Let $P = a\_n(x) D\_x^n + a\_{n-1}(x) D\_x^{n-1} + \ldots + a\_0(x)$ be a linear ordinary differential operator with polynomial (or real analytic) coefficients $a\_j(x)$. Suppose that $a\_n(x)$ doesn't vanish on an interval $(a,b)$ and that $u$ is a [weak solution](http://en.wikipedia.org/wiki/Weak_solution) of $P u = 0$ on $(a,b)$. It can be concluded from the [elliptic regularity theorem](http://en.wikipedia.org/wiki/Elliptic_operator#Elliptic_regularity_theorem) that $u$ is in fact real analytic and a classical solution to the differential equation.


In other words, the operator $P$ is [analytically hypoelliptic](http://en.wikipedia.org/wiki/Hypoelliptic_operator).


However, appealing to the elliptic regularity theorem seems a bit much for the above case of an ordinary differential operator. Does anyone know a (canonical) reference for this (probably much more classical) case? I browsed through several books on differential equations but the closest I could find was in G. Folland's *Fourier analysis and its applications* where he mentions this fact (with just $C^\infty$ smoothness) on the top of page 344 without reference or proof.


*Update:* Thanks for the answers so far! While very helpful, I think I didn't make it clear what I was looking for. For an audience which may not be familiar with the theory of elliptic operators, what would be the proper way to (in one sentence plus a reference) justify that a weak solution in the case at hand is automatically real analytic? Currently, I have to make reference to a textbook on PDEs which treats elliptic regularity even though the problem for linear ODEs seems so much simpler.


Thank you!



Analytic hypoellipticity of linear ordinary differential operators
The answers already given are quite complete and say it all, but maybe the OP was in search of a very simple explanation of what is happening. Let me try.


You will agree that it is not necessary to work in the full generality of an n-th order equation; if we can do it for the first order case, it is easy to generalize by induction. Moreover, we can assume that the highest order coefficient is equal to 1 (just divide by it). So essentially the question is: let $u$ be a weak solution of the equation
$$ u' + a(x) u = f(x) $$
on an open interval, with $ a(x) $ and $ f(x) $ analytic. Then we want to prove that $u$ is also analytic. A further reduction is possible: multiply both sides by $\exp (A(x)) $ where $A'=a$ and call $ v = e^A u $, $g=e^A f$. Then we are reduced to proving the same thing for the simple equation
$$ v ' = g. $$
Is $v$ analytic if $g$ is analytic? A further reduction is possible! just call $ w=v-G $,
where $G'=g$. Nice trick eh? we are reduced to the even simpler equation
$$ w' = 0 $$
and our serach will be over if we prove the fundamental fact that any weak solution of $w'=0$ must be a constant function. Notice that the same chain of arguments applies if the coefficients are $C^\infty$ (you get that $u$ is also $C^\infty$) and if $f$ is just $C^k$.


Now, in the greatest possible generality, if $w$ is any distribution on an open interval, with vanishing derivative, then $w$ must be a constant. This is proved in the following way: by definition, we know that $ w(\phi')=0 $ for any test function $\phi$. Fix a test function $\chi$ with $\int\chi=1$. Let $\psi$ be an arbitrary test function, define
$$ \psi\_1 = \psi-\chi \cdot \int \psi $$
and notice that $\int\psi\_1=0$. This means that $ \psi\_1 $ can be written as the derivative of another test function (guess which one?) $\psi\_1=\psi\_2'$ and hence
$$w(\psi\_1) = w(\psi\_2') = 0 $$
by assumption. This implies
$$ w(\psi) = w(\chi) \cdot \int \psi $$
which in the language of distributions means precisely that $w$ is equal to the constant $w(\chi)$ (notice that $\chi$ is fixed once and for all). That's it.


