# An upper bound of a function using its Taylor series.

Let $$f \in C^{m}(\mathbb{R}^n)$$ of compact support. Then Taylor's theorem give us $$f(x) = \sum_{|\alpha| \leq m} \frac{D^{\alpha} f(x_0)}{\alpha!} (x - x_0)^{\alpha} + \sum_{|\alpha| = m} h_{\alpha}(x) (x - x_0)^{\alpha}$$ using the notation in [https://en.wikipedia.org/wiki/Taylor%27s_theorem#Taylor's_theorem_for_multivariate_functions].

Suppose $$D^{\alpha} f(x_0) = 0$$ for all $$|\alpha| \leq m$$. Take some $$|\beta| < m$$. I want to prove that $$|D^{\beta} f(x)| < |x - x_0|^{m - |\beta|} C \sum_{|\alpha| \leq m} | D^{\alpha} f |_{\infty}$$ for some $$C > 0$$ depending only on $$m$$ and $$n$$.

I am sure this is true but I am having difficulty dealing with the $$h_\alpha$$ function coming from the remainder term of the Taylor expansion.... how can I prove this? Thank you

As you say, it is all about understanding the $$h_\alpha$$ contained in the remainder. The form you wrote is called the Peano form, and I think it is best to use the more explicit Lagrange form which would imply the Peano form. We need the multivariable version.
Let $$f$$ be $$C^m$$ in a ball $$B$$ around $$x_0\in {\mathbb R}^n$$. Then the Lagrange form says $$f(x) = \sum_{|\alpha| where $$\xi$$ lies on the line segment connecting $$x_0$$ to $$x$$. This would imply the Peano form $$f(x) = \sum_{|\alpha|=m} \frac{D^\alpha f(x_0)}{\alpha!} (x-x_0)^\alpha + \sum_{|\alpha|=m} h_\alpha(x) (x-x_0)^\alpha, \quad x\in B,$$ if we let $$h_\alpha(x) = \frac{D^\alpha f(\xi)-D^\alpha f(x_0)}{\alpha!},$$ since $$h_\alpha(x)\to 0$$ as $$x\to x_0$$ by $$f\in C^m$$.
Applying this to $$D^\beta f$$ for a fixed $$\beta$$ with $$|\beta|. Under your condition that $$D^\alpha f(x_0)=0$$ for $$|\alpha|\leq m$$, we have \begin{align*} D^\beta f(x) &= \sum_{|\alpha| Therefore if $$f\in C^m({\mathbb R})$$ has compact support, we have $$|D^\beta f(x)|\leq |x-x_0|^{m-|\beta|} \sum_{|\alpha|=m}\|{D^{\alpha} f(\xi)}\|_\infty,$$ since $$\frac{1}{\alpha!}\leq 1$$. Note that we can take $$C=1$$ and the sum to be only over $$\alpha$$ with $$|\alpha|=m$$.