# Are generic smooth functions analytic?

It is well-known that generic continuous functions are differentiable almost nowhere. I was somewhat surprised to learn in my functional analysis course that the same is true of $\alpha$-Holder functions for $0<\alpha<1$. I am aware that there exist functions that are smooth but analytic almost nowhere, so I've been wondering if this too is the generic situation. My intuition is thoroughly conflicted so I'm wondering if anyone has any insight.

By generic I mean a set whose complement is of first category. Any similar definition would be fine.

Please note: I am fully aware that there exist smooth functions that are nowhere analytic. My question--which as of August 2, 2016 has not been resolved--is whether a generic smooth function in this topological sense is analytic everywhere/nowhere/somewhere.

• The domain can be the reals, a compact interval, or anything else that's convenient. – Funktorality Apr 24 '16 at 21:17
• Analytic a.e. on its whole domain – Funktorality Apr 24 '16 at 21:17
• where can I find an intuitive description of "generic continuous or smooth function" ? – reuns Apr 24 '16 at 21:41
• @user1952009 "generic" is a topological property of a set that says it is in some way the typical case. en.m.wikipedia.org/wiki/Generic_property – Funktorality Apr 25 '16 at 0:08
• This answer (especially the first link provided) gives an answer for genericity in the Baire sense. The strongest result is that (I paraphrase) the "every point of a Baire-typical $C^\infty$ function is a point at which the Taylor series has zero radius of convergence." – Willie Wong Aug 3 '16 at 2:47

Generically, no. For every sequence of real numbers (a set which has obviously a very large cardinality), there exists a smooth function such that the Taylor series of the function at the origin is that sequence. Since there are many more sequences of real numbers than sequences corresponding to convergent Taylor series, it follows that generically smooth functions are not analytic (for example just consider the fact that a necessary condition for the convergence of the Taylor series is that the coefficients approach 0, whereas most sequences of real numbers which have limits don't approach 0, and in fact most sequences don't even have limits).

(Beginning of Rigorous Argument)

https://en.wikipedia.org/wiki/Non-analytic_smooth_function

Look at Theorem 2 of the attached link: http://math.bard.edu/belk/math351/FunctionSpaces.pdf

Since the pointwise limit (as functions from $\mathbb{N} \to \mathbb{R}$) of any sequence of real-valued sequences with limit zero is necessarily again a sequence with limit zero, it follows as a corollary of Theorem 2 that $c_0$ is a closed subset of $\mathbb{R}^{\infty}$ under the standard product topology on that space, i.e. $c_0$ is its own closure. The interior of $c_0$ in the product topology, however, is necessarily empty, since any open subset of $\mathbb{R}^{\infty}$ must contain a basis set; however $c_0$ cannot contain any basis set of the product topology, since any sequence converging to 0 has unrestricted values for any finite number of coordinates (since the only restriction is on its behavior in the limit).

Therefore $c_0$ is nowhere dense in the product topology of $\mathbb{R}^{\infty}$, and thus meager.

By uniqueness of Taylor series for analytic functions, the space of all functions analytic at the origin corresponds to a proper subspace of $c_0$. In contrast, since uniqueness of Taylor series does not apply for non-analytic smooth functions, $\mathbb{R}^{\infty}$ corresponds to a strict subspace of all smooth functions. Hence the space of all analytic functions is a strict subspace of a space meager in a strict subspace of the space of all smooth functions, hence is itself necessarily meager in the space of all smooth functions.

(End of Rigorous Argument)

For example, if we were to define a measure on the space $c$ such that the measure of the set of all sequences with limit in the Lebesgue-measurable set $A \subset \mathbb{R}$ had measure equal to the Lebesgue measure of $A$, then the set of all analytic functions would correspond to a strict subset of the null-set $c_0$, whereas the space of all smooth functions would not even be contained inside $c$.

In fact, we could construct a measure on the space of all real sequences such that the measure of all sequences with $\liminf z_n =x$ and $\limsup z_n =y$ with $(x,y)\in B \subset \mathbb{R}^2$ has the measure equal to the Lebesgue measure of $B$ (assuming B is Lebesgue-measurable). Then the space $c$, in which the analytic functions are a null set, corresponds to the line $y=x$, and hence has measure 0 (since any line in the plane has Lebesgue measure 0). Yet the space of all smooth functions would still not be a subspace of the entire space $\mathbb{R}^{\infty}$, a set of infinite measure.

EDIT: From the above, it follows that:

The set of functions analytic at the origin is strictly smaller than the set of sequences of real numbers with limit 0 (the space of null sequences $c_0$), which is strictly smaller (by a lot) than the set of all real sequences which have a limit (the space of all convergent sequences $c$), which is again strictly smaller than the set of all real sequences $\mathbb{R}^{\infty}$ (by a lot), which corresponds to a subspace of smooth functions.

As a corollary, to each individual function analytic at the origin $f$ (which is a much larger set than the set of functions analytic on the entire real line), there exists a unique set of non-analytic smooth functions corresponding to $\mathbb{R}^2$.

(First, take all sequences of the form $(a_n +x)$ where x is a real number, and $(a_n)$ is the sequence going to 0 corresponding to the Taylor coefficients of the expansion of $f$ at 0. Then each $(a_n)$ corresponds to a smooth function with Taylor coefficients $(a_n +x)$ at the origin. All but one such sequence has limit $x\not=0$, which means it must correspond to a non-analytic smooth function, since its Taylor series clearly does not converge.

Then for each $(a_n +x)$ we can define for every real number $y$ sequences $(a_n,\frac{y}{x}[a_n+x])$ which do not converge but which have limit superior x and limit inferior y if $x<y$, or $\limsup y$ and $\liminf x$ if $y<x$.

Each such sequence again corresponds to a unique non-convergent series and hence a unique non-analytic smooth function. So there are a LOT LOT more non-analytic smooth functions than analytic smooth functions.

• I don't quite see this cardinality argument. Is this just to show that such a smooth non-analytic function exists? – Funktorality Apr 25 '16 at 0:05
• I don't study set theory, but I believe these spaces all have the cardinality of the continuum (see previous stackexchange posts). Also, it's very easy to see that a generic set can have a smaller cardinality than the whole space (eg. the empty set is always generic in an at most countable space with the discrete topology). – Funktorality Apr 25 '16 at 1:48
• Ok the basis for my claim is that the cardinality of continuous functions on R is the continuum (since a continuous function is determined by its values on Q) so the cardinality of the smooth functions is not greater. – Funktorality Apr 25 '16 at 2:14
• It is not true that $c_0$ is closed in $\Bbb {R}^\Bbb {N}$. – PhoemueX Apr 25 '16 at 8:15
• By Taylor coefficient, do you mean the $a_k$ in $$\sum a_k x^k$$ or in $$\sum \frac{1}{k!} a_k x^k?$$ I think William meant the first case, while @zhw meant the second case. This would explain the apparent contradiction. – Willie Wong Aug 3 '16 at 2:26