TITLE: Almost constant bump function
QUESTION [8 upvotes]: I ran into the following situation and it turned out to be more subtle than it looked.
I have a complete Riemannian manifold $M$ and my objective is to construct a sequence of functions $f:M \to [0,1]$ which have compact support, $L^2$ norm bounded away from zero, but all third derivatives converge uniformly to zero.
Trivial case: if the manifold is compact I can take the constant function $1$.
If the manifold is $\mathbb{R}$ I take a smooth bump function $f:\mathbb{R} \to [0,1]$ and then rescale it to $f_\lambda(x) = f(\lambda x)$.  The third derivatives are $f_\lambda'''(x) = \lambda^3 f'''(\lambda x)$ and for the $L^2$ norm you get:
$$|f_\lambda|^2 = \int f(\lambda x)^2 \mathrm{d} x = \frac{1}{\lambda}\int f(y)^2 \mathrm{d} y = \frac{1}{\lambda}|f|^2$$
So that if $\lambda \to 0$ I get what I want.
The same idea works for $\mathbb{R}^k$.
Can this be done on any complete Riemannian manifold $M$?
Idea:  Embed the manifold isometrically in $\mathbb{R}^k$ using Nash's theorem and use the restriction of a sequence of functions constructed for $\mathbb{R}^k$.
Problem:  Maybe the embedding has very large (second and third order) derivatives.  This can be a problem since $M$ is non-compact and the support of the sequence of functions has to grow.  Also this seems too high-tech for the problem at hand.
Any suggestions?

REPLY [13 votes]: Here is a counter-example.
Take a sequence of round 2-dimensional spheres $M_n$ of radii $r_n=n^{-1/2}$, $n=1,2,\dots$. Join them together into a long connected sum, namely connect each sphere to the next one by a tiny handle. Choose the points on the spheres where handles are attached carefully: on $M_n$, the two points (the connections with $M_{n-1}$ and $M_{n+1}$) should be not too close to each other and not too close to being diametrally opposite. (The distance half the diameter is just fine.)
Since $\sum r_n=\infty$, the resulting Riemannian manifold is complete. Since $\sum r_n^2=\infty$, its area is infinite. Now suppose that $f:M\to\mathbb R$ is a function with $|f'''|<\varepsilon$ along every geodesic. Observe that if $\gamma$ is a closed geodesic of length $\lambda$, then the difference between maximum and minimum of $f$ on $\gamma$ is at most $\varepsilon\lambda^3$. Our manifold has lots of short closed geodesics: any two points on neighboring spheres $M_n$ and $M_{n+1}$ can be connected by a chain of, say, at most 5 closed geodesics of length at most $10(r_n+r_{n+1})$. It follows that $|f(x)-f(y)|\le C \varepsilon r_n^3$ for all $x,y\in M_n\cup M_{n+1}$. Since the area is infinite, $f$ must go to 0 at infinity (to keep the $L^2$ norm bounded), therefore for $x\in M_n$ we have
$$
 |f(x)| \le \sum_{k=n}^\infty C\varepsilon  r_k^3 \sim C\varepsilon n^{-1/2} .
$$
Therefore
$$
 \int_M f^2 \le C\varepsilon^2 \sum r_n^2 n^{-1} =C\varepsilon^2 \sum n^{-2}  .
$$
The last series converges, thus $\|f\|_{L^2} \le C\varepsilon$ for some constant $C$.