Exponential map Does there exist an infinite dimensional Lie group $G$, with Lie algebra $\mathfrak g$, such that the exponential map $exp:\mathfrak g \to G$ is not defined?
If so, can one provide an example of such a group.
 A: Sure.  Let $G$ be the group of diffeomorphisms of the real line.  Then $\frak{g}$ is the Lie algebra of vector fields on the real line.  However, many vector fields on the real line cannot be exponentiated to a $1$-parameter subgroup of $G$.  For example,
$$
X = x^2\frac{\partial\ \ }{\partial x}
$$
is not tangent to any $1$-parameter subgroup of $G$.
A: About Robert's example $Diff(\mathbb R)$ one can argue that it is not a Lie group, since it does not admit charts: For the compact $C^\infty$-topology it is not open in $C^\infty(\mathbb R)$. In the Whitney $C^\infty$-topology it is not locally contractible. 
Aside:
A setting where $Diff(\mathbb R)$ is a Lie group in the category of manifolds based on smooth curves etc (where the finite dimensional ones are exactly the known ones, as are the Banach ones)
is:
Peter W. Michor: A convenient setting for differential geometry and global analysis, I, II. Cahiers Topologie Geometrie Differentielle 25 (1984), 63--109, 113--178.(pdf of I)
(pdf of II)
Definition of regular Lie groups:
We consider a  smooth Lie group $G$ with Lie algebra $\mathfrak g=T_eG$ modelled on convenient 
vector spaces. 
The notion of a regular Lie group is originally due to Omori and collaborators
(see [Omori Maeda Yoshioka 1982], [Omori Maeda Yoshioka 1983]) for Frechet Lie groups, was 
weakened and made more transparent by [Milnor 1984] and carried over to convenient Lie 
groups in (here), 
see also 38.4 of (here).
A Lie group $G$
is called  regular  if the following holds:
$\bullet$ 
For each smooth curve 
$X\in C^{\infty}(\mathbb R,\mathfrak g)$ there exists a curve 
$g\in C^{\infty}(\mathbb R,G)$ whose right logarithmic derivative is $X$, i.e.,
$$
g(0) = e, \qquad
\partial_t g(t) = T_e(\mu^{g(t)})X(t) = X(t).g(t),\quad\text{where }  \mu(a,b)=\mu_a(b)=\mu^b(a) = a.b.
$$
The curve $g$ is uniquely determined by its initial value $g(0)$, if it
exists.
$\bullet$
Put $\operatorname{evol}^r_G(X)=g(1)$ where $g$ is the unique solution required above. 
Then $\operatorname{evol}^r_G: C^{\infty}(\mathbb R,\mathfrak g)\to G$ is required to be
$C^{\infty}$ also. 
Note that for $X$ constant in time, $\operatorname{evol}^r_G(X)=\exp(X)$. So each regular Lie group admits an exponential mapping.
The family of regular Lie groups is remarkably stable under constructions like extensions and quotients.
I do not know a Lie group modeled on convenient vector spaces which is not regular.
A quasi-counter-example is due to Wuestner:
Consider the space of trigonometric rational functions on $S^1$ which are everywhere positive and have no pole, with multiplication. This is not regular since $\exp(X)=e^X$ is real analytic and not trigonometric rational any more. But the modelling space is not convenient, since it is not Mackey sequentially complete for any suitable topology.
