Let $f:R\rightarrow R$. If there exists the finite limit $$\lim\_{(x,y) \rightarrow a \atop x\neq y} \frac{f((y)-f(x)}{y-x}$$ then obviously there is a finite derivative $f'(a)$ and is equal this limit.


What about similar problem for higher order divided differences?


May is it true that existence of finite $$\lim\_{(x\_0,...,x\_n)\rightarrow (a,...,a) \atop x\_i \neq a} [x\_0,...,x\_n;f]$$ implies existence of $f^{(n)}(a)$?


If not is there connetion between high order divided differences and derivatives?



Higher order divided differences and derivatives
The answer is yes:


Assume we have a $\delta$ such that for $|x\_i-a|<\delta$, we have $|[x\_0,...,x\_n;f]-\lim| < \epsilon$. Assume WLOG that $\lim = 0$ (by subtracting off a polynomial of degree $n$). Then if $y\_0, ..., y\_{n-1}$ and $z\_0, ..., z\_{n-1}$ are in the $\delta$-ball around $a$, we can show


$|[y\_0,...,y\_{n-1};f]-[z\_0,...,z\_{n-1};f]| < \epsilon\sum\_{i=0}^{n-1}|y\_i-z\_i|$


by inducting on the number of $i$s such that $y\_i\ne z\_i$. Thus $\lim\_{y\_0,...,y\_{n-1}\rightarrow y}[y\_0,...,y\_{n-1};f]$ exists, and by induction on $n$ is equal to $f^{(n-1)}(y)$. Letting the $y\_i$s in the above approach $y$, and letting the $z\_i$s approach $z$, we see


$|f^{(n-1)}(y)-f^{(n-1)}(z)| \le n\epsilon|y-z|$,


so $\left|\frac{f^{(n-1)}(y)-f^{(n-1)}(z)}{y-z}\right| \le n\epsilon$ for any $y,z$ in a $\delta$-ball around $a$. Taking $\epsilon$ to $0$, we see that $f^{(n)}(a)$ exists and is equal to $0$.


