Let $f$ be a univariate polynomials of degree $d$. Assume that:
- $f(x) \ge 0$ for all $x \in \mathbb{R}$
- $f(0) = \varepsilon >0$
- $f'(0) = 0$
My question is the following: can we deduce lower bounds for $f''(0)$ depending on $d$ and $\varepsilon$?
This seems reasonable for me, since id the second derivative is too negative, then the first derivative becomes quickly negative and thus $f$ decreases in a neighbourhood of $0$. But then this may contradict the non negativity of $f$.
I tried to exploit the Taylor expansion of $f$, but got no result. I believe that a possible solution would be to use twice the Markov brothers' inequality, that can be stated as follows:
Theorem (Markov brothers' inequality). Let $f$ be a polynomial of degree $d$. then: $$\max_{x \in [-t,t]}|f'(x)| \le \frac{2d^2}{2t} \max_{x \in [-t,t]}|f(x)|$$
Edit. After the answer of durianice, I restate my question, taking into account also the norm of $f$ (for instance, the max norm of $f$ on [-1,1]): can we deduce lower bounds for $f''(0)$ depending on $d$, $\varepsilon$ and $||f||$?
Clearly, Markov brothers' inequality already gives a bound, but this bound doesn't take into account the special properties of $f$ that I listed above.
This is answering the question before the edit, which did not require the bound to depend on $\lVert f \rVert$.
Consider $f(x)=ax^4-bx^2+\varepsilon$, where $b>0$ and $a>\frac{b^2}{4\varepsilon}$. Clearly $f(0)>0$ and $f'(0)=0$. Also, since the quadratic discriminant $b^2-4a\varepsilon$ is negative, $f$ has no real root, implying that it is positive on $\mathbb{R}$. Note that $f''(0)=-2b$, which depends only on some arbitrary $b$. It then follows that $|f''(0)|$ can be arbitrarily large, and not bounded by any function depending on $d$ or $\varepsilon$.