $f(t)$ and $g(t)$, and that's exactly what the Fourier transform is: it is the projection of $f(t)$ on the complex exponential $e^{j\omega t}$.

So, to answer your questions:

  1. It is actually an inner product. Due to the equivalence of convolution and correlation, it can be seen as both, evaluated at $\tau=0$.
  2. As already mentioned, the meaning of these terms is an inner product.
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  • $\begingroup$ Note that by thinking of the Fourier transform as an inner product you can see that what you're really doing is a change of basis. $\endgroup$ Commented Jun 14, 2015 at 16:58
  • $\begingroup$ @DanielSank how?can you explain in some detail? $\endgroup$ Commented Jun 14, 2015 at 17:23
  • $\begingroup$ @Matt L. sir, 1.I didn't get the word "inner " here? $\endgroup$ Commented Jun 14, 2015 at 17:31
  • $\begingroup$ @Matt L. 2.Also, how can we say that convolution and correlation are same operations? $\endgroup$ Commented Jun 14, 2015 at 17:35
  • $\begingroup$ @devraj: It's a mathematical term: inner (or dot) product. I didn't mean to say that convolution and correlation are exactly the same operations, but they are equivalent in the sense that one can be expressed by the other, as shown in my answer. $\endgroup$ Commented Jun 14, 2015 at 19:57

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