Why does an anti-symmetric function has zero amplitude at the center of an even length window I am performing FFT on a real odd function and the resultant transform has zero amplitude in the last bin. Essentially if Y= rfft(X), then Y[-1] is always zero. I stumbled on this answer which says


  An anti-symmetric waveform has to be zero at the center of an even length window.


Now, this might be very easy to prove but I cannot think of why this is true. If I use the formula then to the first term should be zero, not $N/2$ th. It's clear to me why it should be real but why should the amplitude be zero for real-valued, odd function. Any help will be appreciated.
A real symmetric function has the following property $$f(t) = f(-t)$$

whereas a real anti-symmetric function has the following property $$f(t) = -f(-t)$$

So at $t=0$ you have $$f(0) = -f(0)$$ which implies that $$f(0) = 0$$

A complex conjugate-symmetric function has the following property $$f(t) = f^*(-t)$$

whereas a complex conjugate-antisymmetric function has the following property $$f(t) = -f^*(-t)$$

So at $t=0$ you have $$f(0) = -f^*(0)$$ which implies that $f(0)$ is real and further $$f(0) = 0$$