Given a function $f$ infinitely differentiable at a point $c$ does there exist a neighborhood of $c$ in which $f$ is infinitely differentiable? Suppose $f$ is a real valued function defined on a subset of the reals and $f$ is infinitely differentiable at $c$.Then is it possible that there does not exist any neighborhood of $c$ in which $f$ is infinitely differentiable.Clearly $c$ cannot be an isolated point of any of the domains of $f^{(k)}$.But I cannot proceed further.
Alternatively can we say that the set $S(f) :=\{c : f$ is infinitely differentiable at $c\}$ is an open set?
 A: This is similar to the idea in Kavi Rama Murthy's answer. First, note that for $k = 0, 1, 2, \dots$, you can construct a function $h_k : [-1, 1] \to \mathbb{R}$ which is $C^k$ but is not $(k+1)$ times differentiable at any point: just take repeated antiderivatives (integrals) of a continuous nowhere differentiable function, e.g. the Weierstrass function. By finding a smooth function $g_k$ with $g_k^{(i)}(-1) = h_k^{(i)}(-1)$ and $g_k^{(i)}(1) = h_k^{(i)}(1)$ for $i = 0, 1, \dots, k$, you can construct $f_k = h_k - g_k$, which has $f_k^{(i)}(-1) = f_k^{(i)}(1) = 0$ for $i = 0, 1, \dots, k$ -- this is a $C^k$ function on $\mathbb{R}$ supported on $[-1, 1]$ which is not $(k+1)$ times differentiable anywhere on $[-1, 1]$.
Now, consider the function $f : [-1, 1] \to \mathbb{R}$ defined as follows. On $(1/(n+1), 1/n)$, we let $f$ be equal to a $C^n$ function supported in this interval which is nowhere $(n+1)$-differentiable on its support (similar to one constructed above), and we scale $f$ down sufficiently so that on this interval, $|f^{(i)}| \leq e^{-(n+1)^2}$ for $i = 0, 1, \dots, n$ (which is possible since these derivatives are all bounded). Finally, define $f(0) = 0$ and $f(x) = f(-x)$ for $x < 0$. It is clear that $f$ is $C^n$ on the intervals $(-1/n, 0)$ and $(0, 1/n)$, and thus it is $C^n$ on $(-1/n, 1/n)$ since $|f^{0}(x)|, |f^{(1)}(x)|, \dots, |f^{(n)}(x)| \leq e^{-1/x^2}$ near $x = 0$. Thus $f$ is infinitely differentiable at $0$. However, it is not infinitely differentiable on any neighborhood of $0$: within $(-1/n, 1/n)$ there are points where it is not $(n+1)$ times differentiable.
