来自贴吧的一个猜想

icesheep

定义 f(x) 的 n 阶对称导数为
\[{f^{\left[ n \right]}}\left( x \right) = \mathop {\lim }\limits_{h \to 0} \sum\limits_{k = 0}^n {C_n^k\frac{{{{\left( { - 1} \right)}^k}f\left[ {x + \left( {n - 2k} \right)h} \right]}}{{{{\left( {2h} \right)}^n}}}} \]

若 f(x) 在[a,b]连续，且在(a,b)上其n阶对称导数处处为零，证明或否定：f(x) 为多项式函数。

icesheep

一些 Schwartz 导数的定理：

(和Fermat引理平行)：f 在 [a,b] 连续，在 (a,b) Schwartz 可导 ，若 f(a) < f(b) 则存在 $\xi $ 使得 ${f^{\left[ 1 \right]}}\left( \xi \right) \geqslant 0$ 。

证明：
取 $f\left( a \right) < \lambda < f\left( b \right)$ ，考虑集合 \[E = \left\{ {x \in \left[ {a,b} \right]|f\left( x \right) > \lambda} \right\}\]
由 f(x) 连续性，知 E 为非空有界开集，记 $\xi = \inf E$ 则有
\[x < \xi  \Rightarrow x \notin E \Rightarrow f\left( x \right) \leqslant \lambda \]
\[\forall n,\exists {x_n} \in E \Rightarrow \xi < {x_n} < \xi + \frac{1}{n}\]
于是\[{f^{\left[ 1 \right]}}\left( \xi \right) = \mathop {\lim }\limits_{h \to 0 + } \frac{{f\left( {\xi + h} \right) - f\left( {\xi - h} \right)}}{{2h}} = \mathop {\lim }\limits_{n \to \infty } \frac{{f\left( {{x_n}} \right) - f\left( {\xi - \left( {{x_n} - \xi } \right)} \right)}}{{2\left( {{x_n} - \xi } \right)}} \geqslant 0\]

(和Rolle定理平行)：f 在 [a,b] 连续，在 (a,b) Schwartz 可导且 f(a)=f(b)=0 则存在  $\xi $ , $\eta $ 使得 ${f^{\left[ 1 \right]}}\left( \xi \right) \geqslant 0$ 且 ${f^{\left[ 1 \right]}}\left( \eta \right) \leqslant 0$ 。

(和Lagrange中值定理平行)：f 在 [a,b] 连续，在 (a,b) Schwartz 可导 则存在  $\xi $ , $\eta $ 使得${f^{\left[ 1 \right]}}\left( \eta \right) \leqslant \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} \leqslant {f^{\left[ 1 \right]}}\left( \xi \right)$

icesheep

n=1 时 ，利用上面的广义Lagrange定理，即可证明其为常值函数。
下面考虑 n=2 的情况：
构造
\[F\left( x \right) = f\left( x \right) - l\left( x \right) + \varepsilon \left( {x - a} \right)\left( {x - b} \right)\]
其中 $l\left( x \right)$ 为过 $\left( {a,f\left( a \right)} \right),\left( {b,f\left( b \right)} \right)$ 的直线段，则有${F^{\left[ 2 \right]}}\left( x \right) = 2\varepsilon$

当 $\varepsilon > 0$ 时，考察 F(x) 的最大值点，知其只能在边界取到，否则设其为 ${x_0}$ 则有\[2F\left( {{x_0}} \right) \geqslant F\left( {{x_0} + h} \right) + F\left( {{x_0} - h} \right)\]
与 ${F^{\left[ 2 \right]}}\left( x \right) = 2\varepsilon > 0$ 矛盾，故 F(x) 的最大值为 0，也即
\[f\left( x \right) - l\left( x \right) + \varepsilon \left( {x - a} \right)\left( {x - b} \right) \geqslant 0\]
\[f\left( x \right) - l\left( x \right) \leqslant \varepsilon \left( {x - a} \right)\left( {b - x} \right)\]
令 $\varepsilon  \to {0^ + }$ 可得 $f\left( x \right) \leqslant l\left( x \right)$ ，类似地考虑 $\varepsilon  < 0$ 时最小值点，可得 $f\left( x \right) \geqslant l\left( x \right)$
故 n=2 时， f(x) 为线性函数。

hbghlyj

Constructing the Grünwald–Letnikov derivative

The formula
\[ f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}} \]
for the derivative can be applied recursively to get higher-order derivatives. For example, the second-order derivative would be:
\[ {\begin{aligned}f''(x)&=\lim _{h\to 0}{\frac {f'(x+h)-f'(x)}{h}}\\&=\lim _{h_{1}\to 0}{\frac {\lim \limits _{h_{2}\to 0}{\dfrac {f(x+h_{1}+h_{2})-f(x+h_{1})}{h_{2}}}-\lim \limits _{h_{2}\to 0}{\dfrac {f(x+h_{2})-f(x)}{h_{2}}}}{h_{1}}}\end{aligned}} \]
Assuming that the $h$'s converge synchronously, this simplifies to:
\[ =\lim _{h\to 0}{\frac {f(x+2h)-2f(x+h)+f(x)}{h^{2}}} \]
which can be justified rigorously by the mean value theorem. In general, we have (see binomial coefficient):
\[ f^{(n)}(x)=\lim _{h\to 0}{\frac {\sum \limits _{0\leq m\leq n}(-1)^{m}{n \choose m}f(x+(n-m)h)}{h^{n}}} \]
Removing the restriction that $n$ be a positive integer, it is reasonable to define:
\[ \mathbb {D} ^{q}f(x)=\lim _{h\to 0}{\frac {1}{h^{q}}}\sum _{0\leq m<\infty }(-1)^{m}{q \choose m}f(x+(q-m)h). \]
This defines the Grünwald–Letnikov derivative.

To simplify notation, we set:
\[ \Delta _{h}^{q}f(x)=\sum _{0\leq m<\infty }(-1)^{m}{q \choose m}f(x+(q-m)h). \]
So the Grünwald–Letnikov derivative may be succinctly written as:
\[ \mathbb {D} ^{q}f(x)=\lim _{h\to 0}{\frac {\Delta _{h}^{q}f(x)}{h^{q}}}. \]