A question on finite Fourier series

Question

Let $\mathcal F(N)$ denote the space of finite Fourier series up to frequency $N > 0$, i.e. $f\in \mathcal F(N)$ if and only if it can be written as

$$f(x) = \sum_{k=0}^N a_k\cos(kx+\theta_k)$$

for some $a_k,\theta_k\in\mathbb R$. I will call $f\in\mathcal F(N)$ full if $a_k\neq 0$ for each $k\in\{0,\dots,N\}$.

Does the following statement hold?

"There exist $N > 0$ and a sequence $f_n\in\mathcal F(N)$ such that $f_n\to f$ where $\mathcal f\in \mathcal F(N)$ is full, and there exists a global maximizer $x_* \in \arg\max_{x\in\mathbb R}f(x)$ such that $x_*$ is the limit of two distinct sequences of global maximizers $x_n\neq y_n\in \arg\max_{x\in\mathbb R}f_n(x)$."

I believe this is possible when the limit $f$ is not full, but the fullness assumption seems strong. Are finite Fourier series too nice for the above statement to hold?

Answer 1

No problem, this is doable already for $N=2$. The idea is as follows: at the point of the maximum first derivative must vanish, and yet we want to separate the maximum into two by adding something small ($f_n-f$). In the space $\mathcal{F}(N)$ all the norms are the same, so if the second derivative of $f$ is non-zero then a small perturbation will not change the local behaviour much. So, the trick is to make $f''$ to be zero at that point, and then make the small change $f_n-f$ have positive second derivative at the point to make what you want.

Now, first we need a full function $f$ with second derivative being zero (and actually third as well since it must be at least a local maximum). Well, it is easy to find it by hand: $f(x) = 1 + 4\cos(x) - \cos(2x)$ (extra constant $1$ only to make it full by your definition). Then $\max(f) = f(0) = 4$. On the other hand, for the sequence $f_n(x) = 1 + (4-\frac{1}{n})\cos(x) - \cos(2x)$ you can show that it attains maximum at two points slightly to the left and slightly to the right of $0$, just as you wanted. And of course $f_n\to f$ under any reasonable definition of convergence you like.