$f$ is smooth and periodic function,$\exists \lambda$ such that $f^{(4)}=\lambda f$ prove:$\exists \lambda$ such that $\lambda= (\frac{2\pi n}{T})^4$

Question

Given $f:\mathbb{R} \rightarrow \mathbb{C}$ is smooth and periodic with period $T>0$ and exists $\lambda \in \mathbb{C}$ such that $f^{(4)}(x)=\lambda f(x)$ for any $x \in \mathbb{R}$.

prove: exists $\lambda$ such that $\lambda= (\frac{2\pi n}{T})^4$

I think i have to use the fact that $\hat{f^{(4)}} = n^4 \hat{f}$ but all i was able get from it is: exists $\lambda$ such that $\lambda = n^4$

I can see that if we choose to $T = 2\pi$ we actually need to prove that $\lambda = n^4$.

But how do i prove for any $T>0$ ?

and one more question: why $\mathbb{C}$, and where is complex numbers coming into play when solving this question ?

(smooth is infinitely differentiable )

Answer 1

Hint: Let the Fourier expansion of $f$ be $f(x)=\sum_{n=0}^\infty a_n e^{i\frac{2n\pi}{T}x}$ and you will get the answer.