Smooth continuation of a real function

Somehow I always knew we could continue a smooth real function to a larger domain uniquely. I can't seem to find such a result however and I am probably confusing it with the analytic continuation of a complex function. Thinking about this I am growing more and more convinced that the smooth continuation of a real function is not unique. For example I think I vaguely recall that in functional analysis we could construct multiple different functions, all of which had some desired asymptotic properties at infinity. Of course this is just on the level of intuition, but it seems to contradict the uniqueness of the continuation. Anyway, better to ask and be sure, so I should do this (somewhat) clearly...

Given intervals $A\subset B\subset\mathbb{R}$ and a smooth function $f:A\longrightarrow\mathbb{R}$, can we find a smooth function $g:B\longrightarrow\mathbb{R}$, such that $g$ is unique and $g|_A=f$?

• As pointed out by Levap in his answer, no. There is, however, the concept of real analytical functions. For these any local continuation is unique, like in the case of complex analytic functions. Aug 13, 2017 at 13:30
• what is "smooth" Aug 13, 2017 at 13:30
• @GEdgar Smooth means $C^{\infty}.$ That is, $f$ has derivatives of all orders.
– mfl
Aug 13, 2017 at 13:36
• In this generality, of course the answer is no. We could for example define $f(x)=1/x$ in $A=(0,1)$ and let $B = (-1,1).$
– zhw.
Aug 13, 2017 at 14:57

No. Let's take for example $A = (1,\infty)$ and $B = (-1,\infty)$. Consider the function $f(x) = e^{-\frac{1}{x^2}}$ on $A$. Two different smooth extensions to $B$ are given by $$g_1(x) = \begin{cases} e^{-\frac{1}{x^2}} & x > 0, \\ 0 & -1 < x \leq 0. \end{cases}$$ and $$g_2(x) = \begin{cases} e^{-\frac{1}{x^2}} & x \in (-1,\infty) \setminus \{ 0 \}, \\ 0 & x = 0. \end{cases}$$
In fact, you can construct infinitely many other extensions of $f$. This example comes from the theory of bump functions.