If $\forall n \in \mathbb Z_{\ge0} \ $ and $\forall x \in \mathbb R$, we know that $\big|f^{(n)}(x)\big|\le \big|p(x)\big|$, then $f=0$.

Question

If $p(x)$ is an odd degree polynomial such as $\forall n \in \mathbb Z_{\geq 0}$ and $\forall x \in \mathbb R$ we know that $$\big|f^{(n)}(x)\big|\le \big|p(x)\big|\,.$$ I need to show that $\forall x \in \mathbb R \ $ $f(x)=0$.

My thoughts till now: I tried to use Taylor polynomial but it didn't help. and I really need help.

Thanks in advance.

Answer 1

Let $M_R=\max_{|x| \le R}|p(x)|$. We have $\frac{|f^{(n)}(x)|}{n!} \le \frac{M_R}{n!}, |x| \le R$.

This immediately implies that $f$ is analytic on $(-R,R)$ and its Taylor series there at $0$ has a radius of convergence at least $R$.

But $f^{(n)}(0)=0$ since $p$ odd hence $f$ is identically zero on $(-R,R)$. As $R>0$ arbitrary we are done!

(If we are given that $f$ has odd degree only we apply the above with any real zero and easy modifications by changing the center of the Taylor series at the zero of $p$.)

Answer 2

Since $p$ is odd we have $p(0)=0$, hence $f^{(n)}(0)=0$ for all $n\ge 0$.

Fix $x\in\mathbb{R}$.

We want to show $f(x)=0$.

Let $b$ be an upper bound for $|p(t)|$ on the interval $-|x|\le t\le |x|$.

Applying Taylor's formula we have $$ f(x) = R_k(x) $$ for all nonnegative integers $k$, hence to show $f(x)=0$, it suffices to show $$ \lim_{k\to\infty}R_k(x)=0 $$ Using the integral formula for the remainder, we get \begin{align*} |R_k(x)| &= \left|\,\int_0^x \frac{f^{(k+1)}(t)}{k!}(x-t)^k\,dt\,\right|\\[4pt] &= \frac{1}{k!}\,\left|\,\int_0^x f^{(k+1)}(t)(x-t)^k\,dt\,\right|\\[4pt] &\le \frac{1}{k!}\,\left|\,\int_0^x \left(\left|f^{(k+1)}(t)\right|\right)\left(\left|(x-t)^k\right|\right)\,dt\,\right|\\[4pt] &\le \frac{1}{k!}\,\left|\,\int_0^x (|p(t)|)(|x|^k)\,dt\,\right|\\[4pt] &\le b{\,\cdot}\frac{|x|^k}{k!}\,\left|\,\int_0^x 1\,dt\,\right|\\[4pt] &= b{\,\cdot}|x|{\,\cdot}\frac{|x|^k}{k!} \end{align*} which approaches $0$ as $k$ approaches infinity since $b{\,\cdot}|x|$ is constant and $$ \qquad\qquad\, \lim_{k\to\infty}\frac{|x|^k}{k!}=0\;\;\;\left[\,\text{since $k!\ge \Bigl(\frac{k}{3}\Bigr)^k$}\,\right] $$ Thus we have $$ \lim_{k\to\infty}R_k(x)=0 $$ hence $f=0$.

Answer 3

I imagine that $f$ is analytic in $\mathbb{R}$...

Let $x_0\in\mathbb{R}$ such that $p(x_0)=0$ (I imagine that $p(x)$ odd means that its higher degree term is odd). Since $f$ is analytic, it can be expressed as

$$f(x)=\sum\limits_{n=0}^\infty (x-x_0)^n\frac{f^{(n)}(x_0)}{n!},\quad \forall x\in\mathbb{R}$$

Now, as $|f(x_0)|\leq |p(x_0)|=0$, we have that the first coefficient in the series of Taylor is zero. Then, we express the Taylor series of the derivative

$$f'(x)=\sum\limits_{n=0}^\infty (x-x_0)^n\frac{f^{(n+1)}(x_0)}{n!},\quad \forall x\in\mathbb{R},$$

Again, as $|f'(x_0)|\leq |p(x_0)|=0$, then the second coefficient of the Taylor's serie of $f(x)$ is also zero.

Following the idea, we can show that there is not a coefficient in the Taylor series of $f$ different from zero (here we must be careful in justifying it with the infinity). So, $f$ must be zero.

PD: I just use your idea of using Taylor's series.