Lower bound Sum of Complex Exponentials using derivatives

Question

I wish to either prove or disprove the following conjecture.
Conjecture. Let $f(x)$ be a trigonometric polynomial. Specifically, $$ f(x) = \sum_{j=1}^{J}\alpha_j e^{2\pi i k_j x}, $$ where $k_j \in \mathbf{R}$ for all $j$. Additionally, suppose that $f(x_0) = 0$ for some $x_0 \in [0,1]$. Then there exists positive constants $c', d, \delta$ such that $$ |f(x)| \geq c'\,|x-x_0|^d $$ for all $x\in [x_0-\delta,x_0+\delta]$.

Here are my current ideas.

Ideas. We know that $f(x)$ will equal its Taylor series. Hence, let us consider it's Taylor series. First observe the following lemma.

Lemma. Let $f(x)$ be a trigonometric polynomial defined as above. Additionally, suppose that $f(x_0) = 0$ for some $x\in [0,1]$. Then there exists some $n \leq J-1$ such that $f^{(n)}(x_0) \ne 0$.

Proof Idea. It suffices to consider the system such that $f^{(n)}(x_0) = 0$ for $n=0,\ldots, J$. One will obtain a Vandermonde-like system which has linearly independent columns (when considering the augmented matrix). Hence, for all those derivatives to be zero would require $f \equiv 0$, which is a contradiction.

Potential Proof of the Conjecture. Let us assume that $d \leq J-1$ is the smallest positive integer such that $f^{(d)}(x_0) \ne 0.$ Therefore, we have that

$$ f(x) = \frac{f^{(d)}(x_0)}{d!}(x-x_0)^d + R_{d+1}(x). $$ Recall the Lagrange form of the Remainder. $$ R_{d+1}(x) = \frac{f^{(d+1)}(c)}{d!}(x-x_0)^{d+1} $$ for some $c$ between $x$ and $x_0$. Therefore, we have that $$ \begin{align*} f(x) &= \frac{f^{d}(x_0)}{d!}(x-x_0)^d + \frac{f^{d+1}(c)}{(d+1)!}(x-x_0)^{d+1} \\ &= (x-x_0)^d \left(\frac{f^{d}(x_0)}{d!} + \frac{f^{d+1}(c)}{(d+1)!}(x-x_0)\right). \end{align*} $$ Therefore, it suffices to bound $$ \left|\frac{f^{d}(x_0)}{d!} + \frac{f^{d+1}(c)}{(d+1)!}(x-x_0)\right| $$ from below. Note that the aforementioned function only has a zero when $x = x_0 - \frac{d f^{d}(x_0)}{f^{d+1}(c)}$. Also note that $\frac{d f^{d}(x_0)}{f^{d+1}(c)}$ may not even be real-valued. However, if we let $\delta < \left|\frac{d f^{d}(x_0)}{f^{d+1}(c)}\right|$, we can guarantee that $$ \left|\frac{f^{d}(x_0)}{d!} + \frac{f^{d+1}(c)}{(d+1)!}(x-x_0)\right| $$ obtains a nonzero minimum for $x\in [x_0-\delta,x_0+\delta]$. Let $c' = \min \left|\frac{f^{d}(x_0)}{d!} + \frac{f^{d+1}(c)}{(d+1)!}(x-x_0)\right|$ for $x\in [x_0-\delta,x_0+\delta]$. Hence,

$$ |f(x)| \geq c' |x-x_0|^d $$ for $x \in [x_0-\delta,x_0+\delta]$.

Questions. Does this proof work? Or is there a glaring mistake somewhere? Is this already known? In my cursory literature search I couldn't quite find this exact result, but I didn't look very hard. Any advise would be appreciated.

Answer 1

It follows immediately from your lemma, since Taylor's theorem implies that the vanishing of the function value and the first $n$ derivatives at $x_0$ is equivalent to $\lim\limits_{x \to x_0} \frac{f(x)}{(x-x_0)^n} = 0$

Answer 2

The problem I see is that $\delta$ is not well defined in terms of $f$ and $x_0$, because $c$ depends on $x$. But you can modify this because you know that $f^{d+1}$ is continuous, so you can substitute $|f^{d+1}(c)|$ by the supremum of $|f^{d+1}|$ over a small interval around $x_0$, thus, $$ \delta <\min{1,\frac{|f^{d}(x_0)|}{\sup_{s\in I} |f^{d+1}(s)|}}, $$ where $I=(x_0-1,x_0+1)$ (in general, you can substitute $1$ by $\varepsilon$ depending on the interval on which the function is $C^{d+1}$).

The rest looks fine. The function is of one real variable, so I think Lagrange’s form of the remainder still works.