# Can we control the minimum of a continuous signal $x$ when some Fourier coefficients are constant?

Let us fix a sequence of real numbers $$\{a_k\}_{k=-n}^n$$ and $$\gamma\in \mathbb{R}$$. Is there any $$2\pi$$-periodic continuous signal $$x :\mathbb{R}\to \mathbb{R}$$ such that the following points simultaneously hold?

1. $$\gamma \leq x_{\min}$$ (where $$x_{\min}$$ is the minimum of $$x$$).
2. If $$|k|\leq n$$, the Fourier coefficients $$\mathcal{F}(x)[k]=a_k$$.

If YES, how can we find the closed form of such a function in terms of $$\{a_k\}_{k=-n}^n$$ and $$\gamma\in \mathbb{R}$$?

• What do you know (or postulate) for $\mathcal{F}(x)[k]$ for $|k| > n$ ?. If that's zero $X[k]$ uniquely determines $x(t)$. If this is arbitrary, than you can't make any statement about $x_{min}$ Jan 6, 2023 at 13:47

Let's say $$n=1$$ and let's choose $$\{a_k\} = \{ -1, 2, 1 \}$$ with $$\gamma = \pi$$.
Then using the exponential form of the Fourier series: $$x(t) = \sum_{k=-N}^{+N} a_k e^{\jmath n t}$$ we get for our example coefficients $$x(t) = -e^{-\jmath t} + 2 + e^{\jmath t} = 2 + 2\jmath \sin(t)$$ making the assumption that Hilmar implies in the comments that $$a_k = 0$$ for $$|k| \gt n$$.
Now $$\min_t \Re\{x(t)\} = 2 \\ \min_t \Im\{x(t)\} = -2 \\$$ so your requirement can't hold.
If we relax the assumption that $$a_k = 0$$ for $$|k| \gt n$$, that is we are free to choose the $$a_k$$ for $$|k| \gt n$$ to be some arbitrary values, then I still don't think it holds for this specific example because we'd need to shift the whole signal up by $$\pi$$ over its entire duration. The only way to do that is to change the $$a_0$$ coefficient.