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 $\gamma\in \mathbb{R}$?

  • 2
    $\begingroup$ 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 $x_{min}$ $\endgroup$
    – Hilmar
    Jan 6, 2023 at 13:47

1 Answer 1


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.

So I don't think it's possible to do what you want.


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