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Thinking about a problem, i had got this idea which appears true, but i feel it is a direct consequence of a well known theorem, the name of which i would like to ask here.

The idea can be put in the form of a theorem as below.

If a function $f : \mathbb{R} \to \mathbb{R}$ is infinitely differentiable at a point $x = x_o$ then given any arbitrarily large $r \in \mathbb{N}$ we can find an $\epsilon \in \mathbb{R}$ such that $f$ is $\mathcal{C}^r$ in the interval $(x_o-\epsilon,x_o+\epsilon)$.

I'd like to know the validity of this statement and if it is valid, from which theorem does it follow ?

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What is your definition of being infinitely differentiable at a point? – Plop Apr 27 '11 at 13:22
This is valid, essentially by definition, so it is very unlikely to have a name. For example, the second derivative of $f$ at $a$ is defined to be the limit as $h$ approaches $0$ of $(f'(a+h)-f'(a))/h$. For this limit to exist, $f'$ must exist in a neighborhood of $a$. – André Nicolas Apr 27 '11 at 13:25
@Plop : by infinitely differentiable at a point $x_0$ i mean $f$ has derivatives of all orders at the point $x = x_o$. – Rajesh Dachiraju Apr 27 '11 at 13:31
Not all functions are differentiable, so that's not a definition. I am trying to get you to realize that you cannot define the $k+1$th derivative of $f$ at $x_0$ unless $f^{(k)}$ is defined on a neighborhood of $x_0$. – Plop Apr 27 '11 at 14:41
If $f$ is $k$ times differentiable on $]x_0-\epsilon,x_0+\epsilon[$, it is $\mathcal{C}^{k-1}$ on that interval. – Plop Apr 27 '11 at 16:28

1 Answer 1

up vote 3 down vote accepted

The original post seems to be generating an excessively long string of comments. I will try to explain things using the more generous editing facilities of the Answers format.

Take any function $g(x)$. Let us think about what we mean when we write $$\lim_{x \to a}\ g(x)=b$$ Very crudely, we mean that whenever $x$ is close enough to $a$ (but not necessarily equal to $a$), $g(x)$ is close to $b$. In order for the phrase "$g(x)$ is close to $b$" to have meaning, $g(x)$ must exist, that is, $g(x)$ must be defined at $x$, it must make sense at $x$. So we conclude that if $\lim_{x\to a} \ g(x)=b$, then in particular $g(x)$ must make sense, must be a number, for all $x$ in some interval about $a$, except possibly at $x=a$.

Now let us look at derivatives. In general, denote the $n$-th derivative of $f$ by $f^{(n)}$ (the $0$-th derivative of $f$ is defined to be $f$).

The $(k+1)$-th derivative of $f$ at $a$ is defined as follows. $$f^{(k+1)}(a)=\lim_{x\to a}\frac{f^{(k)}(x)-f^{(k)}(a)}{x-a}$$

Look at this definition. It mentions $f^{(k)}(a)$, so in particular, in order for $f^{(k+1)}(a)$ to exist, the very meaning of $f^{(k+1)}(a)$ forces $f^{(k)}(a)$ to exist. Also, the limit could not exist unless $f^{(k)}(x)$ existed for all $x$ close enough to $a$ but not equal to $a$. So if $f^{(k+1)}(a)$ exists, then $f^{(k)}(x)$ must exist for all $x$ close enough to $a$, including $x=a$.

To sum up, analysis of the very definition (meaning) of differentiation shows that if $f$ is $k+1$ times differentiable at $a$, then $f$ must be $k$ times differentiable in some neighbourhood of $a$.

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On the other hand, there are functions $f$ such that for any $n$, there is a polynomial $P$ of degree $n$ such that $|f(x) - P(x)| = O(|x|^{n+1})$ as $|x| \to 0$, just as if $f^{(k)}(0)$ existed for all $k$. Yet $f$ need not be differentiable at any $x \ne 0$. – Hans Engler Apr 27 '11 at 19:38
@Hans Engler: Indeed, the definition of differentiability at a point could have been, and perhaps should have been, different from what it is. But in any case, one is ordinarily interested mainly in differentiability in an interval. – André Nicolas Apr 27 '11 at 20:05

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