This question about the intuition behind the scaling property of the Fourier transform made me wonder about the corresponding notion for a Fourier series.
The Fourier transform of $f(ax)$ is $\frac{1}{|a|} \mathcal{F(\frac{u}{a})}$. If $a>1$ then the graph of $f(ax)$ is $f$ compressed and so its Fourier transform has frequencies that are higher. However they are scaled down in magnitude.
On the other hand take $f(x)= \cos x$. Its fourier series is trivially itself, with the coefficient of $\cos x$ being $1$. Scaling it to $f(ax) = \cos ax$ where $a>1$ is an integer still has a trivial fourier series, and the coefficient of $\cos ax$ is $1$. This is unlike the Fourier transform where the "coefficients" of each frequency get scaled as the formula shows: $\frac{1}{|a|} \mathcal{F(\frac{u}{a})}$.
Is there a conceptual way to explain this discrepancy?
Best Answer
Let $a\neq 0$. For a function $f$ define $m_af$ by $(m_af)(x) = f(ax)$. Then the scale factor that appears in a Fourier transform of $m_af$ is directly related to the norm of $m_a$ since $$||m_af||_2 = ||m_a||\cdot ||f||_2$$ and a Fourier transform on $\mathbb{R}$ or $S^1$ is (up to a normalization factor) unitary. Now on $\mathbb{R}$ we have $||m_a|| = |a|^{-\frac{1}{2}}$ while on $S^1$ (and $a \in \mathbb{N_{>0}}$) $||m_a||=1$.
As an amusing aside, if you take the Fourier transform of $\cos$ as a tempered distribution then $$\mathcal{F}(m_a\cos) = \tfrac{1}{2|a|}(m_{a^{-1}}\delta_1 + m_{a^{-1}}\delta_{-1})= \tfrac{1}{2|a|}(|a|\delta_a + |a|\delta_{-a}) = \tfrac{1}{2}(\delta_a + \delta_{-a})$$ by the transformation property of the Dirac distribution as noted in a comment. So in this case there is a factor $|a|^{-1}$ as usual although it is not directly visible.