Almost constant bump function 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?
 A: My guess would be that the problem of curvature of the embedding is actually intrinsic to the problem. Heuristically, third (covariant?) derivatives of $f$ will be related to the curvature of $M$, and to globally bound $f'''$ you'll probably need to bound the curvature.
For example, if you assume that $M$ has bounded geometry (i.e. positive injectivity radius and Riemannian curvature and its derivatives bounded globally) then you should be able to make your idea work because the curvature of the embedding will be bounded. Maybe only bounded curvature (and its derivative) is needed.
A: 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 $C$.
