Local Taylor Expansion at each point

Question

for $f\in \mathcal{C}^{\beta}$, denote by $F$ the local Taylor expansion of $f$ of order $\beta$ at each point

Does this mean that, for some fixed point $x$, we have $$F(a) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2}+...+\frac{f^{(\beta)}(a)(x-a)^{\beta}}{\beta !}?$$ Because in that case, doesn't the definition of $F$ depend on the point $x$ which is chosen? Can anyone tell, from the context, what the author means?

See http://www.hairer.org/notes/Regularity.pdf, page 12 (last two lines) for context

Answer 1

It turns out that, for $f\in\mathcal{C}^{\beta}$, $F(x)$ is defined to be $$\sum\limits_{i=0}^{i=\lfloor \beta\rfloor}\frac{f^{(i)}(x)}{i!}X^i.$$

This way, as $f\in\mathcal{C}^{\beta}$, we do indeed have $$\|F(x)-\Gamma_{xy}F(y)\|_{\lfloor \beta\rfloor}=|\frac{f(x)-f(y)}{(\lfloor \beta \rfloor)!}|\lesssim |x-y|^{\beta-\lfloor\beta\rfloor}.$$

For lower orders, \begin{align*}\|F(\mathbf{x})-\Gamma_{\mathbf{xy}}F(\mathbf{y})\|_{\alpha}&=\left|\sum_{|\mathbf{i}|=\alpha}\frac{f^{(\mathbf{i})}(\mathbf{x})}{\mathbf{i}!}-\sum\limits_{|\mathbf{i}|=\alpha}^{\lfloor\beta\rfloor}\frac{f^{(\mathbf{i})}(\mathbf{y})}{\mathbf{i}!}\sum_{|l|=0}^{|i|}\binom{i}{l}({x-y})^{{i-l}}\right|\\ &=\left|\sum_{|\mathbf{i}|=\alpha}\frac{f^{(\mathbf{i})}(\mathbf{x})}{\mathbf{i}!}-\sum\limits_{|\mathbf{i}|=\alpha}^{\lfloor\beta\rfloor}\frac{1}{\mathbf{i}!}\sum_{|j|=0}^{\lfloor\beta\rfloor-|i|}\frac{f^{(\mathbf{i+j})}(\mathbf{y})}{j!}({x-y})^{{i-l}}\right|\\ &=\left|\sum_{|\mathbf{i}|=\alpha}\frac{1}{\mathbf{i}!}\left(f^{(\mathbf{i})}(\mathbf{x})-\sum\limits_{|\mathbf{i}|=\alpha}^{\lfloor\beta\rfloor}\frac{1}{\mathbf{i}!}\sum_{|j|=0}^{\lfloor\beta\rfloor-|i|}\frac{f^{(\mathbf{i+j})}(\mathbf{y})}{j!}({x-y})^{{i-l}}\right)\right|\\ &\lesssim\left|\mathbf{x-y}\right|^{\lfloor\beta\rfloor-\alpha+1}\\ &\lesssim\left|\mathbf{x-y}\right|^{\beta-\alpha},\end{align*}

Answer 2

The full quote is

for $f \in \mathcal C^\beta$, denote by $F$ the element in $\mathcal D^\beta$ given by the local Taylor expansion of $f$ of order $\beta$ at each point.

This is important because $\mathcal D^\beta$ is actually a much bigger set than you think: an element $F \in \mathcal D^\beta$ assigns to each $x \in \mathbf{R}^d$ a polynomial $F_x$ of degree less than $\beta$. In particular, the set of all local Taylor expansions of $f$ of order $\beta$, at all points of $\mathbf{R}^d$, is an element of $\mathcal D^\beta$.

The full definition of $\mathcal D^\beta$ is Definition 2.9 on p.6. It also includes a condition that says that an element $F \in \mathcal D^\beta$ can't just assign arbitrary unrelated polynomials to each point $x \in \mathbf R^d$: the polynomials $F_{x_0}(x - x_0)$ and $F_{x_1}(x - x_1)$ have to be approximately consistent with each other.

This is exactly what a Taylor series does. For example, if $f(x) = \sin x$ and we take $\beta=6$, then $F$ will assign to the point $0$ the polynomial $$F_0(t) = t - \frac{t^3}{6} + \frac{t^5}{120}$$ and to the point $\pi$ the polynomial $$F_\pi(t) = -t + \frac{t^3}{6} -\frac{t^5}{120}.$$ These are not exactly equal: $F_\pi(x - \pi)$ is a complicated polynomial along the lines of $$\left(\pi - \frac{\pi^3}{6} + \frac{\pi^5}{120}\right) - x + \cdots + \frac{\pi x^4}{24} - \frac{x^5}{120}$$ which is not the same polynomial as $F_0(x-0)$. However, they're approximately consistent: because both of them are approximations to $f(x)$, as $\beta$ increases, the coefficients will get closer to each other.