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Here $n$ is a positive integer and $m$ is an integer index of a Fourier coefficient. Split a period into $n$ intervals and set $y=nx-2\pi j$ on the $j$th interval. Periodicity and integrability giveFor $0<|m|<n$, $\zeta =e^{2\pi im/n}$ is a root of unity different from one and satisfies ${\zeta }^n=1$, so ${\sum }_{j=0}^{n-1}{\zeta }^j=(1-{\zeta }^n)/(1-\zeta )=0$. ThereforeFor $m=0$ the same substitution gives ${\int }_{\mathbb{T}}f(nx)dx={\int }_{\mathbb{T}}f(x)dx$. If the mean of $f$ is zero, this last mode vanishes as well. Since the linear span of the exponentials with $|m|\le n-1$ is the space of trigonometric polynomials of degree at most $n-1$, every pairing with that space is zero. This is the periodic dilation annihilates low Fourier modes property. The integer-index assumption is essential; the stated Fourier orthogonality would generally be false for noninteger $m$.