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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

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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 $$ f(x) = \sum_{|\alpha|<m} \frac{D^\alpha f(x_0)}{\alpha!} (x-x_0)^\alpha + \sum_{|\alpha|=m}\frac{D^\alpha f(\xi)}{\alpha!} (x-x_0)^\alpha, \quad x\in B, $$ 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 $|\beta|<m$. Under your condition that $D^\alpha f(x_0)=0$ for $|\alpha|\leq m$, we have \begin{align*} D^\beta f(x) &= \sum_{|\alpha|<m-|\beta|} \frac{D^\alpha D^\beta f(x_0)}{\alpha!} (x-x_0)^\alpha + \sum_{|\alpha|=m-|\beta|}\frac{D^\alpha D^\beta f(\xi)}{\alpha!} (x-x_0)^\alpha\\ & = \sum_{|\alpha|=m-|\beta|}\frac{D^{\alpha+\beta} f(\xi)}{\alpha!} (x-x_0)^\alpha, \quad x\in B. \end{align*} 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 $|\alpha|=m$.

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